1. ## [SOLVED] Rate Meaure

An aeroplane at alt of 1 Km flying horizontaly @ 800Km/hr passes over an observer. Find the @ it is apporoching observer when it is 1250Km away from him.

2. Originally Posted by as5tx
An aeroplane at alt of 1 Km flying horizontaly @ 800Km/hr passes over an observer. Find the @ it is apporoching observer when it is 1250Km away from him.
what are you using "@" to mean in the second sentence?

have you drawn a diagram? do you notice that we are dealing with a triangle here?

3. If I assume the question is asking what is the rate of change of the distance between the person on the ground and the airplane when the airplane is a distance of 1250Km away from the person, I would do it like this.

Hmm, I'm not very confident in my answer, but I got that the distance is growing smaller at a rate of 799.999744 km/hr when the plane is 1250 km away from him.

I got it by drawing a right triangle (see attached picture), with the hypotenuse being 1250, and the two end points being the person on the ground and the airplane. Then the height of the triangle was 1km, and the length was $\displaystyle \sqrt{1562499}$ (which is the square root of 1250^2+1)

I then labeled the hyp as A, the length as B, and the height as C.

We are looking for the rate of change in the hypotenuse because it is the distance between the person and the plane, so write a formula that solves for A (Pythagorean theorem)
$\displaystyle A=\sqrt{B^{2}+C^{2}}$

Now C is a constant, because it will be 1 always, but B and A will change with time. So we want to find out how fast A is changing with respect to time, or basically finding $\displaystyle \frac{dA}{dt}$

So differentiate the equation:

$\displaystyle A=\sqrt{B^{2}+C^{2}}$

$\displaystyle \frac{dA}{dt}=\frac{1}{2\sqrt{B^{2}+C^{2}}}*2B*\fr ac{dB}{dt}$

$\displaystyle \frac{dA}{dt}=\frac{B}{\sqrt{B^{2}+C^{2}}}*\frac{d B}{dt}$

$\displaystyle \frac{dA}{dt}=\frac{B}{A}*\frac{dB}{dt}$

Note: You don't get $\displaystyle \frac{dC}{dt}$ because C is the height, 1km, and does not change with time, it is a constant.

Okay, now we just need to plug in values. A is the hyp and equals 1250km, B is the length and is $\displaystyle \sqrt{1562499}km$, C is height, the constant 1km, and $\displaystyle \frac{dB}{dt}$ is the rate at which the length is changing, which is the given of 800km/hr.

$\displaystyle \frac{dA}{dt}=\frac{B}{A}*\frac{dB}{dt}$

$\displaystyle \frac{dA}{dt}=\frac{\sqrt{1562499}}{1250}*800$

using a calculator:
$\displaystyle \frac{dA}{dt}=799.999744$km/hr

So -799.999744km/hr is the instantaneous rate of change of the distance between the person on the ground and the airplane 1km in the air, when the airplane is 1250Km away from the person. (But we write it as positive, because the question is asking how fast the plane is approaching the observer, not how fast the distance is changing.)