#### mags22

##### New member
A ladder 25 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2 feet per second.
(a) Consider the triangle formed by the side of the house, ladder, and the ground. Find the rate at which the area of the triangle is changing when the base of the ladder is 15 feet from the wall.
(b) Find the rate at which the angle between the ladder and the wall of the house is changing when the base of the ladder is 15 feet from the wall.

#### mmm4444bot

##### Super Moderator
Staff member

Hello. Have you seen any related-rates exercises before? A basic example of relating rates is shown in this thread. #### HallsofIvy

##### Elite Member
A ladder 25 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2 feet per second.
(a) Consider the triangle formed by the side of the house, ladder, and the ground. Find the rate at which the area of the triangle is changing when the base of the ladder is 15 feet from the wall.
(b) Find the rate at which the angle between the ladder and the wall of the house is changing when the base of the ladder is 15 feet from the wall.
Call the distance from the base of the ladder to the wall "b". Call the height of the ladder up the wall "h". Since the house is, of course, vertical while the lawn is horizontal, that triangle referred to is a right triangle and we can use the "Pythagorean Theorem": [imath]b^2+ h^2= 25^2[/imath]. Differentiate both sides with respect to t, "time", [imath]2b\frac{db}{dt}+2h\frac{dh}{dt}= 0[/imath].

You are told what b and and what db/dt are. Use the Pythagorean Theorem to determine h. Putting those into the equation, you can solve the equation for the remaining variable, h.

For (b) use trig functions to find the angle as a function of b and h. The "rate at which it is changing" is, of course, its derivative with respect to t.

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