Question

attempt 1: 10 attempts remaining. the resale value ( v(t) ) of a car, in dollars, ( t ) years after purchase is modeled by: ( v(t) = sqrt{-3400t + 44000} ) find the rate at which the cars value is changing after 5 years. (round your answer to the nearest cent.) ( v(5) = )?

Explanation:

Step1: Rewrite the function

We have \( V(t)=\sqrt{- 3400t + 44000}=(-3400t + 44000)^{\frac{1}{2}}\)

Step2: Apply the chain rule

The chain rule states that if \(y = u^n\) and \(u = f(t)\), then \(y^\prime=n\cdot u^{n - 1}\cdot u^\prime\)

Let \(u=-3400t + 44000\) and \(n=\frac{1}{2}\)

First, find the derivative of \(u\) with respect to \(t\): \(u^\prime=-3400\)

Then, find the derivative of \(y = u^{\frac{1}{2}}\) with respect to \(u\): \(y^\prime=\frac{1}{2}u^{-\frac{1}{2}}\)

By the chain rule, \(V^\prime(t)=\frac{1}{2}(-3400t + 44000)^{-\frac{1}{2}}\cdot(-3400)\)

Simplify the expression: \(V^\prime(t)=\frac{- 1700}{\sqrt{-3400t + 44000}}\)

Step3: Evaluate at \(t = 5\)

Substitute \(t = 5\) into \(V^\prime(t)\):

First, calculate the value inside the square root: \(-3400\times5+44000=-17000 + 44000 = 27000\)

Then, \(V^\prime(5)=\frac{-1700}{\sqrt{27000}}\)

Simplify \(\sqrt{27000}=\sqrt{900\times30}=30\sqrt{30}\approx30\times5.477 = 164.31\)

So \(V^\prime(5)=\frac{-1700}{164.31}\approx - 10.35\) (rounded to the nearest cent)

Answer:

\(-10.35\)