Finding the Highest Point of a Particle Moving Along the Y-Axis

[Squeezebox]

Homework Statement

A particle moves along the y-axis so that its velocity v at time t is v(t)=1-(1/tan(e^t))

Find the time t greater than or equal to zero at which the particle reaches its highest point.

The Attempt at a Solution

I set the equation equal to zero and got t=-.242 which doesn't work so I graphed it on a calculator to see what's going on. The graph ended up being a crazy graph with many, many zeros and I have no idea where to go from here.

 

[radou]

Which equation did you set equal to zero?

You have the velocity function, v(t), so you need to obtain the displacement function y(t),
differentiate, and set it equal to zero in order to find the moment of the maximal displacement.

 

[HallsofIvy]

Which equation did you set equal to zero?

You have the velocity function, v(t), so you need to obtain the displacement function y(t),
differentiate, and set it equal to zero in order to find the moment of the maximal displacement.

??The derivative of the displacement function is the velocity! There is no need to "obtain the displacement function".

Squeezebox, tan(x) goes to infinity as t goes to pi/2. e^t will go to pi/2 as t goes to ln(pi/2)= .452. As that happens 1- 1/tan(e^t) goes to 1. After that, for e^t> pi/2, tan(e^t)< 0 again and the object moves back to the left.

 

[radou]

??The derivative of the displacement function is the velocity! There is no need to "obtain the displacement function".

Thanks, my apologies to the OP.

 

[Squeezebox]

??The derivative of the displacement function is the velocity! There is no need to "obtain the displacement function".

Squeezebox, tan(x) goes to infinity as t goes to pi/2. e^t will go to pi/2 as t goes to ln(pi/2)= .452. As that happens 1- 1/tan(e^t) goes to 1. After that, for e^t> pi/2, tan(e^t)< 0 again and the object moves back to the left.

Now it makes some sense. Thanks