Question

If we have $x={{e}^{t}}\sin t,y={{e}^{t}}\cos t$ are parametric equations, then $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ at (1,1) is equal to:
(a) $-\dfrac{1}{2}$
(b) $-\dfrac{1}{4}$
(c) 0
(d) $\dfrac{1}{2}$

Answer 1

Hint: Get the values of $\dfrac{dy}{dx}$ and $\dfrac{dy}{dt}$ using the ‘u.v’ rule of derivative and use the relation $\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}$ . u.v for two functions in product is given as $\dfrac{d}{dx}u.v=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ can be calculated by the relation $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)=\dfrac{d}{dt}\left( \dfrac{dy}{dx} \right)\dfrac{dt}{dx}$ . Use the following identities to get the answer as $\dfrac{d}{dx}{{e}^{x}}={{e}^{x}},\dfrac{d}{dx}\cos x=-\sin x$ and $\dfrac{d}{dx}\sin x=\cos x$ .Use the trigonometric identity of $\tan \left( A-B \right)$ which is given as
$\tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}$

Complete step-by-step solution -
We are given values of variables $x$ and $y$ in term of variable ‘t’ as
$\begin{aligned} & x={{e}^{t}}\sin t.........\left( i \right) \\\ & y={{e}^{t}}\cos t.........\left( ii \right) \\\ \end{aligned}$
And hence, with the help of above equations, we need to calculate the value of $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ at (1, 1). As we know $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ is the derivative of $\dfrac{dy}{dx}$ i.e. $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)..............\left( iii \right)$
And we know the relation $\dfrac{dy}{dx}$ can be calculated by evaluating the terms $\dfrac{dy}{dt}$ and $\dfrac{dx}{dt}$ from the equation(i) and (ii) and use the following relation among them. So, we have
$\dfrac{dy}{dx}=\dfrac{\left( \dfrac{dy}{dt} \right)}{\left( \dfrac{dx}{dt} \right)}..........\left( iv \right)$
So let us calculate the values of $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$ , by differentiating the equation (i) and (ii).
So, value of $\dfrac{dy}{dt}$ can be calculated from equation(i) as
$\dfrac{dx}{dt}=\dfrac{d}{dt}\left( {{e}^{t}}\sin t \right).................\left( v \right)$
As ${{e}^{t}}$ and $\sin t$ are in product, so we need to apply the multiplication rule of differentiation which is given as
$\dfrac{d}{dx}\left( uv \right)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}..........\left( vi \right)$
Hence, suppose $u={{e}^{t}}$ and $v=\sin t$ from the equation(v) and (vi).
So, we get value of $\dfrac{dx}{dt}$ as
$\dfrac{dx}{dt}={{e}^{t}}\dfrac{d}{dt}\left( \sin t \right)+\sin t\dfrac{d}{dt}{{e}^{t}}$
We know
$\begin{aligned} & \dfrac{d}{dx}{{e}^{x}}={{e}^{x}} \\\ & \dfrac{d}{dx}\sin x=\cos x \\\ \end{aligned}$
Hence, we get
$\begin{aligned} & \dfrac{dx}{dt}={{e}^{t}}\cos t+\sin t\left( {{e}^{t}} \right) \\\ & \dfrac{dx}{dt}={{e}^{t}}\cos t+{{e}^{t}}\sin t............\left( vii \right) \\\ \end{aligned}$
Similarly, we can calculate $\dfrac{dy}{dt}$ by differentiating equation(ii) with respect to ‘t’. So, we get
$\dfrac{dy}{dt}=\dfrac{d}{dt}\left( {{e}^{t}}\cos t \right)$
Now, using the equation(vi), where taking u and v as ${{e}^{t}}$ and $\cos t$ .So, we get
$\dfrac{d}{dt}\left( {{e}^{t}}\cos t \right)={{e}^{t}}\dfrac{d}{dt}\cos t+\cos t\dfrac{d}{dt}{{e}^{t}}$
We know
$\dfrac{d}{dx}\cos x=-\sin x$ and $\dfrac{d}{dx}{{e}^{x}}={{e}^{x}}$
So, we get
$\begin{aligned} & \dfrac{dy}{dt}={{e}^{t}}\left( -\sin t \right)+{{e}^{t}}\cos t \\\ & \dfrac{dy}{dt}=-{{e}^{t}}\sin t+{{e}^{t}}\cos t..........\left( viii \right) \\\ \end{aligned}$
Now, we can get values of $\dfrac{dy}{dx}$ from equation(iv), (vii) and (viii) as
$\begin{aligned} & \dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\dfrac{-{{e}^{t}}\sin t+{{e}^{t}}\cos t}{{{e}^{t}}\cos t+{{e}^{t}}\sin t} \\\ & \dfrac{dy}{dx}=\dfrac{{{e}^{t}}\left[ -\sin t+\cos t \right]}{{{e}^{t}}\left[ \sin t+\cos t \right]} \\\ & \dfrac{dy}{dx}=\dfrac{\cos t-\sin t}{\cos t+\sin t} \\\ \end{aligned}$
Let us divide the numerator and denominator of the above equation by cost. So, we get
$\begin{aligned} & \dfrac{dy}{dx}=\dfrac{\dfrac{\cos t-\sin t}{\cos t}}{\dfrac{\cos t+\sin t}{\cos t}} \\\ & \dfrac{dy}{dx}=\dfrac{1-\dfrac{\sin t}{\cos t}}{1+\dfrac{\sin t}{\cos t}} \\\ \end{aligned}$
We know $\dfrac{\sin \theta }{\cos \theta }=\tan \theta$ and $\tan \dfrac{\pi }{4}=1$ .So, we can rewrite the above equation as
$\dfrac{dy}{dx}=\dfrac{\tan \dfrac{\pi }{4}-\tan t}{1+\left( \tan \dfrac{\pi }{4} \right)\left( \tan t \right)}$
We know the trigonometric identity of $\tan \left( A-B \right)$ is given as
$\tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}$
Hence, we can get value of $\dfrac{dy}{dx}$ by comparing with the above relation as
$\dfrac{dy}{dx}=\tan \left( \dfrac{\pi }{4}-t \right).........\left( ix \right)$
Now, from the equation(iii), we get
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)$
Now, we can multiply and divide the RHS of the above equation by ‘dt’. So, we get
$\begin{aligned} & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)\dfrac{dt}{dx} \\\ & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dt}\left( \dfrac{dy}{dx} \right)\dfrac{dt}{dx}..........\left( \text{x} \right) \\\ \end{aligned}$
Hence, we can get value of $\dfrac{dt}{dx}$ from the equation (vii) as
$\begin{aligned} & \dfrac{dt}{dx}=\dfrac{1}{\left( \dfrac{dx}{dt} \right)} \\\ & \dfrac{dt}{dx}=\dfrac{1}{{{e}^{t}}\left[ \sin t+\cos t \right]}.............\left( xi \right) \\\ \end{aligned}$

And value of $\dfrac{d}{dt}\left( \dfrac{dy}{dx} \right)$ can be calculated using the equation (ix) as
$\dfrac{d}{dt}\left( \dfrac{dy}{dx} \right)=\dfrac{d}{dt}\left( \tan \left( \dfrac{\pi }{4}-t \right) \right)$
As we know
$\dfrac{d}{dx}\tan \theta ={{\sec }^{2}}\theta$
Hence, we get
$\begin{aligned} & \dfrac{d}{dt}\left( \dfrac{dy}{dx} \right)={{\sec }^{2}}\left( \dfrac{\pi }{4}-t \right)\times \dfrac{d}{dt}\left( \dfrac{\pi }{4}-t \right) \\\ & \dfrac{d}{dt}\left( \dfrac{dy}{dx} \right)=-1\times {{\sec }^{2}}\left( \dfrac{\pi }{4}-t \right) \\\ & \dfrac{d}{dt}\left( \dfrac{dy}{dx} \right)=-{{\sec }^{2}}\left( \dfrac{\pi }{4}-t \right).........\left( xii \right) \\\ \end{aligned}$
Now, we can get value of $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ from the equations (x), (xi) and (xii) as
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-{{\sec }^{2}}\left( \dfrac{\pi }{4}-t \right)\times \dfrac{1}{{{e}^{t}}\left[ \sin t+\cos t \right]}.........\left( xiii \right)$
Now, we need to find $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ at (1, 1) i.e. at the values of $x=1$ and y = 1.
Now, we know the relation
$\begin{aligned} & x={{e}^{t}}\sin t \\\ & y={{e}^{t}}\cos t \\\ \end{aligned}$
Divide both the equations, we get
$\dfrac{x}{y}=\dfrac{{{e}^{t}}\sin t}{{{e}^{t}}\cos t}=\dfrac{\sin t}{\cos t}$
Put $x=1$ and $y=1$ and replace $\dfrac{\sin t}{\cos t}$ by $\tan t$ using the equation
$\tan \theta =\dfrac{\sin \theta }{\cos \theta }$
So, we get
$\begin{aligned} & 1=\tan t \\\ & \Rightarrow \tan t=1 \\\ \end{aligned}$
We know $\tan t$ will be equal to 1 at $t=\dfrac{\pi }{4}.$ So, we get value of $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ as
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-{{\sec }^{2}}\left( \dfrac{\pi }{4}-\dfrac{\pi }{4} \right)\times \dfrac{1}{{{e}^{\dfrac{\pi }{4}}}\left[ \sin \dfrac{\pi }{4}+\cos \dfrac{\pi }{4} \right]}$
As we know $\sin \dfrac{\pi }{4}=\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$
So, we get
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-{{\sec }^{2}}0\times \dfrac{1}{{{e}^{\dfrac{\pi }{4}}}\left[ \dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}} \right]}$
We know sec0 = 1. Hence, we get
$\begin{aligned} & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-1}{{{e}^{\dfrac{\pi }{4}}}\left( \dfrac{2}{\sqrt{2}} \right)} \\\ & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-\sqrt{2}}{2{{e}^{\dfrac{\pi }{4}}}}..............\left( xiv \right) \\\ \end{aligned}$
Now, from the equation(i) we have
$x={{e}^{t}}\sin t$
As $x=1$ and $t=\dfrac{\pi }{4}$ will satisfy the above equation.
So, we get
$1={{e}^{\dfrac{\pi }{4}}}\sin \dfrac{\pi }{4}$
$\begin{aligned} & 1={{e}^{\dfrac{\pi }{4}}}\dfrac{1}{\sqrt{2}} \\\ & {{e}^{\dfrac{\pi }{4}}}=\sqrt{2} \\\ \end{aligned}$
Now, we can replace ${{e}^{\dfrac{\pi }{4}}}$ by $\sqrt{2}$ in the equation (xiv) and hence, we get
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ at (1, 1) = $\dfrac{-2}{2\sqrt{2}}=\dfrac{-1}{2}$
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ at (1, 1) = $\dfrac{-1}{2}$
So, option(a) is the correct answer.

Note: Another approach for calculating $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ would be given as
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)$
Directly substitute $\dfrac{dy}{dx}=\tan \left( \dfrac{\pi }{4}-t \right)$ to the above expression, we get $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \tan \left( \dfrac{\pi }{4}-t \right) \right)$
Use chain rule of differentiation, we get
$\begin{aligned} & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}={{\sec }^{2}}\left( \dfrac{\pi }{4}-t \right)\dfrac{d}{dx}\left( \dfrac{\pi }{4}-t \right) \\\ & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}={{\sec }^{2}}\left( \dfrac{\pi }{4}-t \right)\left( -\dfrac{dt}{dx} \right) \\\ \end{aligned}$
Now, put value of $\dfrac{dt}{dx}$ to get $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$
One may go wrong, if he/she first put $x=1$ and $y=1$ to the given expression and get $t=\dfrac{\pi }{4},$ so, he/she will get expression as $x=\dfrac{{{e}^{\dfrac{\pi }{4}}}}{\sqrt{2}}$ and $y=\dfrac{{{e}^{\dfrac{\pi }{4}}}}{\sqrt{2}}$ , and hence differentiation of them is 0, and cannot be able to get $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ , which is the blunder mistake can be possible by the students. So, take care and be careful with the questions related to them.