How do you differentiate the following parametric equation: # x(t)=sqrt(t^2+tcos2t), y(t)=t^2sint #?

Question

How do you differentiate the following parametric equation: #   x(t)=sqrt(t^2+tcos2t),  y(t)=t^2sint   #?

Emilia Aguilar · Accepted Answer

#dy/dx=((4tsint+2t^2cost)sqrt(t^2 + t cos2t))/(2 t - 2 t sin2t + cos2 t)#

The derivative of a parametric equation is given by the following formula:

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

#y(t)=t^2sint#

#dy/(dt)=2tsint+t^2cost#

#x(t)=sqrt(t^2+tcos2t)#

#dx/(dt)=(2t - 2t sin2t + cos2t)/(2 sqrt(t^2 + t cos2t))#

#dy/dx=2tsint+t^2cost divide(2 t - 2 t sin2t + cos2 t)/(2 sqrt(t^2 + t cos2 t))#

#=((4tsint+2t^2cost)sqrt(t^2 + t cos2t))/(2 t - 2 t sin2t + cos2 t)#

Ezra Adams · Answer

undefined To differentiate the parametric equations $x(t) = \sqrt{t^2 + t \cos(2t)}$ and $y(t) = t^2 \sin(t)$ with respect to $t$, we use the chain rule for differentiation.

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

$$ \frac{dy}{dt} = \frac{d}{dt} \left( t^2 \sin(t) \right) = 2t \sin(t) + t^2 \cos(t) $$

Therefore, the differentiation of the given parametric equations yields:

$$ \frac{dx}{dt} = \frac{t + \cos(2t) - 2t \sin^2(t)}{\sqrt{t^2 + t \cos(2t)}} $$

$$ \frac{dy}{dt} = 2t \sin(t) + t^2 \cos(t) $$