question an airplane is flying overhead at a constant elevation of 4000 ft. a man is viewing the plane from…

question an airplane is flying overhead at a constant elevation of 4000 ft. a man is viewing the plane from a position 3000 ft from the base of a radio tower. the airplane is flying horizontally away from the man. if the plane is flying at a rate of 500 ft/s, at what rate is the distance between the man and the plane increasing when the plane passes over the radio tower? (enter an exact answer.) provide your answer below: the distance between the man and the plane is increasing at a rate of feet per second.

question an airplane is flying overhead at a constant elevation of 4000 ft. a man is viewing the plane from a position 3000 ft from the base of a radio tower. the airplane is flying horizontally away from the man. if the plane is flying at a rate of 500 ft/s, at what rate is the distance between the man and the plane increasing when the plane passes over the radio tower? (enter an exact answer.) provide your answer below: the distance between the man and the plane is increasing at a rate of feet per second.

Answer

Explanation:

Step1: Establish the relationship

Let $x$ be the horizontal - distance of the plane from the man, and $z$ be the distance between the man and the plane. The elevation of the plane is $y = 4000$ ft (constant). By the Pythagorean theorem, $z^{2}=x^{2}+y^{2}=x^{2}+4000^{2}$.

Step2: Differentiate with respect to time $t$

Differentiating both sides of the equation $z^{2}=x^{2}+4000^{2}$ with respect to $t$, we get $2z\frac{dz}{dt}=2x\frac{dx}{dt}$, which simplifies to $z\frac{dz}{dt}=x\frac{dx}{dt}$.

Step3: Find the values of $x$, $z$ and $\frac{dx}{dt}$

When the plane passes over the radio - tower, $x = 3000$ ft. Since $y = 4000$ ft, using the Pythagorean theorem $z=\sqrt{x^{2}+y^{2}}=\sqrt{3000^{2}+4000^{2}}=\sqrt{9000000 + 16000000}=\sqrt{25000000}=5000$ ft. We are given that $\frac{dx}{dt}=500$ ft/s.

Step4: Solve for $\frac{dz}{dt}$

Substitute $x = 3000$, $z = 5000$ and $\frac{dx}{dt}=500$ into the equation $z\frac{dz}{dt}=x\frac{dx}{dt}$. We have $5000\frac{dz}{dt}=3000\times500$. Then $\frac{dz}{dt}=\frac{3000\times500}{5000}=300$ ft/s.

Answer:

300