Question

If we have an expression $x = {e^t} \times sint,y = {e^t} \times cost,t$ is a parameter, then $\dfrac{{{d^2}y}}{{d{x^2}}}$at $t{\text{ }} = {\text{ }}\pi$ is equal to
1. $2{e^{( - \pi )}}$
2. $- 2{e^{( - \pi )}}$
3. $0$
4. $none{\text{ }}of{\text{ }}these$

Answer 1

We have to find the double derivative of $y$ with respect to $x$. We solve this using product rules and various basic derivative formulas of trigonometric functions , derivatives of exponential functions and derivatives of ${x^n}$ . We firstly derivate the function of $x$ and $y$ with respect to separately by using product rules . Then by simplifying the expression we get the value of $\dfrac{{dy}}{{dx}}$ . And again differentiating $\dfrac{{dy}}{{dx}}$ with respect to $x$ to get the double derivative of $y$ with respect to $x$.

Complete step-by-step solution:
Given :
$x = {e^t} \times sint,y = {e^t} \times cost$
As , we know
Derivative of product of two function is given by the following product rule :
$\dfrac{{d\left[ {f\left( x \right){\text{ }} \times {\text{ }}g\left( x \right)} \right]}}{{dx}}{\text{ }} = {\text{ }}\dfrac{{{\text{ }}d\left[ {f\left( x \right)} \right]}}{{dx}} \times {\text{ }}g{\text{ }} + {\text{ }}f{\text{ }} \times {\text{ }}\dfrac{{d\left[ {g\left( x \right)} \right]}}{{dx}}$
Also , ( derivative of ${x^n} = n \times {x^{(n - 1)}}$ )
( derivative of ${e^x} = {e^x}$ )
( Derivative of $sin{\text{ }}x{\text{ }} = {\text{ }}cos{\text{ }}x$ )
( Derivative of $cos{\text{ }}x{\text{ }} = {\text{ }} - {\text{ }}sin{\text{ }}x$ )
Now , we have to derivate $x$ with respect to $t$.
Using these derivatives , we get
$\dfrac{{dx}}{{dt}} = {e^t}[\sin t + \cos t]$ —— $(1)$
Now , we have to derive $y$ with respect to $t$.
Using these derivatives , we get
$\dfrac{{dy}}{{dt}} = {e^t}[\cos t - \sin t]$ ——— $(2)$
Dividing $(2)$ by $(1)$ , we get
${\text{ }}\dfrac{{dy}}{{dx}} = {\text{ }}\dfrac{{\left[ {\dfrac{{dy}}{{dt}}} \right]}}{{\left[ {\dfrac{{dx}}{{dt}}{\text{ }}} \right]}}$
$\dfrac{{dy}}{{dx}} = [{e^t} \times (cost - sint)]/[{e^t} \times (sint + cost)]$
Cancelling terms , we get
$\dfrac{{dy}}{{dx}}{\text{ }} = {\text{ }}\dfrac{{{\text{ }}\left( {{\text{ }}cos{\text{ }}t{\text{ }} - {\text{ }}sin{\text{ }}t{\text{ }}} \right)}}{{\left( {{\text{ }}sin{\text{ }}t{\text{ }} + {\text{ }}cos{\text{ }}t{\text{ }}} \right)}}$
Also , we know that
Derivative of quotient of two function is given by the following quotient rule :
$\dfrac{{d[\dfrac{{f(x)}}{{g(x)}}]}}{{dx}} = \dfrac{{[\dfrac{{d[f(x)]}}{{dx}} \times g(x) - f(x) \times \dfrac{{d[g(x)]}}{{dx}}]}}{{{{[g(x)]}^2}}}$
Using these derivatives , we get
$\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{[\dfrac{{(sint + cost)( - sint - cost) \times dt}}{{dx}} - \dfrac{{(cost - sint)(cost - sint) \times dt}}{{dx}}]}}{{{{(sint + cost)}^2}}}$
On simplifying , we get
$\dfrac{{{d^2}y}}{{d{x^2}}} = (\dfrac{{dt}}{{dx}}) \times \dfrac{{[ - {{(sint + cost)}^2} - {{(cost - sint)}^2}]}}{{{{(sint + cost)}^2}}}$
Expanding the term of square , using the formula
${(a + b)^2} = {a^2} + {b^2} + 2ab$
${(a - b)^2} = {a^2} + {b^2} - 2ab$
Now ,
$\dfrac{{{d^2}y}}{{d{x^2}}} = - (\dfrac{{dt}}{{dx}}) \times \dfrac{{[(si{n^2}t + co{s^2}t + 2sint \times cost) + (co{s^2}t + si{n^2}t - 2sint \times cost)]}}{{{{(sint + cost)}^2}}}$
Also we know that $(si{n^2}x + co{s^2}x = 1)$
Using the formula and cancelling the terms , we get
$\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{ - 2(\dfrac{{dt}}{{dx}})}}{{{{(sint + cost)}^2}}}$ —— $(3)$
From $(1)$ , we get the value of $\dfrac{{dt}}{{dx}}$
$\dfrac{{dt}}{{dx}} = \dfrac{1}{{\left[ {\dfrac{{dx}}{{dt}}} \right]}}$
$\dfrac{{dt}}{{dx}} = \dfrac{1}{{[{e^t} \times (sint + cost)]}}$
Putting value of $\dfrac{{dt}}{{dx}}$ in $(3)$ , we get
$\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{ - 2}}{{[{e^t} \times {{(sint + cost)}^3}]}}$—— $(3)$
Put $t{\text{ }} = {\text{ }}\pi$
$\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{ - 2}}{{[{e^\pi } \times {{(sin\pi + cos\pi )}^3}]}}$
As we know that $cos{\text{ }}\pi {\text{ }} = {\text{ }} - 1$ and $sin{\text{ }}\pi {\text{ }} = {\text{ }}0$
Using the values of trigonometric functions , we get
$\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{ - 2}}{{[{e^\pi } \times {{(0 - 1)}^3}]}}$
$\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{2}{{{e^\pi }}}$
Further simplifying we get,
$\dfrac{{{d^2}y}}{{d{x^2}}} = 2{e^{( - \pi )}}$
Thus , the value of $\dfrac{{{d^2}y}}{{d{x^2}}}$at$\;t{\text{ }} = {\text{ }}\pi$ is $2 \times {e^{( - \pi )}}$
Hence , the correct option is $\left( 1 \right)$.

Note: We differentiated $y$ with respect to $x$ to find $\dfrac{{dy}}{{dx}}$. We could also write the y in terms of x by substituting the value of $e^t$ in starting of the solution and then we use differentiation formula of trigonometric function :
$\dfrac{{d\left[ {cos{\text{ }}x} \right]}}{{dx}} = {\text{ }} - sin{\text{ }}x$
$\dfrac{{d\left[ {sin{\text{ }}x} \right]{\text{ }}}}{{dx}}{\text{ }} = {\text{ }}cos{\text{ }}x$
$d[{x^n}] = n{x^{(n - 1)}}$
$d[\tan x] = se{c^2}x$
We use the derivative according to the given problem .