🧠:<think>Alright, so I've got this problem about optimizing training schedules for taekwondo athletes. The coach has a differential equation modeling their success rate, and I need to solve it under certain conditions. Let me try to unpack this step by step.First, the differential equation given is:[ frac{dS}{dt} = k cdot sqrt{T(t) - S(t)} ]Where ( S(t) ) is the success rate, ( T(t) ) is the training intensity, and ( k ) is a constant. Sub-problem 1 asks for the general solution when ( T(t) = 50 + 30 sin(omega t) ). Okay, so we're dealing with a non-constant training intensity that varies sinusoidally over time. That makes sense because training intensity often cycles through periods of high and low intensity.So, plugging ( T(t) ) into the differential equation, we get:[ frac{dS}{dt} = k cdot sqrt{50 + 30 sin(omega t) - S(t)} ]Hmm, this is a first-order differential equation, but it's nonlinear because of the square root. Solving nonlinear differential equations can be tricky. Let me think about substitution methods.Let me denote ( u(t) = T(t) - S(t) ). Then, ( S(t) = T(t) - u(t) ). Taking the derivative of both sides:[ frac{dS}{dt} = frac{dT}{dt} - frac{du}{dt} ]But from the original equation, ( frac{dS}{dt} = k cdot sqrt{u(t)} ). So substituting:[ k cdot sqrt{u(t)} = frac{dT}{dt} - frac{du}{dt} ]Rearranging terms:[ frac{du}{dt} = frac{dT}{dt} - k cdot sqrt{u(t)} ]Hmm, that seems a bit more manageable, but it's still nonlinear because of the square root. Let me compute ( frac{dT}{dt} ):Given ( T(t) = 50 + 30 sin(omega t) ), so:[ frac{dT}{dt} = 30 omega cos(omega t) ]So plugging back in:[ frac{du}{dt} = 30 omega cos(omega t) - k sqrt{u(t)} ]This is a Riccati-type equation, I think, but maybe I can make another substitution. Let me set ( v(t) = sqrt{u(t)} ). Then, ( u(t) = v(t)^2 ), so:[ frac{du}{dt} = 2 v(t) frac{dv}{dt} ]Substituting into the equation:[ 2 v frac{dv}{dt} = 30 omega cos(omega t) - k v ]Dividing both sides by ( 2 v ) (assuming ( v neq 0 )):[ frac{dv}{dt} = frac{30 omega cos(omega t)}{2 v} - frac{k}{2} ]Hmm, this is a Bernoulli equation. Bernoulli equations can be linearized by an appropriate substitution. The standard form is:[ frac{dv}{dt} + P(t) v = Q(t) v^n ]In our case, comparing:[ frac{dv}{dt} + frac{k}{2} v = frac{30 omega cos(omega t)}{2} cdot frac{1}{v} ]So, ( n = -1 ), ( P(t) = frac{k}{2} ), and ( Q(t) = frac{30 omega cos(omega t)}{2} ).To linearize, we use the substitution ( w = v^{1 - n} = v^{2} ). Then, ( frac{dw}{dt} = 2 v frac{dv}{dt} ).Wait, let me recall the standard substitution for Bernoulli equations. If we have:[ frac{dv}{dt} + P(t) v = Q(t) v^n ]Then, substituting ( w = v^{1 - n} ) transforms it into a linear equation in ( w ).So, in our case, ( n = -1 ), so ( 1 - n = 2 ). Thus, ( w = v^2 ).Compute ( frac{dw}{dt} = 2 v frac{dv}{dt} ).From our earlier equation:[ 2 v frac{dv}{dt} = 30 omega cos(omega t) - k v^2 ]But ( w = v^2 ), so:[ frac{dw}{dt} = 30 omega cos(omega t) - k w ]Ah, now this is a linear differential equation for ( w(t) )! That's progress.So, the equation is:[ frac{dw}{dt} + k w = 30 omega cos(omega t) ]Now, we can solve this using an integrating factor.The integrating factor ( mu(t) ) is:[ mu(t) = e^{int k , dt} = e^{k t} ]Multiplying both sides by ( mu(t) ):[ e^{k t} frac{dw}{dt} + k e^{k t} w = 30 omega e^{k t} cos(omega t) ]The left side is the derivative of ( w e^{k t} ):[ frac{d}{dt} left( w e^{k t} right) = 30 omega e^{k t} cos(omega t) ]Integrate both sides:[ w e^{k t} = 30 omega int e^{k t} cos(omega t) , dt + C ]Now, we need to compute the integral ( int e^{k t} cos(omega t) , dt ). I remember that this integral can be solved using integration by parts twice and then solving for the integral.Let me recall the formula:[ int e^{a t} cos(b t) , dt = frac{e^{a t}}{a^2 + b^2} (a cos(b t) + b sin(b t)) ) + C ]Yes, that's the standard result. So applying this with ( a = k ) and ( b = omega ):[ int e^{k t} cos(omega t) , dt = frac{e^{k t}}{k^2 + omega^2} (k cos(omega t) + omega sin(omega t)) ) + C ]So, plugging back into our equation:[ w e^{k t} = 30 omega cdot frac{e^{k t}}{k^2 + omega^2} (k cos(omega t) + omega sin(omega t)) ) + C ]Divide both sides by ( e^{k t} ):[ w(t) = frac{30 omega}{k^2 + omega^2} (k cos(omega t) + omega sin(omega t)) ) + C e^{-k t} ]Recall that ( w(t) = v(t)^2 ), and ( v(t) = sqrt{u(t)} ), and ( u(t) = T(t) - S(t) ). So, let's write:[ w(t) = v(t)^2 = u(t) = T(t) - S(t) ]Wait, hold on. Let me retrace:We set ( u(t) = T(t) - S(t) ), then ( v(t) = sqrt{u(t)} ), so ( v(t)^2 = u(t) ). Then, ( w(t) = v(t)^2 = u(t) ). So, actually, ( w(t) = u(t) = T(t) - S(t) ).Wait, that seems conflicting. Let me make sure.Wait, no. Wait, ( u(t) = T(t) - S(t) ). Then, ( v(t) = sqrt{u(t)} ), so ( v(t)^2 = u(t) ). Then, we set ( w(t) = v(t)^2 ), which is equal to ( u(t) ). So, actually, ( w(t) = u(t) ). Therefore, from the equation above:[ w(t) = frac{30 omega}{k^2 + omega^2} (k cos(omega t) + omega sin(omega t)) ) + C e^{-k t} ]But ( w(t) = u(t) = T(t) - S(t) ). So:[ T(t) - S(t) = frac{30 omega}{k^2 + omega^2} (k cos(omega t) + omega sin(omega t)) ) + C e^{-k t} ]Therefore, solving for ( S(t) ):[ S(t) = T(t) - frac{30 omega}{k^2 + omega^2} (k cos(omega t) + omega sin(omega t)) ) - C e^{-k t} ]But ( T(t) = 50 + 30 sin(omega t) ), so:[ S(t) = 50 + 30 sin(omega t) - frac{30 omega}{k^2 + omega^2} (k cos(omega t) + omega sin(omega t)) ) - C e^{-k t} ]This is the general solution. The constant ( C ) can be determined using initial conditions, which is given in Sub-problem 2.So, summarizing, the general solution is:[ S(t) = 50 + 30 sin(omega t) - frac{30 omega}{k^2 + omega^2} (k cos(omega t) + omega sin(omega t)) ) - C e^{-k t} ]That's Sub-problem 1 done.Moving on to Sub-problem 2. We have initial condition ( S(0) = 20 ), ( k = 0.1 ), and ( omega = frac{pi}{6} ). We need to find the specific solution and analyze its long-term behavior.First, let's plug in the known values into the general solution.Given ( k = 0.1 ), ( omega = frac{pi}{6} ), so ( k^2 + omega^2 = (0.1)^2 + left( frac{pi}{6} right)^2 ).Calculating ( k^2 = 0.01 ), ( omega^2 = left( frac{pi}{6} right)^2 approx left( 0.5236 right)^2 approx 0.2742 ). So, ( k^2 + omega^2 approx 0.01 + 0.2742 = 0.2842 ).So, the coefficient ( frac{30 omega}{k^2 + omega^2} approx frac{30 times 0.5236}{0.2842} approx frac{15.708}{0.2842} approx 55.26 ).Wait, let me compute that more accurately.First, ( omega = pi / 6 approx 0.5235987756 ).So, ( 30 omega approx 30 times 0.5235987756 approx 15.70796327 ).Then, ( k^2 + omega^2 = 0.01 + (0.5235987756)^2 approx 0.01 + 0.2741599999 approx 0.28416 ).So, ( frac{30 omega}{k^2 + omega^2} approx frac{15.70796327}{0.28416} approx 55.26 ).So, approximately 55.26.Therefore, the general solution becomes:[ S(t) = 50 + 30 sinleft( frac{pi}{6} t right) - 55.26 left( 0.1 cosleft( frac{pi}{6} t right) + frac{pi}{6} sinleft( frac{pi}{6} t right) right) - C e^{-0.1 t} ]Let me compute the constants inside the brackets:First, ( 0.1 cos(omega t) ) and ( frac{pi}{6} sin(omega t) ).Compute ( 0.1 times 55.26 approx 5.526 ).Compute ( frac{pi}{6} times 55.26 approx 0.5235987756 times 55.26 approx 29.03 ).So, the term becomes:[ 5.526 cosleft( frac{pi}{6} t right) + 29.03 sinleft( frac{pi}{6} t right) ]Therefore, the solution is:[ S(t) = 50 + 30 sinleft( frac{pi}{6} t right) - 5.526 cosleft( frac{pi}{6} t right) - 29.03 sinleft( frac{pi}{6} t right) - C e^{-0.1 t} ]Combine like terms:The sine terms: ( 30 sin(cdot) - 29.03 sin(cdot) = (30 - 29.03) sin(cdot) = 0.97 sin(cdot) ).So:[ S(t) = 50 + 0.97 sinleft( frac{pi}{6} t right) - 5.526 cosleft( frac{pi}{6} t right) - C e^{-0.1 t} ]Now, let's compute the constants more precisely instead of approximating.Wait, maybe I should keep it symbolic until applying the initial condition.So, let's write the general solution again:[ S(t) = 50 + 30 sin(omega t) - frac{30 omega}{k^2 + omega^2} (k cos(omega t) + omega sin(omega t)) - C e^{-k t} ]Plugging in ( k = 0.1 ), ( omega = pi/6 ):First, compute ( frac{30 omega}{k^2 + omega^2} ):Compute numerator: ( 30 times pi/6 = 5 pi approx 15.70796327 ).Compute denominator: ( (0.1)^2 + (pi/6)^2 = 0.01 + (pi^2)/36 approx 0.01 + 0.2741599999 approx 0.28416 ).So, ( frac{30 omega}{k^2 + omega^2} = frac{5 pi}{0.28416} approx 55.26 ).Then, ( k cos(omega t) + omega sin(omega t) = 0.1 cos(omega t) + (pi/6) sin(omega t) ).So, ( frac{30 omega}{k^2 + omega^2} (k cos(omega t) + omega sin(omega t)) = 55.26 (0.1 cos(omega t) + 0.5235987756 sin(omega t)) ).Calculating:( 55.26 times 0.1 = 5.526 )( 55.26 times 0.5235987756 approx 55.26 times 0.5236 approx 29.03 )So, the term is ( 5.526 cos(omega t) + 29.03 sin(omega t) ).Thus, plugging back into ( S(t) ):[ S(t) = 50 + 30 sin(omega t) - 5.526 cos(omega t) - 29.03 sin(omega t) - C e^{-0.1 t} ]Combine the sine terms:( 30 sin(omega t) - 29.03 sin(omega t) = 0.97 sin(omega t) )So:[ S(t) = 50 + 0.97 sinleft( frac{pi}{6} t right) - 5.526 cosleft( frac{pi}{6} t right) - C e^{-0.1 t} ]Now, apply the initial condition ( S(0) = 20 ).Compute ( S(0) ):[ S(0) = 50 + 0.97 sin(0) - 5.526 cos(0) - C e^{0} ]Simplify:( sin(0) = 0 ), ( cos(0) = 1 ), ( e^{0} = 1 ).So:[ 20 = 50 + 0 - 5.526 - C ]Simplify:[ 20 = 50 - 5.526 - C ]Compute ( 50 - 5.526 = 44.474 )So:[ 20 = 44.474 - C ]Solving for ( C ):[ C = 44.474 - 20 = 24.474 ]So, ( C approx 24.474 ).Therefore, the specific solution is:[ S(t) = 50 + 0.97 sinleft( frac{pi}{6} t right) - 5.526 cosleft( frac{pi}{6} t right) - 24.474 e^{-0.1 t} ]To make it more precise, let's use exact expressions instead of approximations.Wait, perhaps I should express the constants more accurately.Let me recompute ( frac{30 omega}{k^2 + omega^2} ) symbolically.Given ( k = 0.1 ), ( omega = pi / 6 ).Compute ( k^2 + omega^2 = 0.01 + (pi^2)/36 ).So,[ frac{30 omega}{k^2 + omega^2} = frac{30 times pi / 6}{0.01 + pi^2 / 36} = frac{5 pi}{0.01 + pi^2 / 36} ]Let me compute this exactly:First, compute denominator:( 0.01 + pi^2 / 36 approx 0.01 + 0.2741599999 approx 0.28416 ).So, ( 5 pi / 0.28416 approx 55.26 ).But perhaps we can keep it as ( frac{5 pi}{0.01 + pi^2 / 36} ) for exactness, but since we have to plug in numbers for the specific solution, maybe it's better to use the approximate decimal values.So, moving on.So, the specific solution is:[ S(t) = 50 + 0.97 sinleft( frac{pi}{6} t right) - 5.526 cosleft( frac{pi}{6} t right) - 24.474 e^{-0.1 t} ]Now, to analyze the long-term behavior as ( t to infty ).Looking at the solution, as ( t to infty ), the term ( e^{-0.1 t} ) tends to zero. So, the transient term disappears.The remaining terms are:[ 50 + 0.97 sinleft( frac{pi}{6} t right) - 5.526 cosleft( frac{pi}{6} t right) ]This is a sinusoidal function with amplitude determined by the coefficients of sine and cosine.We can write this as a single sine (or cosine) function with phase shift.Let me compute the amplitude:The expression ( A sin(theta) + B cos(theta) ) can be written as ( R sin(theta + phi) ), where ( R = sqrt{A^2 + B^2} ) and ( phi = arctan(B / A) ) or something like that.Wait, actually, the formula is:[ A sin(theta) + B cos(theta) = R sinleft( theta + phi right) ]where ( R = sqrt{A^2 + B^2} ) and ( phi = arctanleft( frac{B}{A} right) ) if we take the sine form.Alternatively, it can also be written as ( R cos(theta - phi) ).But in our case, the coefficients are:( A = 0.97 ), ( B = -5.526 ).So, ( R = sqrt{0.97^2 + (-5.526)^2} approx sqrt{0.9409 + 30.535} approx sqrt{31.4759} approx 5.61 ).So, the amplitude is approximately 5.61.Therefore, the long-term behavior is a sinusoidal oscillation around 50 with amplitude ~5.61. So, the success rate will oscillate between approximately ( 50 - 5.61 = 44.39 ) and ( 50 + 5.61 = 55.61 ).But wait, let me verify the exact expression:The oscillatory part is ( 0.97 sin(omega t) - 5.526 cos(omega t) ). So, combining these:The amplitude ( R = sqrt{0.97^2 + 5.526^2} approx sqrt{0.9409 + 30.535} approx sqrt{31.4759} approx 5.61 ).So, yes, the oscillation has an amplitude of about 5.61, centered at 50.Therefore, as ( t to infty ), the success rate ( S(t) ) approaches a sinusoidal function oscillating between approximately 44.39 and 55.61.But wait, let me think again. The term ( -5.526 cos(omega t) ) is quite large compared to the ( 0.97 sin(omega t) ). So, the oscillation is dominated by the cosine term.Alternatively, perhaps it's better to write it as a single sinusoidal function.Let me compute the exact phase shift.Given ( A = 0.97 ), ( B = -5.526 ).The phase angle ( phi ) is given by:[ phi = arctanleft( frac{B}{A} right) = arctanleft( frac{-5.526}{0.97} right) approx arctan(-5.7) approx -1.414 ) radians.So, approximately -81 degrees.Therefore, the oscillatory part can be written as:[ R sinleft( frac{pi}{6} t + phi right) approx 5.61 sinleft( frac{pi}{6} t - 1.414 right) ]But regardless of the phase, the key point is that the success rate oscillates around 50 with an amplitude of about 5.61.Therefore, in the long term, the success rate doesn't settle to a fixed value but continues to oscillate due to the periodic training intensity ( T(t) ).However, the transient term ( -24.474 e^{-0.1 t} ) decays exponentially to zero, meaning that initially, the success rate is influenced by this term, but as time goes on, it becomes negligible, and the oscillatory behavior dominates.So, in summary, the specific solution is:[ S(t) = 50 + 0.97 sinleft( frac{pi}{6} t right) - 5.526 cosleft( frac{pi}{6} t right) - 24.474 e^{-0.1 t} ]And as ( t to infty ), ( S(t) ) approaches a sinusoidal function oscillating around 50 with an amplitude of approximately 5.61.I think that's the analysis.Final AnswerSub-problem 1: The general solution is (boxed{S(t) = 50 + 30 sin(omega t) - frac{30 omega}{k^2 + omega^2} (k cos(omega t) + omega sin(omega t)) - C e^{-k t}}).Sub-problem 2: The specific solution is (boxed{S(t) = 50 + 0.97 sinleft( frac{pi}{6} t right) - 5.526 cosleft( frac{pi}{6} t right) - 24.474 e^{-0.1 t}}), and the success rate oscillates around 50 with an amplitude of approximately 5.61 as (t to infty).</think>

🧠:<think>Alright, so I have this problem about a biotechnologist developing a drug delivery system using nanoparticles. The problem is split into two parts. The first part is solving a partial differential equation (PDE) that models the release rate of the drug, and the second part is determining the time at which the total drug concentration reaches a certain threshold.Starting with the first part: The PDE given is[ frac{partial C}{partial t} = D frac{partial^2 C}{partial x^2} - kC ]with the initial condition ( C(0, x) = C_0 exp(-alpha x^2) ).Hmm, okay. So this is a linear PDE, and it looks like a combination of the diffusion equation and an exponential decay term. I remember that for linear PDEs, especially parabolic ones like the diffusion equation, methods like separation of variables or Fourier transforms can be useful. Since the initial condition is a Gaussian function, which is its own Fourier transform, maybe a Fourier transform approach would work here.Let me recall that the Fourier transform can turn the PDE into an ordinary differential equation (ODE) in the frequency domain, which might be easier to solve. So, let's try that.First, let's define the Fourier transform of ( C(t, x) ) as:[ mathcal{F}{C(t, x)} = hat{C}(t, xi) = int_{-infty}^{infty} C(t, x) e^{-i xi x} dx ]Then, the inverse Fourier transform would be:[ C(t, x) = frac{1}{2pi} int_{-infty}^{infty} hat{C}(t, xi) e^{i xi x} dxi ]Now, let's take the Fourier transform of both sides of the PDE.Starting with the left-hand side:[ mathcal{F}left{frac{partial C}{partial t}right} = frac{partial}{partial t} mathcal{F}{C(t, x)} = frac{partial hat{C}}{partial t} ]For the right-hand side, we have two terms:1. The diffusion term: ( D frac{partial^2 C}{partial x^2} )2. The degradation term: ( -kC )Taking the Fourier transform of the diffusion term:[ mathcal{F}left{D frac{partial^2 C}{partial x^2}right} = D (-i xi)^2 hat{C}(t, xi) = -D xi^2 hat{C}(t, xi) ]And the Fourier transform of the degradation term is straightforward:[ mathcal{F}{-kC} = -k hat{C}(t, xi) ]Putting it all together, the PDE in the frequency domain becomes:[ frac{partial hat{C}}{partial t} = -D xi^2 hat{C} - k hat{C} ]This simplifies to:[ frac{partial hat{C}}{partial t} = -(D xi^2 + k) hat{C} ]This is a first-order linear ODE in ( t ) for each fixed ( xi ). The solution should be straightforward.The general solution for such an ODE is:[ hat{C}(t, xi) = hat{C}(0, xi) e^{-(D xi^2 + k) t} ]Now, we need the initial condition in the frequency domain. The initial condition is ( C(0, x) = C_0 exp(-alpha x^2) ). Let's compute its Fourier transform.Recall that the Fourier transform of ( e^{-a x^2} ) is ( sqrt{frac{pi}{a}} e^{-pi^2 xi^2 / a} ) or something similar. Wait, let me double-check.Actually, the Fourier transform of ( e^{-pi x^2} ) is ( e^{-pi xi^2} ). So, more generally, the Fourier transform of ( e^{-a x^2} ) is ( sqrt{frac{pi}{a}} e^{-pi^2 xi^2 / a} ). Hmm, maybe I should compute it directly.Compute ( hat{C}(0, xi) = int_{-infty}^{infty} C_0 e^{-alpha x^2} e^{-i xi x} dx ).This is ( C_0 int_{-infty}^{infty} e^{-alpha x^2 - i xi x} dx ).This integral is a standard Gaussian integral. The formula is:[ int_{-infty}^{infty} e^{-a x^2 + b x} dx = sqrt{frac{pi}{a}} e^{b^2 / (4a)} ]But in our case, the exponent is ( -alpha x^2 - i xi x ), so ( a = alpha ) and ( b = -i xi ).Therefore,[ hat{C}(0, xi) = C_0 sqrt{frac{pi}{alpha}} e^{(-i xi)^2 / (4 alpha)} ]Simplify the exponent:[ (-i xi)^2 = (-i)^2 xi^2 = (-1)^2 (i^2) xi^2 = (1)(-1) xi^2 = -xi^2 ]So,[ hat{C}(0, xi) = C_0 sqrt{frac{pi}{alpha}} e^{-xi^2 / (4 alpha)} ]Therefore, the solution in the frequency domain is:[ hat{C}(t, xi) = C_0 sqrt{frac{pi}{alpha}} e^{-xi^2 / (4 alpha)} e^{-(D xi^2 + k) t} ]Simplify the exponents:[ hat{C}(t, xi) = C_0 sqrt{frac{pi}{alpha}} e^{-(D t + frac{1}{4 alpha}) xi^2 - k t} ]Now, to find ( C(t, x) ), we need to take the inverse Fourier transform:[ C(t, x) = frac{1}{2pi} int_{-infty}^{infty} hat{C}(t, xi) e^{i xi x} dxi ]Substituting ( hat{C}(t, xi) ):[ C(t, x) = frac{C_0 sqrt{frac{pi}{alpha}}}{2pi} int_{-infty}^{infty} e^{-(D t + frac{1}{4 alpha}) xi^2 - k t} e^{i xi x} dxi ]Simplify constants:[ frac{C_0 sqrt{frac{pi}{alpha}}}{2pi} = frac{C_0}{2 sqrt{pi alpha}} ]So,[ C(t, x) = frac{C_0}{2 sqrt{pi alpha}} e^{-k t} int_{-infty}^{infty} e^{-(D t + frac{1}{4 alpha}) xi^2 + i xi x} dxi ]Now, the integral is again a Gaussian integral. Let me denote:Let ( a = D t + frac{1}{4 alpha} ), and ( b = x ). Then,[ int_{-infty}^{infty} e^{-a xi^2 + b xi} dxi = sqrt{frac{pi}{a}} e^{b^2 / (4a)} ]In our case, ( b = x ), so:[ int_{-infty}^{infty} e^{-(D t + frac{1}{4 alpha}) xi^2 + i xi x} dxi = sqrt{frac{pi}{D t + frac{1}{4 alpha}}} e^{(i x)^2 / (4 (D t + frac{1}{4 alpha}))} ]Simplify the exponent:( (i x)^2 = -x^2 ), so:[ e^{-x^2 / (4 (D t + frac{1}{4 alpha}))} ]Therefore, the integral becomes:[ sqrt{frac{pi}{D t + frac{1}{4 alpha}}} e^{-x^2 / (4 (D t + frac{1}{4 alpha}))} ]Putting this back into the expression for ( C(t, x) ):[ C(t, x) = frac{C_0}{2 sqrt{pi alpha}} e^{-k t} sqrt{frac{pi}{D t + frac{1}{4 alpha}}} e^{-x^2 / (4 (D t + frac{1}{4 alpha}))} ]Simplify constants and exponents:First, let's simplify the constants:[ frac{C_0}{2 sqrt{pi alpha}} times sqrt{frac{pi}{D t + frac{1}{4 alpha}}} = frac{C_0}{2 sqrt{pi alpha}} times sqrt{frac{pi}{D t + frac{1}{4 alpha}}} ]Simplify the square roots:[ sqrt{frac{pi}{D t + frac{1}{4 alpha}}} = frac{sqrt{pi}}{sqrt{D t + frac{1}{4 alpha}}} ]So,[ frac{C_0}{2 sqrt{pi alpha}} times frac{sqrt{pi}}{sqrt{D t + frac{1}{4 alpha}}} = frac{C_0}{2 sqrt{alpha}} times frac{1}{sqrt{D t + frac{1}{4 alpha}}} ]Which simplifies to:[ frac{C_0}{2 sqrt{alpha (D t + frac{1}{4 alpha})}} ]Simplify the denominator inside the square root:[ alpha (D t + frac{1}{4 alpha}) = D t alpha + frac{1}{4} ]So,[ frac{C_0}{2 sqrt{D t alpha + frac{1}{4}}} ]Now, the exponential term:[ e^{-x^2 / (4 (D t + frac{1}{4 alpha}))} = e^{-x^2 / (4 D t + frac{1}{alpha})} ]Putting it all together, the concentration is:[ C(t, x) = frac{C_0}{2 sqrt{D t alpha + frac{1}{4}}} e^{-k t} e^{-x^2 / (4 D t + frac{1}{alpha})} ]Hmm, this seems a bit complicated. Let me see if I can write it in a more compact form.Let me factor out the constants in the denominator of the exponential:Note that ( 4 D t + frac{1}{alpha} = frac{4 D t alpha + 1}{alpha} ). So,[ frac{1}{4 D t + frac{1}{alpha}} = frac{alpha}{4 D t alpha + 1} ]Therefore, the exponential term becomes:[ e^{-x^2 alpha / (4 D t alpha + 1)} ]Similarly, the coefficient:[ frac{C_0}{2 sqrt{D t alpha + frac{1}{4}}} = frac{C_0}{2 sqrt{frac{4 D t alpha + 1}{4}}} = frac{C_0}{2 times frac{sqrt{4 D t alpha + 1}}{2}} = frac{C_0}{sqrt{4 D t alpha + 1}} ]So, putting it all together, the concentration simplifies to:[ C(t, x) = frac{C_0}{sqrt{4 D t alpha + 1}} e^{-k t} e^{- frac{alpha x^2}{4 D t alpha + 1}} ]That looks nicer. So, the solution is a Gaussian function that spreads out over time due to diffusion and decays exponentially due to degradation.So, that should be the solution to the first part.Moving on to the second part: We need to find the time ( t^* ) at which the total amount of drug in the bloodstream reaches a threshold ( M ). The total amount is given by:[ int_{-infty}^{infty} C(t, x) dx ]Given our solution ( C(t, x) ), let's compute this integral.First, write down ( C(t, x) ):[ C(t, x) = frac{C_0}{sqrt{4 D t alpha + 1}} e^{-k t} e^{- frac{alpha x^2}{4 D t alpha + 1}} ]So, the integral becomes:[ int_{-infty}^{infty} frac{C_0}{sqrt{4 D t alpha + 1}} e^{-k t} e^{- frac{alpha x^2}{4 D t alpha + 1}} dx ]Let me factor out the constants with respect to ( x ):[ frac{C_0 e^{-k t}}{sqrt{4 D t alpha + 1}} int_{-infty}^{infty} e^{- frac{alpha x^2}{4 D t alpha + 1}} dx ]This integral is again a Gaussian integral. The standard result is:[ int_{-infty}^{infty} e^{-a x^2} dx = sqrt{frac{pi}{a}} ]In our case, ( a = frac{alpha}{4 D t alpha + 1} ). Therefore,[ int_{-infty}^{infty} e^{- frac{alpha x^2}{4 D t alpha + 1}} dx = sqrt{frac{pi (4 D t alpha + 1)}{alpha}} ]So, substituting back into the expression:[ frac{C_0 e^{-k t}}{sqrt{4 D t alpha + 1}} times sqrt{frac{pi (4 D t alpha + 1)}{alpha}} ]Simplify the expression:First, the ( sqrt{4 D t alpha + 1} ) in the denominator cancels with the same term in the numerator:[ frac{C_0 e^{-k t}}{sqrt{4 D t alpha + 1}} times sqrt{frac{pi (4 D t alpha + 1)}{alpha}} = C_0 e^{-k t} sqrt{frac{pi}{alpha}} ]So, the total amount of drug is:[ Q(t) = C_0 sqrt{frac{pi}{alpha}} e^{-k t} ]Wait, that's interesting. The total amount is independent of ( D ) and ( t ) in the expression, except for the exponential decay term. So, the total drug amount decreases exponentially over time.But wait, that seems a bit counterintuitive because the concentration is spreading out due to diffusion, but the total amount is decreasing only due to degradation. So, the integral of the concentration over all space is just the initial total amount multiplied by ( e^{-k t} ).But let's verify:The initial total amount at ( t = 0 ):[ Q(0) = C_0 sqrt{frac{pi}{alpha}} ]Which is consistent with the integral of the initial condition ( C(0, x) = C_0 e^{-alpha x^2} ):[ int_{-infty}^{infty} C_0 e^{-alpha x^2} dx = C_0 sqrt{frac{pi}{alpha}} ]So, that's correct. Then, as time progresses, the total amount decreases exponentially with rate ( k ).Therefore, the total drug amount at time ( t ) is:[ Q(t) = Q(0) e^{-k t} ]Where ( Q(0) = C_0 sqrt{frac{pi}{alpha}} ).So, to find the time ( t^* ) when ( Q(t^*) = M ), we set:[ Q(0) e^{-k t^*} = M ]Solving for ( t^* ):[ e^{-k t^*} = frac{M}{Q(0)} ]Take natural logarithm on both sides:[ -k t^* = lnleft(frac{M}{Q(0)}right) ]Multiply both sides by ( -1 ):[ k t^* = -lnleft(frac{M}{Q(0)}right) = lnleft(frac{Q(0)}{M}right) ]Therefore,[ t^* = frac{1}{k} lnleft(frac{Q(0)}{M}right) ]Substitute ( Q(0) = C_0 sqrt{frac{pi}{alpha}} ):[ t^* = frac{1}{k} lnleft(frac{C_0 sqrt{frac{pi}{alpha}}}{M}right) ]Alternatively, we can write it as:[ t^* = frac{1}{k} lnleft(frac{C_0}{M} sqrt{frac{pi}{alpha}}right) ]But perhaps it's more straightforward to leave it in terms of ( Q(0) ):[ t^* = frac{1}{k} lnleft(frac{C_0 sqrt{pi / alpha}}{M}right) ]So, that's the time when the total drug amount reaches the threshold ( M ).Wait a second, let me just think if this makes sense. If ( M ) is less than ( Q(0) ), then ( t^* ) is positive, which is correct because the drug amount is decreasing over time. If ( M ) is equal to ( Q(0) ), then ( t^* = 0 ), which is correct. If ( M ) is greater than ( Q(0) ), then ( t^* ) would be negative, which doesn't make physical sense, so we should have ( M leq Q(0) ).Therefore, the time ( t^* ) is given by:[ t^* = frac{1}{k} lnleft(frac{C_0 sqrt{pi / alpha}}{M}right) ]But let me check the algebra again to make sure I didn't make a mistake.We had:[ Q(t) = C_0 sqrt{frac{pi}{alpha}} e^{-k t} ]Set ( Q(t^*) = M ):[ C_0 sqrt{frac{pi}{alpha}} e^{-k t^*} = M ]Divide both sides by ( C_0 sqrt{frac{pi}{alpha}} ):[ e^{-k t^*} = frac{M}{C_0 sqrt{frac{pi}{alpha}}} ]Take natural logarithm:[ -k t^* = lnleft(frac{M}{C_0 sqrt{frac{pi}{alpha}}}right) ]Multiply both sides by ( -1 ):[ k t^* = -lnleft(frac{M}{C_0 sqrt{frac{pi}{alpha}}}right) = lnleft(frac{C_0 sqrt{frac{pi}{alpha}}}{M}right) ]Therefore,[ t^* = frac{1}{k} lnleft(frac{C_0 sqrt{frac{pi}{alpha}}}{M}right) ]Yes, that's correct. So, that's the expression for ( t^* ).So, summarizing:1. The concentration ( C(t, x) ) is given by:[ C(t, x) = frac{C_0}{sqrt{4 D t alpha + 1}} e^{-k t} e^{- frac{alpha x^2}{4 D t alpha + 1}} ]2. The time ( t^* ) when the total drug amount reaches ( M ) is:[ t^* = frac{1}{k} lnleft(frac{C_0 sqrt{frac{pi}{alpha}}}{M}right) ]I think that's the solution.Final Answer1. The concentration is ( boxed{C(t, x) = frac{C_0}{sqrt{4 D t alpha + 1}} e^{-k t} e^{- frac{alpha x^2}{4 D t alpha + 1}}} ).2. The time ( t^* ) is ( boxed{t^* = dfrac{1}{k} lnleft( dfrac{C_0 sqrt{pi / alpha}}{M} right)} ).</think>

🧠:<think>Alright, so I have this problem where a parent is planning a fundraising event for the school's math club. There are two sub-problems here: one about ticket sales and another about a probability game. Let me try to tackle each part step by step.Starting with the first sub-problem: Ticket Sales.The parent has studied past events and found that if they price each ticket at P dollars, they can expect to sell 200 - 5P tickets. The cost to organize the event is fixed at 500. I need to write a function R(P) that represents the revenue from ticket sales minus the cost. Then, I have to determine the price P that maximizes this revenue function and find the maximum revenue.Okay, so first, let's break down what revenue is. Revenue is typically the amount of money made from selling tickets, which would be the price per ticket multiplied by the number of tickets sold. Then, we subtract the fixed cost to get the net revenue or profit.So, the revenue function R(P) would be:R(P) = (Price per ticket) * (Number of tickets sold) - Fixed costGiven that the price per ticket is P, and the number of tickets sold is 200 - 5P. So plugging that in:R(P) = P * (200 - 5P) - 500Let me compute that:R(P) = 200P - 5P² - 500Hmm, that's a quadratic function in terms of P. Quadratic functions have the form ax² + bx + c, and since the coefficient of P² is negative (-5), the parabola opens downward, meaning the vertex is the maximum point.To find the price P that maximizes R(P), I can use the vertex formula. For a quadratic function ax² + bx + c, the vertex occurs at x = -b/(2a). In this case, a = -5 and b = 200.So, P = -200 / (2 * -5) = -200 / (-10) = 20Wait, so P is 20? Let me check that.If P = 20, then the number of tickets sold is 200 - 5*20 = 200 - 100 = 100 tickets.So, revenue from tickets is 20 * 100 = 2000. Then, subtract the fixed cost of 500, so net revenue is 1500.Is that the maximum? Let me see if a slightly different price gives a higher revenue.Suppose P = 19:Number of tickets sold = 200 - 5*19 = 200 - 95 = 105Revenue = 19 * 105 = 1995Subtract cost: 1995 - 500 = 1495, which is less than 1500.What about P = 21:Number of tickets sold = 200 - 5*21 = 200 - 105 = 95Revenue = 21 * 95 = 1995Subtract cost: 1995 - 500 = 1495, same as above.So, yes, P = 20 gives the maximum revenue of 1500.Wait, but let me think again. The function is R(P) = -5P² + 200P - 500. To find the maximum, taking the derivative might be another way.dR/dP = -10P + 200. Setting derivative to zero:-10P + 200 = 0 => 10P = 200 => P = 20. Yep, same result.So, the optimal price is 20, and maximum revenue is 1500.Alright, that seems solid.Moving on to the second sub-problem: Game Probability.One of the games involves a math puzzle. The probability of a participant winning is 1/4. Each participant pays 5 to play, and the prize is 15. Let X be the number of participants. I need to write the expected profit function and determine the number of participants X needed to achieve an expected profit of 100.Okay, let's break this down. Expected profit is the expected revenue minus the expected cost.Each participant pays 5, so the revenue per participant is 5. If X participants play, total revenue is 5X.The cost is the prize money. For each participant, the expected prize payout is the probability of winning multiplied by the prize amount. So, for each participant, expected payout is (1/4)*15 = 3.75.Therefore, for X participants, the expected total payout is 3.75X.So, expected profit is total revenue minus total expected payout:Profit = 5X - 3.75X = 1.25XWait, so the expected profit is 1.25X.We need this profit to be 100. So,1.25X = 100Solving for X:X = 100 / 1.25 = 80So, 80 participants are needed to achieve an expected profit of 100.Let me verify that.Each participant brings in 5, so 80 participants bring in 80*5 = 400.The expected payout is 80*(1/4)*15 = 80*(3.75) = 300.So, profit is 400 - 300 = 100. Yep, that checks out.Alternatively, thinking in terms of expected profit per participant: each participant contributes 5 - 3.75 = 1.25. So, to get 100, you need 100 / 1.25 = 80 participants.That makes sense.Wait, is there another way to model this? Maybe considering variance or something else? But since the question is about expected profit, variance isn't required here. It's just linear expectation.So, I think that's solid.So, to summarize:1. For ticket sales, the optimal price is 20, leading to a maximum revenue of 1500.2. For the game, 80 participants are needed to achieve an expected profit of 100.I don't see any issues with these calculations. The first part is a straightforward quadratic optimization, and the second part is a linear expected value problem. Both seem to be correctly approached.Final Answer1. The optimal ticket price is boxed{20} dollars, resulting in a maximum revenue of boxed{1500} dollars.2. The number of participants needed is boxed{80}.</think>

🧠:<think>Alright, so I have two math problems here related to a Swedish e-commerce owner setting up a multilingual website using AngularJS. The first problem is about calculating the total number of hours required for translating all product descriptions into four languages using a combination of machine and manual translation. The second problem is about determining how many days it will take for a real-time language detection algorithm to reach 95% accuracy, given that it starts at 85% and improves by 1.5% for every 1000 detections, with 10,000 visitors per day.Let me tackle the first problem first. Problem 1: The website has 200 unique product descriptions. For each product, manual translation takes 1 hour per language, while machine translation takes 0.1 hours per language but requires an additional 0.2 hours of review per language. The owner is using machine translation for 75% of the content and manual translation for the remaining 25%. I need to calculate the total number of hours required for translating all product descriptions into all four languages.Okay, so let's break this down. There are 200 product descriptions, each needing translation into four languages: Swedish, English, German, and French. So, for each product, that's four translations. First, let's figure out how many product descriptions will be machine translated and how many will be manually translated. The owner is using machine translation for 75% and manual for 25%. So, 75% of 200 is 150 product descriptions, and 25% is 50 product descriptions.Now, for each of these, we need to calculate the time taken for translation and review (for machine translation). Starting with manual translation: Each manual translation takes 1 hour per language. Since each product needs to be translated into four languages, that's 4 hours per product. So, for 50 product descriptions, that's 50 * 4 = 200 hours.Next, machine translation: Each machine translation takes 0.1 hours per language, but then requires an additional 0.2 hours of review per language. So, for each product, machine translation would take (0.1 + 0.2) hours per language, which is 0.3 hours per language. Since each product is translated into four languages, that's 4 * 0.3 = 1.2 hours per product. For 150 product descriptions, that's 150 * 1.2 = 180 hours.So, adding both manual and machine translation times together: 200 hours (manual) + 180 hours (machine) = 380 hours total.Wait, let me make sure I didn't make a mistake here. So, 75% of 200 is indeed 150, and 25% is 50. For manual, 50 * 4 * 1 = 200. For machine, each product is 4 languages, each taking 0.1 + 0.2 = 0.3 hours, so 4 * 0.3 = 1.2 per product, times 150 is 180. Yeah, that adds up to 380. That seems correct.Problem 2: The owner wants to implement a real-time language detection algorithm that improves over time. The initial accuracy is 85%, and it improves by 1.5% for every 1000 detections. The website gets 10,000 visitors per day, each triggering the algorithm once. How many days until accuracy reaches 95%?Alright, so starting at 85%, needs to reach 95%. That's an improvement of 10%. Each 1000 detections improve accuracy by 1.5%. So, how many 1.5% increments are needed to get from 85% to 95%? That's 10% improvement.So, 10% divided by 1.5% per 1000 detections. Let me compute that: 10 / 1.5 = 6.666... So, approximately 6.666 increments of 1000 detections each. But since you can't have a fraction of a detection, we need to round up to 7 increments.Each increment is 1000 detections. The website gets 10,000 visitors per day, each triggering once, so 10,000 detections per day. Therefore, each day provides 10,000 / 1000 = 10 increments. Wait, hold on. Wait, no, each 1000 detections give 1.5% improvement. So, per day, with 10,000 detections, that's 10,000 / 1000 = 10 sets of 1000 detections. So, each day, the algorithm improves by 1.5% * 10 = 15%.Wait, hold on, that seems high. Wait, no, actually, no. The improvement is 1.5% per 1000 detections, regardless of how many detections happen in a day. So, each day, with 10,000 detections, that's 10,000 / 1000 = 10 times the improvement. So, each day, the algorithm improves by 1.5% * 10 = 15%.But wait, that would mean that each day, the accuracy increases by 15%. Starting at 85%, we need to reach 95%, which is 10% improvement. So, how many days?Wait, but 15% per day is a huge improvement. So, on day 1, accuracy would be 85% + 15% = 100%, which is over 95%. So, actually, it would take just 1 day? That can't be right because 10,000 detections per day, which is 10 times 1000, so 10 * 1.5% = 15% improvement per day. So, starting at 85%, after 1 day, it's 100%, which is more than 95%. So, the answer would be 1 day.But that seems counterintuitive because 10,000 is a large number, but the improvement per 1000 is 1.5%, so per day, it's 15%. So, yeah, 1 day.Wait, but maybe I'm misunderstanding the problem. It says the algorithm improves by 1.5% for every 1000 detections made. So, each detection contributes to improvement, but it's not that each 1000 detections add 1.5% to the accuracy. It's that each 1000 detections, the accuracy increases by 1.5%. So, it's a stepwise improvement every 1000 detections.So, if you have 10,000 detections in a day, that's 10 steps of 1000 detections, each adding 1.5%. So, 10 * 1.5% = 15% improvement per day.So, starting at 85%, after 1 day, it's 85% + 15% = 100%. So, it reaches 95% on day 1.But maybe the question is asking for when it reaches 95%, so it might be before the end of day 1. But since the question is about days, and you can't have a fraction of a day, it would still be 1 day.Alternatively, maybe the improvement is compounded? Or is it additive? The problem says it improves by 1.5% for every 1000 detections. So, it's additive, not multiplicative.So, each 1000 detections, add 1.5% to the accuracy. So, with 10,000 detections, that's 10 * 1.5% = 15% added. So, 85% + 15% = 100%.Therefore, it takes 1 day to reach 100%, which is more than 95%. So, the answer is 1 day.But maybe I'm misinterpreting. Maybe the improvement is 1.5% per 1000 detections, so each detection adds 0.0015%? No, that can't be, because 1.5% per 1000 is 0.0015% per detection, but that would be 10,000 * 0.0015% = 15% per day, same as before.Alternatively, maybe the improvement is 1.5% of the current accuracy for each 1000 detections, which would be multiplicative. So, each 1000 detections, the accuracy is multiplied by 1.015. So, starting at 85%, after 1000 detections, it's 85% * 1.015, then after 2000, 85%*(1.015)^2, etc.In that case, we need to solve for n where 85%*(1.015)^n >= 95%.But the problem says "improves by 1.5% for every 1000 detections made." The wording is a bit ambiguous. It could mean additive or multiplicative. But in most contexts, when they say "improves by X%", it usually means additive unless specified otherwise.But let's check both interpretations.First, additive: each 1000 detections add 1.5% to the accuracy. So, starting at 85%, need to reach 95%, which is 10% improvement. Each 1000 detections gives 1.5%, so 10 / 1.5 = 6.666..., so 7 increments of 1000 detections. Since each day has 10,000 detections, which is 10 increments, so 7 increments would take 7/10 of a day, which is 0.7 days. But since we can't have a fraction of a day, we round up to 1 day.Alternatively, multiplicative: each 1000 detections, accuracy is multiplied by 1.015. So, after n increments, accuracy is 85%*(1.015)^n. We need 85*(1.015)^n >= 95.Divide both sides by 85: (1.015)^n >= 95/85 ≈ 1.1176.Take natural log: n*ln(1.015) >= ln(1.1176).Compute ln(1.1176) ≈ 0.1118.ln(1.015) ≈ 0.0148.So, n >= 0.1118 / 0.0148 ≈ 7.56. So, n ≈ 7.56 increments of 1000 detections. Since each day provides 10 increments, 7.56 / 10 ≈ 0.756 days. Again, since we can't have a fraction of a day, it would take 1 day.But the problem says "improves by 1.5% for every 1000 detections made." The phrasing "improves by" usually implies additive. So, I think additive is the correct interpretation. So, 10% needed, 1.5% per 1000, so 7 increments, which is 7000 detections. At 10,000 per day, that's 0.7 days, so 1 day.But let me think again. If it's additive, each 1000 adds 1.5%, so 10,000 per day adds 15% per day. So, starting at 85%, after 1 day, it's 100%, which is more than 95%. So, it reaches 95% on day 1.Alternatively, if it's multiplicative, it takes about 7.56 increments, which is 7560 detections, which is less than 10,000, so it would take 1 day.Either way, the answer is 1 day.But wait, let me check the math again. If additive, each 1000 adds 1.5%, so 10,000 adds 15%. So, 85% + 15% = 100%. So, on day 1, it's already over 95%. So, the answer is 1 day.Alternatively, if it's multiplicative, 85%*(1.015)^n >=95%. So, n = ln(95/85)/ln(1.015) ≈ ln(1.1176)/ln(1.015) ≈ 0.1118 / 0.0148 ≈7.56. So, 7.56 increments, each 1000 detections. So, 7.56 *1000=7560 detections. Since 10,000 per day, 7560 is 0.756 days, so less than a day. So, it would take 1 day.But the question is about days, so regardless of the interpretation, it's 1 day.Wait, but maybe the improvement is compounded daily? Or is it compounded per detection? The problem says "improves by 1.5% for every 1000 detections made." So, it's per 1000 detections, not per day. So, each time 1000 detections are made, the accuracy increases by 1.5%. So, it's stepwise.So, starting at 85%, after 1000 detections: 86.5%, after 2000: 88%, 3000:89.5%, 4000:91%, 5000:92.5%, 6000:94%, 7000:95.5%. So, at 7000 detections, it's 95.5%, which is above 95%. So, how many days does it take to reach 7000 detections? Since 10,000 per day, 7000 is 0.7 days, so 1 day.But the question is how many days until it reaches 95%. So, on day 1, after 10,000 detections, it's already at 85% + (10,000 /1000)*1.5% =85% +15%=100%. So, it surpasses 95% on day 1.Alternatively, if it's stepwise, each 1000 detections add 1.5%, so at 7000 detections, it's 85% +7*1.5%=85%+10.5%=95.5%. So, 7000 detections needed. At 10,000 per day, that's 0.7 days, so 1 day.So, either way, the answer is 1 day.But let me think again. Maybe the improvement is not additive but multiplicative. So, each 1000 detections, the accuracy is multiplied by 1.015. So, starting at 85%, after 1000 detections:85*1.015=86.225%, after 2000:86.225*1.015≈87.46%, 3000≈88.73%, 4000≈90.03%, 5000≈91.38%, 6000≈92.78%, 7000≈94.24%, 8000≈95.75%. So, at 8000 detections, it's 95.75%, which is above 95%. So, 8000 detections needed. At 10,000 per day, that's 0.8 days, so 1 day.So, regardless of additive or multiplicative, it's 1 day.But the problem says "improves by 1.5% for every 1000 detections made." The phrase "improves by" is usually additive. So, I think additive is correct. So, 10% needed, 1.5% per 1000, so 7 increments, which is 7000 detections. At 10,000 per day, that's 0.7 days, so 1 day.Therefore, the answer is 1 day.But let me just make sure I didn't misinterpret the problem. It says the algorithm improves by 1.5% for every 1000 detections. So, each 1000 detections, it's 1.5% better. So, it's additive. So, 1000:86.5%, 2000:88%, 3000:89.5%, 4000:91%, 5000:92.5%, 6000:94%, 7000:95.5%. So, at 7000 detections, it's 95.5%, which is above 95%. So, 7000 detections needed. At 10,000 per day, that's 0.7 days, so 1 day.Therefore, the answer is 1 day.Wait, but 7000 is less than 10,000, so it would take less than a day. But since the question is about days, and partial days are not counted, it would still be 1 day.Alternatively, if we consider that the improvement happens incrementally, not in steps, then it's a continuous improvement. So, the accuracy as a function of detections is 85% + (1.5% /1000)*detections. So, set that equal to 95%:85 + (1.5/1000)*d =95(1.5/1000)*d=10d=10*(1000)/1.5≈6666.67 detections.So, 6666.67 detections needed. At 10,000 per day, that's 0.6667 days, so 1 day.So, regardless of interpretation, it's 1 day.Therefore, the answer is 1 day.But wait, let me think again. If it's additive, each 1000 adds 1.5%, so 7000 adds 10.5%, which gets us to 95.5%. So, 7000 detections, which is 0.7 days. So, if we can have partial days, it's 0.7 days, but since the question is about days, we have to round up to 1 day.Alternatively, if it's continuous, 6666.67 detections, which is 0.6667 days, still 1 day.So, yeah, the answer is 1 day.But wait, the problem says "how many days will it take for the algorithm’s accuracy to reach 95%?" So, if it reaches 95% during the first day, then it's 1 day. So, the answer is 1 day.Okay, so summarizing:Problem 1: 380 hours.Problem 2: 1 day.Final Answer1. The total number of hours required is boxed{380}.2. It will take boxed{1} day for the algorithm’s accuracy to reach 95%.</think>