an airplane is flying at an elevation of 3200 ft. a man is 2200 ft from the base of a radio tower. the…

an airplane is flying at an elevation of 3200 ft. a man is 2200 ft from the base of a radio tower. the airplane flew over the man heading in the direction of the radio tower. if the plane is flying at a rate of 400 ft/sec, how fast is the distance between the man and the plane increasing when the plane passes directly over the radio tower? round your answer to three decimal places.
Answer
Explanation:
Step1: Establish the distance - relationship
Let $x$ be the horizontal distance of the plane from the man, $y = 3200$ ft be the elevation of the plane (constant), and $z$ be the distance between the man and the plane. By the Pythagorean theorem, $z^{2}=x^{2}+y^{2}=x^{2}+3200^{2}$.
Step2: Differentiate with respect to time $t$
Differentiating both sides of the equation $z^{2}=x^{2}+3200^{2}$ with respect to $t$, we get $2z\frac{dz}{dt}=2x\frac{dx}{dt}$. Then $\frac{dz}{dt}=\frac{x}{z}\cdot\frac{dx}{dt}$.
Step3: Find the values of $x$, $z$ when the plane is over the tower
When the plane passes directly over the radio - tower, $x = 2200$ ft. And $z=\sqrt{x^{2}+y^{2}}=\sqrt{2200^{2}+3200^{2}}=\sqrt{4840000 + 10240000}=\sqrt{15080000}=20\sqrt{37700}\approx3908.964$ ft. We know that $\frac{dx}{dt}=- 400$ ft/sec (negative because $x$ is decreasing as the plane moves towards the tower).
Step4: Calculate $\frac{dz}{dt}$
Substitute $x = 2200$, $z\approx3908.964$, and $\frac{dx}{dt}=-400$ into $\frac{dz}{dt}=\frac{x}{z}\cdot\frac{dx}{dt}$. So $\frac{dz}{dt}=\frac{2200}{3908.964}\times(-400)\approx - 225.129$ ft/sec. The negative sign just indicates the direction of change. The speed (magnitude) is approximately $225.129$ ft/sec.
Answer:
$225.129$