Question

question 9 of 14, step 1 of 1
a passenger airplane, flying at an altitude of 6.8 miles at a ground speed of 565 miles per hour, passes directly over an observer who is on the ground. how fast is the
distance between the observer and the plane increasing 3 minutes later? round any intermediate calculations, if needed, to no less than six decimal places, and round
your final answer to two decimal places.
answer
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Explanation:

Step1: Calculate the horizontal distance

The ground - speed of the plane is \(v = 565\) miles per hour. In \(t=\frac{3}{60}=\frac{1}{20}\) hours (since \(3\) minutes \(=\frac{3}{60}\) hours), the horizontal distance \(x\) that the plane travels is given by the formula \(x = vt\). So, \(x=565\times\frac{1}{20}=28.25\) miles.

Step2: Use the Pythagorean theorem

Let \(y = 6.8\) miles (altitude, constant) and \(z\) be the distance between the observer and the plane. By the Pythagorean theorem \(z=\sqrt{x^{2}+y^{2}}\). Substituting \(x = 28.25\) and \(y = 6.8\), we have \(z=\sqrt{(28.25)^{2}+(6.8)^{2}}=\sqrt{798.0625 + 46.24}=\sqrt{844.3025}\).

Step3: Differentiate the Pythagorean theorem with respect to time

We have \(z^{2}=x^{2}+y^{2}\). Differentiating both sides with respect to time \(t\): \(2z\frac{dz}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}\). Since \(y\) (altitude) is constant, \(\frac{dy}{dt} = 0\). So, \(\frac{dz}{dt}=\frac{x}{z}\cdot\frac{dx}{dt}\).
We know \(x = 28.25\), \(y = 6.8\), so \(z=\sqrt{28.25^{2}+6.8^{2}}\approx29.04\) (from step 2) and \(\frac{dx}{dt}=565\) (ground - speed of the plane).
Substituting the values: \(\frac{dz}{dt}=\frac{28.25}{29.04}\times565\)
\(\frac{dz}{dt}=\frac{28.25\times565}{29.04}=\frac{15961.25}{29.04}\approx549.63\)

Answer:

\(549.63\)