Question Detail
Question
A point moves along \( y = x^2 + 1 \). When \( x = 4 \), the \( x \)-coordinate increases at 5 ft/sec. At what rate is \( y \) changing?
Correct Answer
Option D is the correct answer.
Detailed Explanation
( y = x^2 + 1 \implies \frac{dy}{dt} = 2x \frac{dx}{dt} ). Given ( x=4 ) and ( \frac{dx}{dt}=5 ), then ( \frac{dy}{dt} = 2(4)(5) = 40 \text{ ft/sec} ).
Hint
Differentiate both sides with respect to time \( t \).