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**stapel** Draw the post and the man. Draw the ground. Draw the slanty line from the top of the post, across the top of the man's head, and down to the ground. This gives you two nested (and thus similar) right triangles.

Label the post as "12", the man as "6", and the ground between the post and the man as "x", and note that dx/dt = 5.

Label the ground between the man and the tip of the triangle as "y". You need to find dy/dt.

The similar triangles give you the proportion 6/y = 12/(x + y). Use this to find the value of y when x = 5.

Differentiate the proportion with respect to time "t". Plug in the known values, and solve for the value of dy/dt.

Draw the upside-down cone. Draw a horizontal line indicating the water level. Draw the vertical "height" line. Note that, regarding one side or half of the cone's cross-sectional view, you now have two nested (and thus similar) right triangles.

Label the total height as "12" and the radius across the top as "6"; note that dV/dt = 10. Label the water height as "h" and the radius of the water's height as "r".

The similar triangles give you the proportion 12/6 = h/r. Use this to find the value of "r" when h = 3.

Solve the proportion for r in terms of h. Plug the result into the formula the formula for the volume of the cone. Differentiate with respect to time "t". Plug in all the known values, and solve for dh/dt.

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