Here is the problem: A person 2 meters tall walks on level ground toward a flagpole of height 26 meters at the rate of 1.8 meters per second. Find the rate of change of the angle of elevation of the top of the flagpole from the top of the person`s head when the person is 16 meters from the pole. (The correct answer is .05 radians/second. I just don`t know how to get it.)

- stapel_eliz
**Posts:**1687**Joined:**Mon Dec 08, 2008 4:22 pm-
**Contact:**

stellar wrote:A person 2 meters tall walks on level ground toward a flagpole of height 26 meters at the rate of 1.8 meters per second. Find the rate of change of the angle of elevation of the top of the flagpole from the top of the person`s head when the person is 16 meters from the pole.

Draw a vertical line for the pole and another for the person. Draw a horizontal line from the top of the person's line over to the pole's line.

Label the part of the pole above this horizontal line as "24", and draw the slanty line from the top of the pole to the top of the person's head, thus forming a right triangle. Label the angle on the guy's head as

You are given that the horizontal distance x is changing at a rate of dx/dt = -1.8. The height y of the pole is obviously not changing, so dy/dt = 0.

The angle's change is what you need to find. You have information on the "opposite" and "adjacent" sides, so relate this (using the tangent) to the angle. Differentiate, and see what you can find!

If you get stuck, please reply showing how far you have gotten. Thank you!

I did this:

tan(theta)=h/x

sec^2(theta)(d-theta/dt)=((dh/dt)x-h(dx/dt))/x^2

I know x=16,h=24, so sec(theta)=sqrt(52)/4, sec^2(theta)=52/16=13/4:

(13/4)(d-theta/dt)=(0-24(-1.8))/(16^2)

d-theta/dt=(43.2/16^2)(4/13)=.0519230769

that matches the answer, but did I do it right? thanks.

tan(theta)=h/x

sec^2(theta)(d-theta/dt)=((dh/dt)x-h(dx/dt))/x^2

I know x=16,h=24, so sec(theta)=sqrt(52)/4, sec^2(theta)=52/16=13/4:

(13/4)(d-theta/dt)=(0-24(-1.8))/(16^2)

d-theta/dt=(43.2/16^2)(4/13)=.0519230769

that matches the answer, but did I do it right? thanks.

- stapel_eliz
**Posts:**1687**Joined:**Mon Dec 08, 2008 4:22 pm-
**Contact:**

Thanks for confirming!