Question

1. superman is in level flight 6 miles above the ground. his flight plan takes him directly over and past avon high school. how fast is he flying when the distance between him and ahs is exactly 10 miles and this distance is increasing at the rate of 40 mph?

Explanation:

Step1: Set up variables

Let \(x\) be the horizontal distance from the school, \(y = 6\) (constant height), and \(z\) be the distance between Superman and the school. By the Pythagorean theorem \(z^{2}=x^{2}+y^{2}\). Since \(y = 6\), we have \(z^{2}=x^{2}+36\).

Step2: Differentiate with respect to time \(t\)

Differentiating both sides of \(z^{2}=x^{2}+36\) with respect to \(t\) gives \(2z\frac{dz}{dt}=2x\frac{dx}{dt}\). We know \(\frac{dz}{dt} = 40\) mph.

Step3: Find \(x\) when \(z = 10\)

When \(z = 10\), from \(z^{2}=x^{2}+36\), we substitute \(z = 10\) and solve for \(x\): \(100=x^{2}+36\), so \(x^{2}=64\) and \(x = 8\) (we take the positive value since distance is non - negative).

Step4: Solve for \(\frac{dx}{dt}\)

Substitute \(z = 10\), \(x = 8\), and \(\frac{dz}{dt}=40\) into \(z\frac{dz}{dt}=x\frac{dx}{dt}\). We get \(10\times40=8\times\frac{dx}{dt}\). Then \(\frac{dx}{dt}=\frac{10\times40}{8}\).

Answer:

\(50\) mph