Question

If $x = a\left( {\cos 2t + 2t\sin 2t} \right)$ and $y = a\left( {\sin 2t - 2t\cos 2t} \right)$. Find the second order derivative.

Answer 1

Hint: The given pair equations are in the parameterized form where both dependent and independent variables are expressed in terms of a parameter $t$. Determine the parametric derivative of $x$ with respect to $t$ and derivative of $y$ with respect to $t$. Use the values of $\dfrac{{dx}}{{dt}}$ and $\dfrac{{dy}}{{dt}}$ obtained during the calculation to determine the value of $\dfrac{{{d^2}y}}{{d{x^2}}}$.

Complete step-by-step answer:
If $f\left( x \right)$ and $g\left( x \right)$ are two differentiable functions then by product rule,
$\dfrac{d}{{dx}}\left[ {f\left( x \right)g\left( x \right)} \right] = g\left( x \right)\dfrac{d}{{dx}}f\left( x \right) + f\left( x \right)\dfrac{d}{{dx}}g\left( x \right)$
The first given parametric equation is $x = a\left( {\cos 2t + 2t\sin 2t} \right)$.
Differentiating both sides with respect to $t$ using product rule if needed,
\[ \Rightarrow \dfrac{{dx}}{{dt}} = \dfrac{d}{{dt}}\left[ {a\left( {\cos 2t + 2t\sin 2t} \right)} \right]\]
As $a$ is constant take it out because differentiation of constant is 0.
\[ \Rightarrow \dfrac{{dx}}{{dt}} = a\left( { - 2\sin 2t + 4t\cos 2t + 2\sin 2t} \right)\]
Simplify the expression,
\[ \Rightarrow \dfrac{{dx}}{{dt}} = 4at\cos 2t\]..........….. (1)
The second given parametric equation is $y = a\left( {\sin 2t - 2t\cos 2t} \right)$.
Differentiating both sides with respect to $t$ using product rule if needed,
\[ \Rightarrow \dfrac{{dy}}{{dt}} = \dfrac{d}{{dt}}\left[ {a\left( {\sin 2t - 2t\cos 2t} \right)} \right]\]
As $a$ is constant take it out because differentiation of constant is 0.
\[ \Rightarrow \dfrac{{dy}}{{dt}} = a\left( {2\cos 2t + 4t\sin 2t - 2\cos 2t} \right)\]
Simplify the expression,
\[ \Rightarrow \dfrac{{dy}}{{dt}} = 4at\sin 2t\].........….. (2)
We know that,
$\dfrac{{dy}}{{dx}} = \dfrac{{\dfrac{{dy}}{{dt}}}}{{\dfrac{{dx}}{{dt}}}}$
Substitute the values of $\dfrac{{dy}}{{dt}}$ and $\dfrac{{dx}}{{dt}}$ obtained from equation (1) and (2),
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{4at\sin 2t}}{{4at\cos 2t}}$
Cancel out the common terms,
$ \Rightarrow \dfrac{{dy}}{{dx}} = \tan t$
Differentiate again with respect to $x$,
$ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{d}{{dx}}\left( {\tan 2t} \right)$
Use the formula $\dfrac{{dy}}{{dx}} = \dfrac{{\dfrac{{dy}}{{dt}}}}{{\dfrac{{dx}}{{dt}}}}$,
$ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{\dfrac{d}{{dt}}\left( {\tan t} \right)}}{{\dfrac{{dt}}{{dx}}}}$
Substitute the value of $\dfrac{{dx}}{{dt}}$ from equation (1),
$ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{{{\sec }^2}t}}{{4at\cos t}}$
Substitute $\dfrac{1}{{\cos t}} = \sec t$,
$\therefore \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{{{\sec }^3}t}}{{4at}}$

Hence, the second derivative is $\dfrac{{{{\sec }^3}t}}{{4at}}$.

Note: Parametric equations are used when an equation cannot be expressed either in implicit or explicit form. That is exactly why we can express $\dfrac{{dy}}{{dx}}$ in terms of $\dfrac{{dy}}{{dt}}$ and $\dfrac{{dx}}{{dt}}$.