A plane flying horizontally at an altitude of 2 mi and a speed…

Here is a calculus question I'm apparently struggling with.

A plane flying horizontally at an altitude of $2~\mathrm{mile}$ and a speed of $520~\mathrm{mile\ h^{-1}}$ passes directly over a radar station. Find the rate at which the distance from the plane to the station is increasing when it is $5~\mathrm{mile}$ away from the station. (Round your answer to the nearest whole number.)

I used the equation $z^2$=$x^2$=$y^2$ and found $z=\sqrt{ 29}$, and $\frac{dz}{dt}$=$\frac{2600}{\sqrt{29}}$ which is $483~\mathrm{mile\ h^{-1}}$, but web assign claims its wrong. Am I mistaken somewhere? Thanks!

1 Answer

We probably agree on $2z\frac{dz}{dt}=2x\frac{dx}{dt}$.

When the distance from the station, which is $z$, is $5$, we have $x=\sqrt{21}$. Now calculate.

• Why is it √21 and not √29? – soxrok2212 Nov 3 '15 at 2:53
• The straight line distance is $5$. This is the hypotenuse. Thus $x^2+y^2=25$, But $y=2$. So $x^2=21$. – André Nicolas Nov 3 '15 at 2:55
• So then 520x5=2600, and 2600/√21 = 567, but I'm still being told I'm incorrect. – soxrok2212 Nov 3 '15 at 2:58
• You are solving incorrectly. The answer is $\frac{\sqrt{21}}{5}(520)$. – André Nicolas Nov 3 '15 at 3:02
• Note that you are after $\frac{dz}{dt}$. This is $\frac{x}{z}\cdot\frac{dx}{dt}$. – André Nicolas Nov 3 '15 at 3:03