How do you differentiate the following parametric equation: # x(t)=tsqrt(t^2-1), y(t)= sqrt(t^2-e^(t) #?

Question

How do you differentiate the following parametric equation: #   x(t)=tsqrt(t^2-1),  y(t)= sqrt(t^2-e^(t)  #?

Charles Arocha · Accepted Answer

#(dy)/(dx)=((2t-e^t)sqrt(t^2-1))/((t^2-1+t)sqrt(t^2-e^t))#

When equation are given in parametric form such as

#x(t)=tsqrt(t^2-1)# and #y(t)=sqrt(t^2-e^t)#

Then #(dy)/(dx)=((dy)/(dt))/((dx)/(dt))#

#(dy)/(dt)=1/2xx1/sqrt(t^2-e^t)xx(2t-e^t)#

#(dx)/(dt)=1xxsqrt(t^2-1)+txx1/2xx1/sqrt(t^2-1)#

= #sqrt(t^2-1)+t/2xx1/sqrt(t^2-1)#

Hence #(dy)/(dx)=(1/(2sqrt(t^2-e^t))xx(2t-e^t))/(sqrt(t^2-1)+t/2xx1/sqrt(t^2-1))#

= #((2t-e^t)sqrt(t^2-1))/((t^2-1+t)sqrt(t^2-e^t))#

Evelyn Agard · Answer

undefined To differentiate the given parametric equations $ x(t) = t\sqrt{t^2 - 1} $ and $ y(t) = \sqrt{t^2 - e^t} $, we can use the chain rule for differentiation.

For $ x(t) $:
$$ \frac{dx}{dt} = \frac{d}{dt} (t\sqrt{t^2 - 1}) $$
$$ = \sqrt{t^2 - 1} + t\left(\frac{1}{2}\right)\frac{1}{\sqrt{t^2 - 1}}(2t) $$
$$ = \sqrt{t^2 - 1} + \frac{t^2}{\sqrt{t^2 - 1}} $$

For $ y(t) $:
$$ \frac{dy}{dt} = \frac{d}{dt} \left(\sqrt{t^2 - e^t}\right) $$
$$ = \frac{1}{2\sqrt{t^2 - e^t}}(2t - e^t) $$
$$ = \frac{t - e^t}{2\sqrt{t^2 - e^t}} $$

So, the derivatives of the given parametric equations are:

$$ \frac{dx}{dt} = \sqrt{t^2 - 1} + \frac{t^2}{\sqrt{t^2 - 1}} $$
$$ \frac{dy}{dt} = \frac{t - e^t}{2\sqrt{t^2 - e^t}} $$