Parametric and derivative Calculus Question

[sarahjohnson]

(image too small on here so I uploaded it elsewhere)

http://postimg.org/image/lo19upwq3/

Hi, so the above image is what I have accomplished so far. But I am not sure how to find the:

- when tangent line is vertical (first and second time?)

- what is max/min horiz velocity

- equation for speed of object

- max and min for speed

For the tangent line to be vertical I think I should set the y' to 0? But I am not sure. And I think the method to figure out the max and min for speed should be the same for the max and min for horiz velocity right?

I don't need the exact answers but just an explanation how to do them would be appreciated. Thank You

 

[DrPhil]

View attachment 3033(image too small on here so I uploaded it elsewhere)

http://postimg.org/image/lo19upwq3/

Hi, so the above image is what I have accomplished so far. But I am not sure how to find the:

- when tangent line is vertical (first and second time?)

- what is max/min horiz velocity

- equation for speed of object

- max and min for speed

For the tangent line to be vertical I think I should set the y' to 0? But I am not sure. And I think the method to figure out the max and min for speed should be the same for the max and min for horiz velocity right?

I don't need the exact answers but just an explanation how to do them would be appreciated. Thank You

Vertical slope means the horizontal velocity (and thus the denominator of the slope) is zero:

\(\displaystyle \displaystyle \pi - \sin \left( 4\pi\ t + \frac{\pi}{2} \right)(4\pi) = 0 \)

The first zero will involve an angle in the 1st quadrant, and the 2nd will be in the second quadrant.

You need to differentiate the horizontal velocity component and set to zero to find min and max.

The speed is the magnitude of the velocity, the square root of sum of squares of the components, or

\(\displaystyle \displaystyle |v|^2 = \left[ \pi - \sin \left( 4\pi\ t + \frac{\pi}{2} \right)(4\pi) \right]^2 +
\left[ \cos \left( 4\pi\ t + \frac{\pi}{2} \right)(4\pi) \right]^2 \)

The only simplification I see in that is sin^2(x) + cos^2(x) = 1.

Differentiate \(\displaystyle |v|^2\) to find min and max

 

[sarahjohnson]

Vertical slope means the horizontal velocity (and thus the denominator of the slope) is zero:

\(\displaystyle \displaystyle \pi - \sin \left( 4\pi\ t + \frac{\pi}{2} \right)(4\pi) = 0 \)

The first zero will involve an angle in the 1st quadrant, and the 2nd will be in the second quadrant.

You need to differentiate the horizontal velocity component and set to zero to find min and max.

The speed is the magnitude of the velocity, the square root of sum of squares of the components, or

\(\displaystyle \displaystyle |v|^2 = \left[ \pi - \sin \left( 4\pi\ t + \frac{\pi}{2} \right)(4\pi) \right]^2 +
\left[ \cos \left( 4\pi\ t + \frac{\pi}{2} \right)(4\pi) \right]^2 \)

The only simplification I see in that is sin^2(x) + cos^2(x) = 1.

Differentiate \(\displaystyle |v|^2\) to find min and max

I'm sorry but I'm having some trouble isolating t in the trig function. So I subtract pi from both sides and then divided by the 4pi. Then arcsin of both sides and solved for t. and it's wrong?

 

[DrPhil]

I'm sorry but I'm having some trouble isolating t in the trig function. So I subtract pi from both sides and then divided by the 4pi. Then arcsin of both sides and solved for t. and it's wrong?

what you say should be right. Did you get the sign right?

\(\displaystyle \displaystyle \pi - \sin \left( 4\pi\ t + \frac{\pi}{2} \right)(4\pi) = 0\)

\(\displaystyle \displaystyle \sin \left( 4\pi\ t + \frac{\pi}{2} \right) = \frac{1}{4} \)

\(\displaystyle \displaystyle 4\pi\ t + \frac{\pi}{2} = \sin^{-1} \left( \frac{1}{4} \right) \)

\(\displaystyle \displaystyle t = \dfrac{\sin^{-1} \left( \frac{1}{4} \right) - \frac{\pi}{2}}{4\pi} \)

There are two angles with sine 1/4, one in the first quadrant and one in the 2nd:
..........\(\displaystyle \sin^{-1}\left(\frac{1}{4}\right)\) and.... \(\displaystyle \pi - \sin^{-1}\left(\frac{1}{4}\right) \)
One of those two gives a negative value of t, so since t begins at zero, you have to add one cycle, or \(\displaystyle \Delta t = 1/2\). Then which is the first and which is the second to occur?