Answer to Question #155284 in Calculus for Pia

Question

Find dy/dx and d²y/dx² without eliminating the parameter.

a.) x= e^(2t) , y= t In(t)

b.) x= sinh(t) , y= cosh(t)

c.) x= 1-t^2 , y= 1+t

Answer 1

We shall use the formulas:

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}},\ \ \frac{d^2y}{dx^2}=\frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}$

a.) $x= e^{2t} , y= t\ln(t)$

$\frac{dy}{dx}=\frac{\ln(t)+t\frac{1}{t}}{2e^{2t}}=\frac{\ln(t)+1}{2e^{2t}}$

$\frac{d^2y}{dx^2}=\frac{1}{2e^{2t}}\cdot \frac{\frac{1}{t}2e^{2t}-(\ln(t)+1)4e^{2t}}{4e^{4t}}= \frac{1-2t(\ln(t)+1)}{4te^{4t}}$

b.) $x= \sinh(t) , y= \cosh(t)$

$\frac{dy}{dx}=\frac{\sinh(t)}{\cosh(t)}=\tanh(t)$

$\frac{d^2y}{dx^2}=\frac{1-\tanh^2(t)}{\cosh(t)}$

c.) $x= 1-t^2 , y= 1+t$

$\frac{dy}{dx}=\frac{1}{-2t}=-\frac{1}{2t}$

$\frac{d^2y}{dx^2}=\frac{1}{-2t}\cdot\frac{1}{2t^2}=-\frac{1}{4t^3}$