Question

a plane flying horizontally at an altitude of 1 mile and a speed of 540 mi/h passes directly over a radar station. find the rate at which the distance from the plane to the station is increasing when it has a total distance of 5 miles away from the station. (round your answer to the nearest whole number.) mi/h 6. -/1 points if a snowball melts so that its surface area decreases at a rate of 5 cm²/min, find the rate (in cm/min) at which the diameter decreases when the diameter is 9 cm. (round your answer to three decimal places.) cm/min

Explanation:

Step1: Set up the relationship for the plane - radar problem

Let $x$ be the horizontal distance of the plane from the point directly above the radar station, and $y$ be the distance from the plane to the radar station. Given the altitude of the plane is $1$ mile, by the Pythagorean theorem, $y^{2}=x^{2}+1$.

Step2: Differentiate both sides with respect to time $t$

Using implicit - differentiation, $2y\frac{dy}{dt}=2x\frac{dx}{dt}$. Then $\frac{dy}{dt}=\frac{x}{y}\cdot\frac{dx}{dt}$.

Step3: Find the value of $x$ when $y = 5$

Since $y^{2}=x^{2}+1$ and $y = 5$, we have $25=x^{2}+1$, so $x=\sqrt{25 - 1}=\sqrt{24}=2\sqrt{6}$. And we know that $\frac{dx}{dt}=540$ mi/h.

Step4: Calculate $\frac{dy}{dt}$

Substitute $x = 2\sqrt{6}$, $y = 5$, and $\frac{dx}{dt}=540$ into $\frac{dy}{dt}=\frac{x}{y}\cdot\frac{dx}{dt}$. So $\frac{dy}{dt}=\frac{2\sqrt{6}}{5}\times540 = 432\sqrt{6}\approx1057$ mi/h.

Step5: Set up the relationship for the snow - ball problem

The surface area of a sphere $A = 4\pi r^{2}$, and since $d = 2r$ (where $d$ is the diameter), $A=\pi d^{2}$.

Step6: Differentiate both sides with respect to time $t$

Using implicit - differentiation, $\frac{dA}{dt}=2\pi d\frac{dd}{dt}$.

Step7: Solve for $\frac{dd}{dt}$

We know that $\frac{dA}{dt}=- 5$ cm²/min and $d = 9$ cm. Rearranging the equation $\frac{dA}{dt}=2\pi d\frac{dd}{dt}$ gives $\frac{dd}{dt}=\frac{\frac{dA}{dt}}{2\pi d}$.

Step8: Calculate $\frac{dd}{dt}$

Substitute $\frac{dA}{dt}=-5$ and $d = 9$ into the formula: $\frac{dd}{dt}=\frac{-5}{2\pi\times9}=-\frac{5}{18\pi}\approx - 0.088$ cm/min.

Answer:

For the plane - radar problem: $1057$ mi/h
For the snow - ball problem: $-0.088$ cm/min