1. ## Possible Related Rates Problem w/ Ladder

Hello everyone,

I did the following SAT problem correctly using Pythagoras' Theorem and algebra, but I would like to try to do it with calculus. Is it possible to do this problem as a related rates problem?

Thank you.

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1. A 25-foot ladder is placed against a vertical wall of a building, with the bottom of the ladder standing on concrete 7 feet from the base of the building. If the top of the ladder slips down 4 feet, then the bottom of the ladder will slide out ___ feet.

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Let: $y$ = height of the ladder on wall

$x$ = distance of the base of the ladder to the wall

Therefore:

$\frac{dy}{dt} = 4$ $\frac{dx}{dt} = ?$

$x^2 + y^2 = 25$

$\frac{d}{dx}(x^2 + y^2 = 25)$

$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$

$2x\frac{dx}{dt} + 8y = 0$

$\frac{dx}{dt} = \frac{-4y}{x}$

I am not sure what to substitute as x and y here.

2. Originally Posted by scherz0
Hello everyone,

I did the following SAT problem correctly using Pythagoras' Theorem and algebra, but I would like to try to do it with calculus. Is it possible to do this problem as a related rates problem?

Thank you.

---

1. A 25-foot ladder is placed against a vertical wall of a building, with the bottom of the ladder standing on concrete 7 feet from the base of the building. If the top of the ladder slips down 4 feet, then the bottom of the ladder will slide out ___ feet.

---

Let: $y$ = height of the ladder on wall

$x$ = distance of the base of the ladder to the wall

Therefore:

$\frac{dy}{dt} = 4$ $\frac{dx}{dt} = ?$

$x^2 + y^2 = 25$

$\frac{d}{dx}(x^2 + y^2 = 25)$

$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$

$2x\frac{dx}{dt} + 8y = 0$

$\frac{dx}{dt} = \frac{-4y}{x}$

I am not sure what to substitute as x and y here.
you don't have to use it..

note that the length of the ladder is 25, and the foot of the ladder is 7 units away from the base of the building..

thus, from the base of the building to the top of the ladder is given by
$y = \sqrt{25^2 - 7^2}$

so, if the ladder slips down 4 units, the distance from the base of the building from the bottom of the ladder is given by
$x =\sqrt{25^2 - (y-4)^2}$

thus, the bottom of the ladder moves away x - 7 units from the base of the building..

3. Originally Posted by scherz0
Hello everyone,

I did the following SAT problem correctly using Pythagoras' Theorem and algebra, but I would like to try to do it with calculus. Is it possible to do this problem as a related rates problem?
I'm pretty sure not. There are no rates here to relate.

4. All right, thank you!