Question

If $x=a{{\cos }^{2}}t,y=a{{\sin }^{2}}t$, find $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$.

Answer 1

Hint:Simplify the equation above by eliminating the variable t and using the trigonometric identity ${{\cos }^{2}}t+{{\sin }^{2}}t=1$. Then differentiate the simplified function twice to find the second derivative of the given function using the fact that differentiation of any function of the form $y=a{{x}^{n}}$ is $\dfrac{dy}{dx}=an{{x}^{n-1}}$.

Complete step-by-step answer:
We have the function $x=a{{\cos }^{2}}t,y=a{{\sin }^{2}}t$. We have to find the second derivative of the given function.
This function is given in parametric form. We will first simplify the given function to eliminate the variable t.
Adding the two equations, we get $x+y=a({{\cos }^{2}}t+{{\sin }^{2}}t)$.
We know that ${{\cos }^{2}}t+{{\sin }^{2}}t=1$.
Hence, we get $x+y=a$ as the function.
We will now differentiate it twice to evaluate the second derivative.
We will begin by differentiating the given function with respect to x on both sides of the equation.
Thus, we have $\dfrac{d}{dx}\left( x+y \right)=\dfrac{d}{dx}a$.
We can rewrite the above equation as $\dfrac{d}{dx}\left( x \right)+\dfrac{d}{dx}\left( y \right)=\dfrac{d}{dx}a.....\left( 1 \right)$.
We know that differentiation of a constant is zero with respect to any variable. Thus, we have $\dfrac{d}{dx}a=0.....\left( 2 \right)$.
We know that differentiation of any function of the form $y=a{{x}^{n}}$ is $\dfrac{dy}{dx}=an{{x}^{n-1}}$.
Substituting $a=1,n=1$ in the above equation, we get $\dfrac{d}{dx}\left( x \right)=1.....\left( 3 \right)$.
To find the value of $\dfrac{d}{dx}\left( y \right)$, multiply and divide the equation by $dy$.
Thus, we have $\dfrac{d}{dx}\left( y \right)=\dfrac{d}{dy}\left( y \right)\times \dfrac{dy}{dx}=1\times \dfrac{dy}{dx}=\dfrac{dy}{dx}.....\left( 4 \right)$.
Substituting the value of equation $\left( 2 \right),\left( 3 \right),\left( 4 \right)$ in equation $\left( 1 \right)$, we get $\dfrac{d}{dx}\left( x \right)+\dfrac{d}{dx}\left( y \right)=\dfrac{d}{dx}a$.
So, we have $1+\dfrac{dy}{dx}=0$.
Simplifying the above equation, we have $\dfrac{dy}{dx}=-1$.
We will differentiate the above function again to find the second derivative as $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)$.
Thus, we have $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)=\dfrac{d}{dx}\left( -1 \right)$.
We know that differentiation of a constant is zero with respect to any variable.
So, we have $\dfrac{d}{dx}\left( -1 \right)=0$.
Thus, we have $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)=\dfrac{d}{dx}\left( -1 \right)=0$.
Hence, the second derivative of the given function is $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=0$.

Note: The function is given in parametric form. We can also solve this question by differentiating each function of x and y with respect to the variable t twice and then dividing the two functions to get the double derivative of the required function, instead of eliminating the parameter t. However, this method will be time consuming.