Question

The second order derivative of $a{\sin ^3}t$ with respect to $a{\cos ^3}t$ at $t = \dfrac{\pi }{4}$ is?
A)$2$
B)$\dfrac{1}{{12a}}$
C)$\dfrac{{4\sqrt 2 }}{{3a}}$
D)$\dfrac{{3a}}{{4\sqrt 2 }}$

Answer 1

Here we are asked to find the second derivative of a function with respect to another function. Since both the functions are dependent on $t$, we can find derivatives of them with respect to $t$. Then dividing we get the first derivative in the question. Again by differentiating and substituting the value for $t$, we get the required second derivative.

Useful formula:
For functions $x,y$ and some parameter $t$,
$\dfrac{{dy}}{{dx}} = \dfrac{{\dfrac{{dy}}{{dt}}}}{{\dfrac{{dx}}{{dt}}}}$
Also, for any $x,t$ and $u$ as a function of $x$,we have
$\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}$, where $n$ is any number.
$\dfrac{{d(\sin t)}}{{dt}} = \cos t$
$\dfrac{{d(\cos t)}}{{dt}} = - \sin t$
$\dfrac{{d(\tan t)}}{{dt}} = {\sec ^2}t$
$\dfrac{{du(x)}}{{dt}} = \dfrac{{du}}{{dx}} \times \dfrac{{dx}}{{dt}}$
Also, we have, $\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }},\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$
$\sec t = \dfrac{1}{{\cos t}}$

Complete step-by-step answer:
To find the second order derivative of $a{\sin ^3}t$ with respect to $a{\cos ^3}t$,
Let $y = a{\sin ^3}t$, $x = a{\cos ^3}t$
So, we have to find $\dfrac{{{d^2}y}}{{d{x^2}}}$.
We have $\dfrac{{dy}}{{dx}} = \dfrac{{\dfrac{{dy}}{{dt}}}}{{\dfrac{{dx}}{{dt}}}}$
Consider $y = a{\sin ^3}t$
Differentiating both sides with respect to $t$, we get,
$\dfrac{{dy}}{{dt}} = a \times 3{\sin ^2}t \times \dfrac{{d(\sin t)}}{{dt}}$ $\sec t = \dfrac{1}{{\cos t}}$
(Since $\dfrac{{du(x)}}{{dt}} = \dfrac{{du}}{{dx}} \times \dfrac{{dx}}{{dt}}$ and $\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}$)
Also $\dfrac{{d(\sin t)}}{{dt}} = \cos t$
$\Rightarrow \dfrac{{dy}}{{dt}} = 3a{\sin ^2}t\cos t - - - (i)$
Now consider $x = a{\cos ^3}t$
Differentiating both sides with respect to $t$, we get,
$\dfrac{{dx}}{{dt}} = a \times 3{\cos ^2}t \times \dfrac{{d(\cos t)}}{{dt}}$
(Since $\dfrac{{du(x)}}{{dt}} = \dfrac{{du}}{{dx}} \times \dfrac{{dx}}{{dt}}$ and $\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}$)
Also $\dfrac{{d(\cos t)}}{{dt}} = - \sin t$
$\Rightarrow \dfrac{{dx}}{{dt}} = 3a{\cos ^2}t \times - \sin t$
$\Rightarrow \dfrac{{dx}}{{dt}} = - 3a{\cos ^2}t\sin t - - - (ii)$
Now we have, $\dfrac{{dy}}{{dx}} = \dfrac{{\dfrac{{dy}}{{dt}}}}{{\dfrac{{dx}}{{dt}}}}$
Substituting using $(i)\& (ii)$ we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{3a{{\sin }^2}t\cos t}}{{ - 3a{{\cos }^2}t\sin t}}$
Simplifying we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{{\sin t}}{{\cos t}}$
$\Rightarrow \dfrac{{dy}}{{dx}} = - {\text{tan t}}$
Again differentiating both sides with respect to $t$ we get,
$\dfrac{{{d^2}y}}{{d{x^2}}} = - {\sec ^2}t \times \dfrac{{dt}}{{dx}}$ (since $\dfrac{{d(\tan t)}}{{dt}} = {\sec ^2}t$)
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{ - {{\sec }^2}t}}{{\dfrac{{dx}}{{dt}}}}$
Substituting for $\dfrac{{dx}}{{dt}} = - 3a{\cos ^2}t\sin t$ we get,
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{ - {{\sec }^2}t}}{{ - 3a{{\cos }^2}t\sin t}}$
Since $\sec t = \dfrac{1}{{\cos t}}$, squaring both sides we have, ${\sec ^2}t = \dfrac{1}{{{{\cos }^2}t}}$
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{1}{{3a{{\cos }^2}t\sin t \times {{\cos }^2}t}}$
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{1}{{3a{{\cos }^4}t\sin t}}$
We need to find $\dfrac{{{d^2}y}}{{d{x^2}}}$ at $t = \dfrac{\pi }{4}$.
$\Rightarrow {\left( {\dfrac{{{d^2}y}}{{d{x^2}}}} \right)_{t = \dfrac{\pi }{4}}} = \dfrac{1}{{3a{{\cos }^4}\dfrac{\pi }{4}\sin \dfrac{\pi }{4}}}$
We have, $\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }},\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$
Substituting we get,
$\Rightarrow {\left( {\dfrac{{{d^2}y}}{{d{x^2}}}} \right)_{t = \dfrac{\pi }{4}}} = \dfrac{1}{{3a{{\left( {\dfrac{1}{{\sqrt 2 }}} \right)}^4}\left( {\dfrac{1}{{\sqrt 2 }}} \right)}}$
$\Rightarrow {\left( {\dfrac{{{d^2}y}}{{d{x^2}}}} \right)_{t = \dfrac{\pi }{4}}} = \dfrac{1}{{3a\dfrac{1}{4} \times \dfrac{1}{{\sqrt 2 }}}}$
Simplifying we get,
$\Rightarrow {\left( {\dfrac{{{d^2}y}}{{d{x^2}}}} \right)_{t = \dfrac{\pi }{4}}} = \dfrac{{4\sqrt 2 }}{{3a}}$
$\therefore$ The answer is option C.

Note: We could use the above method here because both the given functions were dependent on a common variable. Otherwise taking derivatives about a function is a tedious task. The value of $t$ must be substituted after finding the second derivative. It is wrong if we substitute it in the first derivative itself and then differentiating the second time.