How Fast is the Triangle's Area Changing When the Board is 9 Feet from the Wall?

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A 15-foot board leaning against a wall creates a right triangle with the wall and ground, where the area is defined as A = 1/2 * x * sqrt(225 - x^2. The bottom of the board slides away from the wall at 3 feet per second, and the problem seeks to determine how fast the area of the triangle is changing when the bottom is 9 feet from the wall. To find the rate of change of the area, the relationship A = 1/2 * x * sqrt(225 - x^2) is differentiated with respect to time, incorporating the rate of change of x. Evaluating this derivative at the specified distance provides the solution to the problem.
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What is the y prime of:
A 15 foot board rests against a vertical wall. If the bottom of the board slides away from the wall at the rate of 3 feet per second, how fast is the area of the triangle formed by the board, the wall and the ground changing at the instant the bottom of the board is 9 feet from the wall?

I got this far:
y= square root of (225 - x^2)

I'm stuck after this step.

Can the answer for this be [(-9 yprime /2) + 18] square feet per second.
 
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Hint x in terms of t. The displacement x away from the wall is equal to what in terms of t? It will be easy to then sub in this for x.

Area of a triangle is 1/2 * base * height
Base is x height is sqrt(225-x^2) right?

so A = 1/2 * x * sqrt(225-x^2)

However you need to plug in t to this I just didn't want to give away that much work.

Differentiate with respect to t in the new equation you find using the product rule (gets a little messy). Since 3t = 9, t = 3 goes into this new derivative and bam you got your answer.
 
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I'm not sure what you mean by the question "what is the y prime of...?". If anything, the question seems to be asking for A'(t), where A is the area of the right triangle formed by the ladder, the wall, and the ground.

Draw a diagram, label the horizontal distance between the bottom of the ladder and wall x, and the corresponding vertical distance y

Given:

\frac{dx}{dt} = 3 ft/s

This is a related rates problem. A is a function of x, so given the rate at which x changes with time, you should be able to find the rate at which A changes with time using the functional relationship:

A = \frac{1}{2}xy = \frac{1}{2}x\sqrt{225 - x^2}

You identified this relationship correctly! :smile:

Now, please relate the rates:

\frac{dA}{dt} = \frac{dA}{dx}\frac{dx}{dt}

= \frac{1}{2}\frac{d}{dx}(x\sqrt{225 - x^2})(3 \text{ft/s})

Once you solve that, evaluate it at the value specified. I think that's right. It's kinda late here, so if any of this is bogus, please let me know.

EDIT: right, vsage, I forgot that you need x in terms of t.
 
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