Question

An airplane flying at an altitude of 6 miles passes directly over a radar antenna. When the airplane is 10 miles away (s = 10), the radar detects that the distance s is changing at a rate of 290 miles per hour. What is the speed of the airplane?

Answer 1

Answer:

Explanation:

Given

altitude of the Plane $h=6\ miles$

When Airplane is $s=10\ miles$ away

Distance is changing at the rate of $\frac{\mathrm{d} s}{\mathrm{d} t}=290\ mph$

From diagram we can write as

$h^2+x^2=s^2$

differentiate above equation w.r.t time

$2h\frac{\mathrm{d} h}{\mathrm{d} t}+2x\frac{\mathrm{d} x}{\mathrm{d} t}=2s\frac{\mathrm{d} s}{\mathrm{d} t}$

as altitude is not changing therefore $\frac{\mathrm{d} h}{\mathrm{d} t}=0$

$0+x\frac{\mathrm{d} x}{\mathrm{d} t}=s\frac{\mathrm{d} s}{\mathrm{d} t}$

at $s=10\ miles\ and\ h=6\ miles$

substitute the value we get $x=\sqrt{10^2-6^2}=8\ miles$

$8\times \frac{\mathrm{d} x}{\mathrm{d} t}=10\times 290$

$\frac{\mathrm{d} x}{\mathrm{d} t}=362.5\ mph$

Answer 2

The speed of the airplane is 335 mile per hour.

From the diagram attached:

Using Pythagoras theorem:

S² = x² + 25   (1)

Differentiating both sides of the equation with respect to t:

$2S\frac{dS}{dt}=2x\frac{dx}{dt}\\\\\frac{dx}{dt}=\frac{S}{x} \frac{dS}{dt}\\\\\\But\ \frac{dS}{dt} =290\ mph, hence:\\\\\frac{dx}{dt}=290\frac{S}{x} \\\\\\When\ S=10:\\\\S^2=x^2+25\\\\10^2=x^2+25\\\\x^2=75\\\\x=5\sqrt{3} \\\\\\\\\frac{dx}{dt}=290\frac{S}{x} \\\\\frac{dx}{dt}=290\frac{10}{5\sqrt{3} } \\\\\\\frac{dx}{dt}=335\ mph$

The speed of the airplane is 335 mile per hour.

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