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

Question

Quizplus · Accepted Answer

To find the function $r(t)$ given its derivative $r'(t) = \sec^2 t - 4$, we integrate $r'(t)$ with respect to $t$. The integral of $\sec^2 t$ is $\tan t$, and the integral of $-4$ is $-4t$, so $r(t) = \tan t - 4t + C$, 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, $r(t) = \tan t - 4t$.

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

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