An airplane flies over an airport at an altitude of 10000 meters and at a speed of 900 km/hr. Find the rate at which the actual distance from the airport is increasing 2 minutes after the airplane was directly over the airport.

I'm getting a weird answer. Someone please help. The correct answer is 854 km/hr.


Let $x$ be the horizontal distance from the airport to the plane. Let $y$ be the vertical distance, and let $r$ be the diagonal (that is, $r = \sqrt{x^2 + y^2})$.

Thus, we have: $$\begin{align} \frac{dr}{dt} &= \frac{d}{dt}(x^2 + y^2)^{1/2} \\ &= \frac{1}{2}(x^2 + y^2)^{-1/2}\cdot\frac{d}{dt}(x^2 + y^2)\\ &= \frac{1}{2}(x^2 + y^2)^{-1/2}\cdot(2x\frac{dx}{dt} + 2y\frac{dy}{dt}) \end{align}$$

Based on the speed of the plane, we know that: $$\frac{dx}{dt} = 900\frac{\text{km}}{\text{hr}}$$

Because the height of the plane, $y$ is constant, we know that $\frac{dy}{dt} = 0$.

At $2$ minutes after the flyover, we know that the horizontal distance $x$ is described by: $$x = \underbrace{900\frac{\text{km}}{\text{hr}}}_{\text{rate}}\cdot \underbrace{\frac{1}{30}\text{hr}}_{\text{time}} = 30\text{km}$$

Also, $y = 10\text{km}$.

Thus: $$\begin{align} \frac{dr}{dt} &= \frac{1}{2}(\color{red}{x}^2 + \color{blue}{y}^2)^{-1/2}\cdot(2\color{red}{x}\color{green}{\frac{dx}{dt}} + 2\color{blue}{y}\color{purple}{\frac{dy}{dt}}) \\ &= \frac{1}{2}(\color{red}{30}^2 + \color{blue}{10}^2)^{-1/2}\cdot(2\color{red}{(30)}\color{green}{(900)} + 2\color{blue}{(10)}\color{purple}{(0)}) \\ &= \frac{1}{2}(1000)^{-1/2}\cdot(54000) \\ &\approx 854\frac{\text{km}}{\text{hr}}\end{align}$$

  • 1
    $\begingroup$ The biggest thing to watch out for in this problem are the unit conversions... make sure you're always working with kilometers and hours--never meters or minutes. $\endgroup$
    – apnorton
    Feb 18 '14 at 16:33

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