Best solutions to - "A plane flies directly over a radar station..."

High School
Geometry
Trigonometry
Right triangles and trigonometry

Mabel Breault

Answered question

2021-12-30

A plane flies directly over a radar station while traveling at $500 \frac{m i}{h}$ passes directly over a radar station. Find the rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station.
Finish solving the problem

Answer & Explanation

Jenny Bolton

Beginner2021-12-31Added 32 answers

Note that y is the distance from the plane to the station the problem is referring to.

Differentiate with respect to time t the equation that relates everything.

x^{2} + 1^{2} = y^{2}

2 x \frac{d x}{d t} + 0 = 2 y \frac{d y}{d t}

x \frac{d x}{d t} = y \frac{d y}{d t}

\frac{x}{y} \frac{d x}{d t} = \frac{d y}{d t}

We are given that a

\frac{d x}{d t} = 500 \frac{m i}{h}

. We also need to find x at the moment

y = 2 m i

x^{2} + 1^{2} = 2^{2}

x^{2} = 4 - 1 = 3

x = \sqrt{3}

(ignore negative root)
Plug the values into the differentiated equation.

\frac{d y}{d t} = \frac{x}{y} \frac{d x}{d t}

\frac{\sqrt{3}}{2} \cdot 500

250 \sqrt{3} \frac{m i}{h}

levurdondishav4

Beginner2022-01-01Added 38 answers

P is the planes

Vasquez

Expert2022-01-07Added 669 answers

The appropriate diagram here is a right triangle with a short vertical leg of length 1 representing the distance between the station and a point 1 mi directly above it, a longer horizontal leg of variable length x representing the distance from the point above to the position of the plane at time t, and a hypotenuse connecting the plane’s
position with the station (call this distance y). We're told that $\frac{d x}{d t} = 500$ . By the Pythagorean Theorem $1 + x^{2} = y^{2}$ . Differentiating with respect to t and using our rules and formulas
$2 \frac{d y}{d t} = \frac{d y^{2}}{d y} \frac{d y}{d t} = \frac{d y^{2}}{d t} = \frac{d (1 + x^{2})}{d t}$
$= \frac{d 1}{d t} + \frac{d x^{2}}{d t} = 0 + \frac{d x^{2}}{d x} \frac{d x}{d t} = 2 x \frac{d x}{d t}$
Now setting y=2 in the equation $1 + x^{2} = y^{2}$ gives $x = \sqrt{3}$ . So at this point,
$4 \frac{d y}{d t} = 2 \sqrt{3} \frac{d x}{d t} = 2 \sqrt{3} 500$
so that $\frac{d y}{d t} = 250 \sqrt{3}$ (in miles per hour).

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