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{-3000t + 43000} find the rate at which the cars value is changing after 6 years. (round your answer to the nearest cent.) v(6) = ? submit answer next item

Explanation:

Step 1: Rewrite the function

We have \( V(t)=\sqrt{-3000t + 43000}=(-3000t + 43000)^{\frac{1}{2}} \)

Step 2: Apply the chain rule

The chain rule states that if \( y = u^n \), then \( y^\prime=nu^{n - 1}u^\prime \). Let \( u=-3000t + 43000 \) and \( n=\frac{1}{2} \)
First, find the derivative of \( u \) with respect to \( t \): \( u^\prime=\frac{d}{dt}(-3000t + 43000)=- 3000 \)
Then, find the derivative of \( V(t) \) with respect to \( t \) using the chain rule:
\( V^\prime(t)=\frac{1}{2}(-3000t + 43000)^{\frac{1}{2}-1}\times(-3000) \)
Simplify the exponent: \( \frac{1}{2}-1=-\frac{1}{2} \)
So \( V^\prime(t)=\frac{1}{2}(-3000t + 43000)^{-\frac{1}{2}}\times(-3000)=\frac{- 1500}{\sqrt{-3000t + 43000}} \)

Step 3: Evaluate at \( t = 6 \)

Substitute \( t = 6 \) into \( V^\prime(t) \):
First, calculate the value inside the square root: \( -3000\times6 + 43000=-18000 + 43000 = 25000 \)
Then, \( V^\prime(6)=\frac{-1500}{\sqrt{25000}} \)
Simplify \( \sqrt{25000}=\sqrt{25\times1000}=50\sqrt{10}\approx50\times3.1623 = 158.115 \)
\( V^\prime(6)=\frac{-1500}{158.115}\approx - 9.49 \)

Answer:

\( V^\prime(6)\approx\boxed{-9.49} \)