How do you find #dy/dx# for the curve #x=t*sin(t)#, #y=t^2+2# ?

Question

14242 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2014-08-28T02:11:36

To find the derivative of a parametric function, you use the formula:

#dy/dx = (dy/dt)/(dx/dt)#, which is a rearranged form of the chain rule.

To use this, we must first derive #y# and #x# separately, then place the result of #dy/dt #over #dx/dt#.

#y=t^2 + 2#

#dy/dt = 2t# (Power Rule)

#x=tsin(t)#

#dx/dt = sin(t) + tcos(t)# (Product Rule)

Placing these into our formula for the derivative of parametric equations, we have:

#dy/dx = (dy/dt)/(dx/dt) = (2t)/(sin(t)+tcos(t))#