Application of derivative: 4m long ladder leans against the wall and being pulled horizontally....

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KamCh

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Hey guys, had a test today with a derivative application question. Can you please tell me if I got it right.

4m long ladder leans against the wall and being pulled horizontally away from the wall at the rate of 0.5 m/s. How fast area of right triangle changes when top of the ladder is 1m above the ground? Sorry for bad image quality.
WIN_20190404_00_24_11_Pro (2).jpg
 
I can't make heads nor tails out of the image you uploaded. I would let \(x\) be the horizontal distance of the bottom of the ladder from the wall and \(y\) be the vertical distance along the wall the top of the ladder is above the ground. Both are measured in meters. And so the area of the triangle formed by the ladder, wall and ground is:

[MATH]A=\frac{1}{2}xy[/MATH]
Differentiating with respect to time \(t\) (in seconds), we obtain:

[MATH]\d{A}{t}=\frac{1}{2}\left(\d{x}{t}y+x\d{y}{t}\right)[/MATH]
Now, we know by the theorem of Pythagoras:

[MATH]x^2+y^2=4^2\implies x^2=4^2-y^2[/MATH]
Differentiating with respect to time:

[MATH]2x\d{x}{t}+2y\d{y}{t}=0\implies \d{y}{t}=-\frac{x}{y}\d{x}{t}[/MATH]
And so we have:

[MATH]\d{A}{t}=\frac{1}{2}\d{x}{t}\left(y-\frac{4^2-y^2}{y}\right)=\frac{1}{y}\d{x}{t}\left(y^2-8\right)[/MATH]
Plugging in the given data (\(y=1,\,\d{x}{t}=\frac{1}{2}\)), we find:

[MATH]\d{A}{t}=\frac{1}{1}\cdot\frac{1}{2}\left(1^2-8\right)=-\frac{7}{2}\,\frac{\text{m}^2}{\text{s}}[/MATH]
Is that the result you obtained?
 
Last edited:
MarkFL said:
I can't make heads nor tails out of the image you uploaded. I would let \(x\) be the horizontal distance of the bottom of the ladder from the wall and \(y\) be the vertical distance along the wall the top of the ladder is above the ground. Both are measured in meters. And so the area of the triangle formed by the ladder, wall and ground is:

[MATH]A=\frac{1}{2}xy[/MATH]
Differentiating with respect to time \(t\) (in seconds), we obtain:

[MATH]\d{A}{t}=\frac{1}{2}\left(\d{x}{t}y+x\d{y}{t}\right)[/MATH]
Now, we know by the theorem of Pythagoras:

[MATH]x^2+y^2=4^2\implies x^2=4^2-y^2[/MATH]
Differentiating with respect to time:

[MATH]2x\d{x}{t}+2y\d{y}{t}=0\implies \d{y}{t}=-\frac{x}{y}\d{x}{t}[/MATH]
And so we have:

[MATH]\d{A}{t}=\frac{1}{2}\d{x}{t}\left(y-\frac{4^2-y^2}{y}\right)=\frac{1}{y^2}\d{x}{t}\left(y^2-8\right)[/MATH]
Plugging in the given data (\(y=1,\,\d{x}{t}=\frac{1}{2}\)), we find:

[MATH]\d{A}{t}=\frac{1}{1^2}\cdot\frac{1}{2}\left(1^2-8\right)=-\frac{7}{2}\,\frac{\text{m}^2}{\text{s}}[/MATH]
Is that the result you obtained?
Click to expand...
Thank you so much!
 
KamCh said:
Thank you so much!
Click to expand...

I made a minor error, but it didn't affect the numerical result. I should have obtained:

[MATH]\d{A}{t}=\frac{1}{y}\d{x}{t}\left(y^2-8\right)[/MATH]
I previously had \(y^2\) in the denominator of the first factor on the RHS. Sorry for the confusion. :)
 
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