Find the function with the given derivative whose graph passes through the point P.
-$$r ^ { \prime } ( t ) = \sec ^ { 2 } t - 4 , P ( 0,0 )$$
A) $r ( t ) = \tan t - 4 t$
B) $r ( t ) = \sec t \tan t - 4 t - 1$
C) $r ( t ) = \sec t - t - 6$
D) $r ( t ) = \sec t - 4 t - 4$

Question

Quizplus · Accepted Answer

To find the function $r(t)$ given its derivative $r'(t) = \sec^2 t - 4$, we integrate $r'(t)$. The integral of $\sec^2 t$ is $\tan t$, and the integral of $-4$ with respect to $t$ is $-4t$. Thus, $r(t) = \tan t - 4t + C$, where $C$ is the constant of integration. To find $C$, we use the point $P(0,0)$, which gives $0 = \tan(0) - 4(0) + C$, leading to $C = 0$. Therefore, $r(t) = \tan t - 4t$, which matches choice A.

Find the Function with the Given Derivative Whose Graph Passes r′(t)=sec⁡2t−4,P(0,0)r ^ { \prime } ( t ) = \sec ^ { 2 } t - 4 , P ( 0,0 )r′(t)=sec2t−4,P(0,0)

Find the Function with the Given Derivative Whose Graph Passes $r ^ { \prime } ( t ) = \sec ^ { 2 } t - 4 , P ( 0,0 )$