Question

If we have given \[ x=\tan t\text{ and }y=3\sec t\text{, then the value of }\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\text{ at }t=\dfrac{\pi }{4}\text{ is}\]
\[\begin{array}{*{35}{l}}
   \text{A) }\dfrac{3}{2\sqrt{2}} \\
   \text{B) }\dfrac{1}{3\sqrt{2}} \\
   \text{C) }\dfrac{1}{6} \\
   \text{D) }\dfrac{1}{6\sqrt{2}} \\
\end{array}\]

Answer 1

Hint: In this, we will find the value of \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\text{ at }t=\dfrac{\pi }{4}\] when\[x=\tan t\text{ and }y=3\sec t\]. At first we will derivatives of x and y with respect to t and then find \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\] by using chain rule. Then compute the value at \[t=\dfrac{\pi }{4}\].

Complete step by step answer:
Given that\[x=\tan t\text{ and }y=3\sec t\].
Differentiating x with respect to t, we get
$\dfrac{dx}{dt}=\dfrac{d\left( \tan t \right)}{dt}$
 Since, $\dfrac{d\left( \tan x \right)}{dx}={{\sec }^{2}}x$
$\Rightarrow \dfrac{dx}{dt}={{\sec }^{2}}t.....(1)$
Differentiating y with respect to t, we get
$\dfrac{dy}{dt}=\dfrac{d\left( 3\sec t \right)}{dt}=\dfrac{3d\left( \sec t \right)}{dt}$
Since, $\dfrac{d\left( \sec x \right)}{dx}=\sec x\cdot \tan x$
$\Rightarrow \dfrac{dy}{dt}=3\sec t\cdot \tan t.....(2)$
Now,$\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}},\text{ }\dfrac{dx}{dt}\ne 0.....(3)$
Since, \[t=\dfrac{\pi }{4}\text{, }\dfrac{dx}{dt}={{\sec }^{2}}\left( \dfrac{\pi }{4} \right)={{\left( \sqrt{2} \right)}^{2}}=2\]
Hence, equation (3) holds as $\dfrac{dx}{dt}\ne 0\text{ at }t=\dfrac{\pi }{4}.$
Using equation (1) and equation (2) in equation (3), we get
$\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\dfrac{3\sec t\cdot \tan t}{{{\sec }^{2}}t}$
By cancelling sect from numerator and denominator, we get
$\dfrac{dy}{dx}=\dfrac{3\tan t}{\sec t}$
Since,$\tan t=\dfrac{\sin t}{\cos t}\text{ and }\sec t=\dfrac{1}{\cos t}.$
$\dfrac{dy}{dx}=\dfrac{3\tan t}{\sec t}=\dfrac{3\left( \dfrac{\sin t}{\cos t} \right)}{\dfrac{1}{\cos t}}.$
By cancelling $\dfrac{1}{\cos t}$ from numerator and denominator, we get
$\dfrac{dy}{dx}=3\sin t.....(4)$
By chain rule and using equation (4),
$\dfrac{d{{y}^{2}}}{{{d}^{2}}x}=\dfrac{d}{dt}\left( \dfrac{dy}{dx} \right)\cdot \dfrac{dt}{dx}=\dfrac{d\left( 3\sin t \right)}{dt}\cdot \dfrac{1}{\dfrac{dx}{dt}}$
By equation (1), we get
\[\dfrac{d{{y}^{2}}}{{{d}^{2}}x}=\dfrac{3d\left( \sin t \right)}{dt}\cdot \left( \dfrac{1}{{{\sec }^{2}}t} \right)\]
Since, \[\dfrac{d\left( \sin x \right)}{dx}=\cos x\]
\[\Rightarrow \dfrac{d{{y}^{2}}}{{{d}^{2}}x}=3\cos t\cdot \left( \dfrac{1}{{{\sec }^{2}}t} \right)\]
$\dfrac{1}{\sec t}=\cos t\Rightarrow \dfrac{1}{{{\sec }^{2}}t}={{\cos }^{2}}t$
\[\Rightarrow \dfrac{d{{y}^{2}}}{{{d}^{2}}x}=3\cos t\cdot \left( {{\cos }^{2}}t \right)=3{{\cos }^{3}}t\]
At \[t=\dfrac{\pi }{4}\Rightarrow \dfrac{d{{y}^{2}}}{{{d}^{2}}x}=3{{\left( \cos \left( \dfrac{\pi }{4} \right) \right)}^{3}}=3{{\left( \dfrac{1}{\sqrt{2}} \right)}^{3}}=\dfrac{3}{2\sqrt{2}}\]
\[\dfrac{d{{y}^{2}}}{{{d}^{2}}x}=\dfrac{3}{2\sqrt{2}}\].

So, the correct answer is “Option A”.

Note: In this problem one should know all formulas of derivative and what is chain rule?. Keep in mind that we did not calculate the double derivative of x and y with respect to t and then find the double derivative of y with respect to x using double derivatives of x and y.