Question

A curve is represented parametrically by the equations $x=f\left(t\right)=a^{\ln \left(b^t\right)}$ and $y=g\left(t\right)=b^{-\ln \left(a^t\right)};a,b>0$ and $a\ne 1,b\ne 1$ where $t\in R$.

The value of $\frac{d^{2}y}{d{x}^2}$ at the point where $f\left(t\right)=g\left(t\right)$ is

• A. $0$
• B. $\frac{1}{2}$
• C. $1$
• D. $2$

Answer 1

The correct option is D $2$

$x=f\left(t\right)=a^{\ln \left(b^t\right)}\Rightarrow x=a^{t\ln b}={\left(a^{\ln b}\right)}^t$
$y=g\left(t\right)=b^{-\ln \left(a^t\right)}\Rightarrow y=b^{-t\ln a}={\left(b^{\ln a}\right)}^{-t}\Rightarrow y={\left(a^{\ln b}\right)}^{-t}\Rightarrow y=a^{-t\ln b}=\frac{1}{x}\Rightarrow xy=1\Rightarrow x\frac{dy}{dx}+y=0\Rightarrow \frac{dy}{dx}=-\frac{y}{x}=-\frac{1}{x^2}$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{2}{x^3}$

When,
$f\left(t\right)=g\left(t\right)\Rightarrow x=y=1\Rightarrow t=0$
Putting $t=0,x=1$
$\frac{d^{2}y}{d{x}^2}=\frac{2}{1^3}=2$