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

Question

1774 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-08T04:03:47

#x'(t) = 2t - 4sin(2t)#

#y'(t) = sint + tcost#

We're asked to find the derivative of a set of parametric equations.

We differentiate each individual equation as normal:

x(t)

We can differentiate term by term, and factor out the constant,#2# in the second term:

#x'(t) = d/(dt) [x^2] + 2d/(dt) [cos(2t)]#

Use the power rule:

#d/(dt) [t^n] = nt^(n-1)#

where #n = 2#:

#= color(red)(2t) + 2d/(dt) [cos(2t)]#

Using the chain rule:

#d/(dt) [cos(2t)] = (dcosu)/(du)(du)/(dt)#

where #u = 2t# and #d/(du) [cosu] = -sinu#:

#= 2t + 2d/(dt)[2t]*-sin(2t)#

Simplifying:

#= 2t - 2d/(dt)[2t]sin(2t)#

Factor out the constant, #2#:

#= 2t - 4d/(dt)[t]sin(2t)#

The derivative of #x# is #1#

#color(red)(= 2t - 4sin(2t)#

y(t)

Use the product rule:

#d/(dt) [uv] = v(du)/(dt) + u(dv)/(dt)#

where #u = t# and #v = sint#:

#y'(t) = td/(dt) [sint] + d/(dt) [t] sint#

The derivative of #sint# is #cost#:

#= d/(dt)[t] sint +t cost#

The derivative of #t# is #1#:

#color(blue)(= sint + tcost#