Question

section 2.6: chain rule (homework)
score: 130/170 answered: 13/17
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question 14
0/10 pts 4 99 details
a product is introduced to the market. the weekly profit (in dollars) of that product decays exponentially as function of the price that is charged (in dollars) and is given by
$p(x)=85000cdot e^{-0.04cdot x}$
suppose the price in dollars of that product, $x(t)$, changes over time $t$ (in weeks) as given by
$x(t)=52 + 0.61cdot t^{2}$
find the rate that profit changes as a function of time, $p(t)$ dollars/week
how fast is profit changing with respect to time 3 weeks after the introduction. dollars/week
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Explanation:

Step1: Apply the chain - rule

The chain - rule states that if $P(x)$ is a function of $x$ and $x(t)$ is a function of $t$, then $\frac{dP}{dt}=\frac{dP}{dx}\cdot\frac{dx}{dt}$. First, find $\frac{dP}{dx}$. Given $P(x) = 85000e^{-0.04x}$, by the derivative formula of the exponential function $\frac{d}{dx}(e^{ax})=ae^{ax}$, we have $\frac{dP}{dx}=85000\times(- 0.04)e^{-0.04x}=-3400e^{-0.04x}$.

Step2: Find $\frac{dx}{dt}$

Given $x(t)=52 + 0.61t^{2}$, by the power - rule $\frac{d}{dt}(t^{n})=nt^{n - 1}$, we get $\frac{dx}{dt}=1.22t$.

Step3: Calculate $\frac{dP}{dt}$

By the chain - rule $\frac{dP}{dt}=\frac{dP}{dx}\cdot\frac{dx}{dt}$. Substitute $\frac{dP}{dx}=-3400e^{-0.04x}$ and $\frac{dx}{dt}=1.22t$ into it. Since $x = 52+0.61t^{2}$, we have $P^{\prime}(t)=-3400e^{-0.04(52 + 0.61t^{2})}\times1.22t=-4148te^{-2.08-0.0244t^{2}}$.

Step4: Evaluate $P^{\prime}(t)$ at $t = 3$

Substitute $t = 3$ into $P^{\prime}(t)$. First, calculate the exponent: $-2.08-0.0244\times3^{2}=-2.08 - 0.0244\times9=-2.08-0.2196=-2.2996$. Then, $P^{\prime}(3)=-4148\times3\times e^{-2.2996}$. Since $e^{-2.2996}\approx0.101$, $P^{\prime}(3)=-12444\times0.101=-1256.844$.

Answer:

$P^{\prime}(t)=-4148te^{-2.08 - 0.0244t^{2}}$; $P^{\prime}(3)\approx - 1256.84$ dollars/week