Related Rates

A

Asutrick

New member
Joined
Feb 25, 2020
Messages
1
Question: A plane flying horizontally at an altitude of 1 mi and a speed of 530 mph passes ditectly over a radar system. Find the rate at which the distance from the plane to the station is increasing when it is 2 mi away from the station.

I worked this 2 ways and got wrong amswers both time.

First solution: 212 mph
(H^2)+(x^2)=y^2
Sqrt((h^2)+(x^2))=y
((H^2)+(x^2))^-1/2=y using chain rule I got:
(1/(h^2)+(x^2))(dx/t)(x)=y'
(1/(1^2)+(2^2))(530)(2)=y'
212=y'

Second solution:474
(H^2)+(x^2)=y^2
2h(dh/t)+2x(dx/dt)=2y(dy/t)
2(1)(0)+2(2)(530)=(2sqrt(5))(dy/t)
dy/t=474

What did I do wrong? And what is the correct steps for getting the answer right?
 
Couple of things. Just like algebra, start by DEFINING things. And remember that calculus is about functions.

[MATH]s = \text {distance from station to airplane in miles.}[/MATH]
[MATH]t = \text {time in hours.}[/MATH]
You want to find [MATH]\dfrac{ds}{dt}.[/MATH] To do that, you need to find the function of t that determines s.

You correctly saw that the distance itself depends on the Pythagorean Theorem, where

[MATH]x = \text {horizontal distance of airplane from station to airplane in miles, and}[/MATH]
[MATH]y = \text {vertical distance of airplane above ground in miles.}[/MATH]
You saw, but had no notation to record, that

[MATH]s = \sqrt{x^2 + y^2}.[/MATH]
As the most brilliant cat in the universe pointed out, y = 1. And y squared is 1. So

[MATH]s = \sqrt{x^2 + 1}.[/MATH]
But to get s as a function of t, you need x as a function of t.

What next?

Moral: if you want to find a rate, you need a function of TIME. So define your variables and functions before you do the mechanics. Otherwise you are likely to differentiate the wrong thing.
 
You should know:

[ds/dt] = [ds/dx] * [dx/dt]

[dx/dt] is given for your problem.
 
You must log in or register to reply here.
Top