Trig Derivative: Understanding the Simplified Solution

[yoleven]

Homework Statement

r(t)=(cos t +tsint)i + (sin t -tcost)j + 3k

Homework Equations

dr/dt=v(t)

The Attempt at a Solution

when I take the derivative of r(t) I get;
v(t)=(-sin t + tcost)i +(cos t +tsint)j

the book says;
v(t)=(t cos t)i +(tsint)j
could some one tell me why? Where did the -sint and the cos t go?
I guess it is an identity but I don't see it.

 

[SpringPhysics]

Homework Statement

r(t)=(cos t +tsint)i + (sin t -tcost)j + 3k

Homework Equations

dr/dt=v(t)

The Attempt at a Solution

when I take the derivative of r(t) I get;
v(t)=(-sin t + tcost)i +(cos t +tsint)j

the book says;
v(t)=(t cos t)i +(tsint)j
could some one tell me why? Where did the -sint and the cos t go?
I guess it is an identity but I don't see it.

Recall the chain rule to differentiate t sin t and - t cost t.

Example: to differentiate (t-1)t^2

= 1(t^2) + (t-1)(2)t
= t^2 + 2t(t-1)

 

[slider142]

Note that the rule in question is the product rule, not the chain rule.

 

[206PiruBlood]

Are you familiar with the product rule for derivatives? \frac{d}{dt}(uv)=uv^{'}+u^{'}v

beat me to it :(

 

[SpringPhysics]

My bad, *product rule.

 

[yoleven]

thank you.