The total stopping distance T of a vehicle is where T is in feet and x is the speed in miles per hour.Use differentials to approximate the percent change in total stopping distance as speed changes from x = 22 to x = 27 miles per hour.Round your answer to one decimal place.
A)51.8%
B)22.1%
C)42.7%
D)78.0%
E)122.7%

Question

The total stopping distance T of a vehicle is  where T is in feet and x is the speed in miles per hour.Use differentials to approximate the percent change in total stopping distance as speed changes from x = 22 to x = 27 miles per hour.Round your answer to one decimal place. ​
A)51.8% 
B)22.1% 
C)42.7% 
D)78.0% 
E)122.7%

Quizplus · Accepted Answer

Using differentials, we have:
dT ≈ (∂T/∂x)dx 
where dT is the change in total stopping distance and dx is the change in speed.
Thus, the percent change in total stopping distance is:
[(∂T/∂x)dx/T] × 100%
At x = 22 mph, we have:
T(22) = 67.6 ft 
At x = 27 mph, we have:
T(27) = 94.3 ft 
Therefore, dx = 5 mph, and we need to calculate (∂T/∂x) at x = 22 mph.
∂T/∂x = 0.1x^2 + 2x + 20
∂T/∂x|_(x=22) = 74.4 ft/mph
Hence, the percent change in total stopping distance is approximately:
[(74.4)(5)/(67.6)] × 100% ≈ 42.7%
Therefore, the best choice is C.

The Total Stopping Distance T of a Vehicle Is