Answer to Question #155703 in Calculus for Pia

Question

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

a.) x= acosh(t) , y= bsinh(t)

b.) x= acos(t) , y= bsin(t)

Answer 1

To find $\frac{dy}{dx}$ , $\frac{d^2y}{dx^2}$ will be used formula $\frac{dy}{dx}=\frac{dy}{dt}/\frac{dx}{dt}$ . Such formula are takeen the standard chain rule 

$\frac{dy}{dt} =\frac{ dy}{dx }\frac{ dx}{dt}$

a) Let's take derivatives of the $x$ and $y$ with respect to the independent variable $t$

$x=a\cosh{t}$

$y=b\sinh{t}$

$\frac{dx}{dt}=a\sinh{t}$

$\frac{dy}{dt}=b\cosh{t}$

Substitute derivatives into formula $\frac{dy}{dx}=\frac{dy}{dt}/\frac{dx}{dt}$ :

$\frac{dy}{dx}=\frac{b\cosh{t}}{a\sinh{t}}=\frac{b}{a}\coth{t}$

To find $\frac{d^2y}{dx^2}$ , firstly take the derivative of the function for $\frac{dy}{dx}$ (as $y'$ ) :

$y'=\frac{d}{dt}(\frac{dy}{dx})=\frac{b}{a}(1-\coth^2{t})=-\frac{b}{a}\sinh^{-2}{t}$

$\frac{d^2y}{dx^2}=\frac{dy'}{\frac{dx}{dt}}= \frac{b}{a}(1-\coth^2{t})/(a\sinh{t})=-\frac{b}{a^2}\sinh^3{t} ​$

b) Let's take derivatives of the $x$ and $y$ with respect to the independent variable $t$

$x=a\cos{t}$

$y=b\sin{t}$

$\frac{dx}{dt}=-a\sin{t}$

$\frac{dy}{dt}=b\cos{t}$

Substitute derivatives into formula $\frac{dy}{dx}=\frac{dy}{dt}/\frac{dx}{dt}$ :

$\frac{dy}{dx}=-\frac{b\cos{t}}{a\sin{t}}=-\frac{b}{a}\cot{t}$

To find $\frac{d^2y}{dx^2}$ , firstly take the derivative of the function for $\frac{dy}{dx}$ :

$y'=\frac{d}{dt}(\frac{dy}{dx})=-\frac{b}{a}\csc^2{t}$

$\frac{d^2y}{dx^2}=\frac{dy'}{\frac{dx}{dt}}= -\frac{b}{a}\csc^2{t}/(-a\sin{t})=\frac{b}{a^2}\sin^3{t} ​$