$\text { Find the second derivative } \frac { d ^ { 2 } y } { d x ^ { 2 } } \text { of the parametric equations } x = 4 \cos \theta , y = 4 \sin \theta \text {. }$
A) $- \frac { \tan ^ { 3 } \theta } { 4 }$
B) $- \frac { \csc ^ { 3 } \theta } { 4 }$
C) $\frac { \csc ^ { 3 } \theta } { 4 }$
D) $- \frac { \sec ^ { 3 } \theta } { 4 }$
E) $\frac { \sec ^ { 3 } \theta } { 4 }$

Question

$\text { Find the second derivative } \frac { d ^ { 2 } y } { d x ^ { 2 } } \text { of the parametric equations } x = 4 \cos \theta , y = 4 \sin \theta \text {. }$ 
A) $- \frac { \tan ^ { 3 } \theta } { 4 }$
B) $- \frac { \csc ^ { 3 } \theta } { 4 }$
C) $\frac { \csc ^ { 3 } \theta } { 4 }$
D) $- \frac { \sec ^ { 3 } \theta } { 4 }$
E) $\frac { \sec ^ { 3 } \theta } { 4 }$

Quizplus · Accepted Answer

To find the second derivative $ \frac { d ^ { 2 } y } { d x ^ { 2 } } $ for the parametric equations $ x = 4 \cos \theta $ and $ y = 4 \sin \theta $, we first find $ \frac{dy}{dx} $ and then $ \frac{d^2y}{dx^2} $. First derivative:$$ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{4\cos\theta}{-4\sin\theta} = -\cot\theta $$Second derivative involves differentiating $ \frac{dy}{dx} $ with respect to $\theta$ and then dividing by $dx/d\theta$:$$ \frac{d^2y}{dx^2} = \frac{d(-\cot\theta)/d\theta}{dx/d\theta} = \frac{\csc^2\theta}{-4\sin\theta} = -\frac{\csc^3\theta}{4} $$Thus, the correct answer is $- \frac { \csc ^ { 3 } \theta } { 4 }$.

A) −tan⁡3θ4- \frac { \tan ^ { 3 } \theta } { 4 }−4tan3θ​

A) $- \frac { \tan ^ { 3 } \theta } { 4 }$