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Charlene's question at Yahoo! Answers regarding related rates
- Thread starter MarkFL
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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.
Since we are given:
(1) \(\displaystyle \left|\frac{dx}{dt} \right|=v\,\therefore\,\frac{dx}{dt}=\pm v\)
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:
\(\displaystyle \tan(\theta)=\frac{h}{x}\)
Now, implicitly differentiating with respect to time $t$, we obtain:
\(\displaystyle \sec^2(\theta)\frac{d\theta}{dt}=-\frac{h}{x^2}\cdot\frac{dx}{dt}\)
Multiplying through by \(\displaystyle \cos^2(\theta)\) and using (1) we have:
\(\displaystyle \frac{d\theta}{dt}=\pm\frac{hv\cos^2(\theta)}{x^2}\)
Now, from our diagram, we see that:
\(\displaystyle \cos(\theta)=\frac{x}{s}\)
and so we find:
\(\displaystyle \frac{d\theta}{dt}=\pm\frac{hv\left(\frac{x}{s} \right)^2}{x^2}=\pm\frac{hv}{s^2}\)
As we are only asked for the magnitude of the rate of change in $\theta$ with respect to time, we may write:
\(\displaystyle \left|\frac{d\theta}{dt} \right|=\frac{hv}{s^2}\)
Now, using the given data:
\(\displaystyle h=8,\,v=450,\,s=25\)
we find, in radians per hour:
\(\displaystyle \left|\frac{d\theta}{dt} \right|=\frac{8\cdot450}{25^2}=\frac{144}{25}\)
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.
Since we are given:
(1) \(\displaystyle \left|\frac{dx}{dt} \right|=v\,\therefore\,\frac{dx}{dt}=\pm v\)
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:
\(\displaystyle \tan(\theta)=\frac{h}{x}\)
Now, implicitly differentiating with respect to time $t$, we obtain:
\(\displaystyle \sec^2(\theta)\frac{d\theta}{dt}=-\frac{h}{x^2}\cdot\frac{dx}{dt}\)
Multiplying through by \(\displaystyle \cos^2(\theta)\) and using (1) we have:
\(\displaystyle \frac{d\theta}{dt}=\pm\frac{hv\cos^2(\theta)}{x^2}\)
Now, from our diagram, we see that:
\(\displaystyle \cos(\theta)=\frac{x}{s}\)
and so we find:
\(\displaystyle \frac{d\theta}{dt}=\pm\frac{hv\left(\frac{x}{s} \right)^2}{x^2}=\pm\frac{hv}{s^2}\)
As we are only asked for the magnitude of the rate of change in $\theta$ with respect to time, we may write:
\(\displaystyle \left|\frac{d\theta}{dt} \right|=\frac{hv}{s^2}\)
Now, using the given data:
\(\displaystyle h=8,\,v=450,\,s=25\)
we find, in radians per hour:
\(\displaystyle \left|\frac{d\theta}{dt} \right|=\frac{8\cdot450}{25^2}=\frac{144}{25}\)