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Product Rule of x = r cos()

  • #1

Homework Statement



r = r(t)
[tex]\theta[/tex] = [tex]\theta[/tex](t)

x = r cos([tex]\theta[/tex])

dx/dt =dr/dt cos([tex]\theta[/tex]) - r sin([tex]\theta[/tex]) d[tex]\theta[/tex]/dt





The Attempt at a Solution



Where does the d[tex]\theta[/tex]/dt come from at the end of the derivative? I know I'm using product rule here because r and theta are both functions of t. But, the derivative of cos is just -sin. Why would there be a d[tex]\theta[/tex]/dt at the end?
 

Answers and Replies

  • #2
gabbagabbahey
Homework Helper
Gold Member
5,002
6
. But, the derivative of cos is just -sin. Why would there be a d[tex]\theta[/tex]/dt at the end?
No, [itex]\frac{d}{d\theta}\cos\theta=-\sin\theta[/itex] but [itex]\frac{d}{dt}\cos\theta=\left(\frac{d}{d\theta}\cos\theta\right)\left(\frac{d\theta}{dt}\right)[/itex] via the chain rule. :wink:
 
  • #3
33,152
4,838

Homework Statement



r = r(t)
[tex]\theta[/tex] = [tex]\theta[/tex](t)

x = r cos([tex]\theta[/tex])

dx/dt =dr/dt cos([tex]\theta[/tex]) - r sin([tex]\theta[/tex]) d[tex]\theta[/tex]/dt





The Attempt at a Solution



Where does the d[tex]\theta[/tex]/dt come from at the end of the derivative? I know I'm using product rule here because r and theta are both functions of t. But, the derivative of cos is just -sin. Why would there be a d[tex]\theta[/tex]/dt at the end?
Chain rule.
d/dt(cos(theta)) = -sin(theta)*d(theta)/dt
 
  • #4
oooooooooooooohhhhhhhhhhhh! man I'm kinda disappointed I didn't see that one : ( oh well. Thank very much guys!
 

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