MHB Charlene's question at Yahoo Answers regarding related rates

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The discussion focuses on calculating the rate of change of the angle theta as an airplane flies at a constant altitude of 8 miles and a speed of 450 mph. The relationship between the angle, altitude, and distance from the radar is established using the tangent function. By differentiating the tangent function with respect to time, the formula for the rate of change of theta is derived as |dθ/dt| = hv/s². Substituting the given values of altitude, speed, and distance results in a final calculation of |dθ/dt| = 144/25 radians per hour. This provides a clear solution to Charlene's question regarding the rate of change of theta.
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Here is the question:

Rate of change of theta?

An airplane is flying at a constant altitude of 8 miles over a radar at a rate of 450mph. At what rate is the angle theta changing when s=25?

I have posted a link there to this topic so the OP can see my work.
 
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Hello Charlene,

I would first draw a diagram to represent the problem. I will assume the quantity $s$ represents the distance from the radar station directly to the plane. $R$ represents the location of the radar station, and $P$ the location of the plane. I have let $h$ represent the constant altitude of the plane. All linear measures are in miles.

View attachment 970

Since we are given:

(1) $$\left|\frac{dx}{dt} \right|=v\,\therefore\,\frac{dx}{dt}=\pm v$$

where $v$ is the speed of the plane, and we know $h$, we want to relate $\theta$, $x$ and $h$. We may do so by using the definition of the tangent function as follows:

$$\tan(\theta)=\frac{h}{x}$$

Now, implicitly differentiating with respect to time $t$, we obtain:

$$\sec^2(\theta)\frac{d\theta}{dt}=-\frac{h}{x^2}\cdot\frac{dx}{dt}$$

Multiplying through by $$\cos^2(\theta)$$ and using (1) we have:

$$\frac{d\theta}{dt}=\pm\frac{hv\cos^2(\theta)}{x^2}$$

Now, from our diagram, we see that:

$$\cos(\theta)=\frac{x}{s}$$

and so we find:

$$\frac{d\theta}{dt}=\pm\frac{hv\left(\frac{x}{s} \right)^2}{x^2}=\pm\frac{hv}{s^2}$$

As we are only asked for the magnitude of the rate of change in $\theta$ with respect to time, we may write:

$$\left|\frac{d\theta}{dt} \right|=\frac{hv}{s^2}$$

Now, using the given data:

$$h=8,\,v=450,\,s=25$$

we find, in radians per hour:

$$\left|\frac{d\theta}{dt} \right|=\frac{8\cdot450}{25^2}=\frac{144}{25}$$
 

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