John's Question: Air Pollution Rate of Change w/Distance

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SUMMARY

The air pollution rate of change with distance is modeled by the equation y + 2xy + x²y = 600. By isolating y, the function becomes y = 600(x + 1)⁻². Differentiating this function with respect to x yields dy/dx = -1200(x + 1)⁻³. Evaluating this derivative at x = 10 results in a pollution rate of approximately -0.9016, confirming option C as the correct answer.

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MarkFL
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Here is the question:

The air pollution y(in parts per million) x miles away is given by y+2xy+x^2y=600.?


The air pollution y(in parts per million) x miles away is given by y+2xy+x^2y=600. Find the rate of pollution 10 miles away.
A. 1.0020
B. -1.997
C. -0.9016
D. 3.0924
E. None of the above

I have no idea how to do this but it has something to do with derivative because that's what were doing

I have posted a link there to this thread so the OP may view my work.
 
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Hello John,

We are given:

$$y+2xy+x^2y=600$$

If we solve for $y$, we find:

$$y\left(x^2+2x+1 \right)=600$$

$$y(x+1)^2=600$$

$$y=600(x+1)^{-2}$$

Differentiating with respect to $x$, we obtain:

$$\frac{dy}{dx}=-1200(x+1)^{-3}$$

Hence:

$$\left.\frac{dy}{dx} \right|_{x=10}=-\frac{1200}{1331}\approx-0.9016$$

Thus, C.) is the correct answer.
 

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