$$\text { Find the value of } d ^ { 2 } y / d x ^ { 2 } \text { at the point defined by the given value of } t \text {. }$$
-$$x = t + \cos t , y = 2 - \sin t , t = \frac { \pi } { 6 }$$
A) $- 2$
B) $- 4$
C) $- \sqrt { 3 }$
D) $\frac { 1 } { 4 }$

Question

Quizplus · Accepted Answer

Using the given parametric equations, we can find the first and second derivatives:

$$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-\cos t}{- \sin t - 1}$$

$$\frac{d^2y}{dx^2} = \frac{d}{dx}(\frac{dy}{dx}) = \frac{d}{dt}(\frac{dy}{dx}) \cdot \frac{dt}{dx}$$

We can find $\frac{dt}{dx}$ by taking the reciprocal of $\frac{dx}{dt}$:

$$\frac{dx}{dt} = 1 - \sin t \quad \Rightarrow \quad \frac{dt}{dx} = \frac{1}{1 - \sin t}$$

Now,

$$\frac{d}{dt}(\frac{dy}{dx}) = \frac{d}{dt}(\frac{-\cos t}{- \sin t - 1}) = \frac{\sin^2 t}{(\sin t + 1)^3}$$

Putting it all together:

$$\frac{d^2y}{dx^2} = \frac{d}{dt}(\frac{dy}{dx}) \cdot \frac{dt}{dx} = \frac{\sin^2 t}{(\sin t + 1)^3} \cdot \frac{1}{1 - \sin t}$$

Plugging in $t = \frac{\pi}{6}$ gives:

$$\frac{d^2y}{dx^2} = \frac{1}{8} \cdot (-2) = -\frac{1}{4}$$

Thus, the answer is $\boxed{	ext{B}}$.

Find the value of d2y/dx2 at the point defined by the given value of t. \text { Find the value of } d ^ { 2 } y / d x ^ { 2 } \text { at the point defined by the given value of } t \text {. } Find the value of d2y/dx2 at the point defined by the given value of t.

$\text { Find the value of } d ^ { 2 } y / d x ^ { 2 } \text { at the point defined by the given value of } t \text {. }$