Rate of change: find rate length is changing at moment

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SUMMARY

The discussion focuses on calculating the rate at which the side length of an equilateral triangle is increasing when placed atop a rectangle. Given that the rectangle's base length is 20 mm and its height is seven times the base, the combined area of the triangle and rectangle is increasing at a rate of 8 mm²/sec. The derived formula for the area leads to the conclusion that the side length of the triangle is increasing at a rate of 0.188 mm/sec when the base of the rectangle is 20 mm.

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Homework Statement



An equilateral triangle with side length x sits atop a rectangle with a base length of x, with the base of the triangle coinciding with the top base of the rectangle. the height of the rectangle is 7 times the length of its base. If the combined area of the figures is increasing at a rat of 8 mm^2/sec, find the rate at which a side of the triangle is lengthening at the moment the base of the rectangle is 20 mm.

The Attempt at a Solution



so i started by making a sketch. (I attached a sketch of the figures). then, i found the Area and then dA/dt:

A=lw+1/2 bh
=(7x)(x)+ 1/2 (x)((x*sqrt3)/2)
=7x^2 + (x^2*sqrt3)/4

dA/dt= 14x dx/dt + (x*sqrt3)/2 dx/dt

then, plug in x=20 mm, and dA/dt=8 mm^2/min:

8= 40 dx/dt + 2.474 dx/dt
8= dx/dt (40+2.474)
8= dx/dt (42.474)
dx/dt = 0.188 mm^2/sec
 

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That's great - did you have a question?

Note - for isomorphic change, $$A\propto x^2 \Rightarrow \frac{dA}{dt} \propto 2x\frac{dx}{dt}$$ ... which could save you a lot of trouble.
 
Last edited:

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