Calc 3: Find the value of t in [0,2pi] where the speed is maximum of

Question

Calc 3: Find the value of t in [0,2pi] where the speed is maximum of <t-sint,1-cost>

kainui · Answer

So first off, you have to note two things. The derivative of position is velocity and the absolute value of velocity is speed. Try to get to an expression for speed and I'll help you figure out how to find the max speed on the interval.

anonymous · Answer

Ok, so R'(t)=<1+cos(t),1-sin(t)>.

anonymous · Answer

$$Speed = \sqrt{(1+\cos(t))^{2}}+(1-\sin(t))^{2}$$

anonymous · Answer

is this correct?

kainui · Answer

The derivative of sine is cosine and the derivative of cosine is negative sine. You've swapped the signs on both in your derivative.

Also, the derivative of the constant in your y-coordinate is 0.

 After that you have the idea right, and you can simplify it further with the identity sin^2(x)+cos^2(x)=1.

anonymous · Answer

Opps! You're right. Should be [(1+sin(t))^2+(1+cos(t))^2]^1/2

anonymous · Answer

What would be the next step?