Related Rates problems

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December 6th 2008, 01:27 PM
fattydq

Related Rates problems

I am greatly struggling with related rates problems. I'll give an example and perhaps someone can walk me through it.

At noon, ship A is 70 km west of ship B. Ship A is sailing south at 25 km/h and ship B is sailing north at 15 km/h. How fast is the distance between the ships changing at 6:00 PM?

I don't even know where to begin really, according to my notes I should begin by writing down the given info and what I'm trying to find and drawing a picture of applicable, but after doing all this I still have no idea how to go about finding the answer. I know that da/dt would be 25 and db/dt would be 15, but I don't know how I would use that information at all. I also know that a triangle will be involved, but again I don't know how to use that information to solve this problem.
December 6th 2008, 02:58 PM
polymerase

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Ok so bottom line, based on the diagram, you want to find $\frac{dz}{dt}$ when $y=240$ - that's your question

So at 6pm, 6 hrs have passed; therefore, ship A has travelled 150km south (y1) and ship B has travelled 90 km north (y2). y= y1+y2=240km

x = 70km = CONSTANT

AT THE INSTANT of 6pm, z would be $\sqrt{240^2+70^2}=250$ km

Therefore, to solve this question, you use a modified version of pythagorean's theorem: $z^2=x^2+(y_1+y_2)^2$
Note: $y_1$ and $y_2$ are variables, NOT constants.

$z^2=x^2+(y_1+y_2)^2 \to 2z\frac{dz}{dt}=2x\frac{dx}{dt}+2(y_1+y_2)\left(\f rac{dy_1}{dt}+\frac{dy_2}{dt}\right)$

Plug in all your numbers and solve... should get $\frac{192}{5}$ km/h
December 6th 2008, 03:08 PM
fattydq

I'm still very very confused. I understood everything up to the final two lines, the equation involving z^2=etc. I have no iea how you came up with that to solve it? I understand that the distance between the ships at 6pm is 250 km but I don't understand how to use that information to find the rate of change between the ships at that time.
December 6th 2008, 05:12 PM
polymerase

Quote:

Originally Posted by fattydq

I'm still very very confused. I understood everything up to the final two lines, the equation involving z^2=etc. I have no iea how you came up with that to solve it? I understand that the distance between the ships at 6pm is 250 km but I don't understand how to use that information to find the rate of change between the ships at that time.

you need to find a relationship between z, x, and y...there's a right angle triangle... $z^2=y^2+x^2$ seem fimilar to you?
December 6th 2008, 05:44 PM
fattydq

Yes of course, but he already used the pythagorean theorem to find the 250km, I don't see how using the pythagorean theorem could possibly lead to the answer of the fraction he provided
December 6th 2008, 06:14 PM
skeeter

Quote:

Originally Posted by fattydq

I am greatly struggling with related rates problems. I'll give an example and perhaps someone can walk me through it.

At noon, ship A is 70 km west of ship B. Ship A is sailing south at 25 km/h and ship B is sailing north at 15 km/h. How fast is the distance between the ships changing at 6:00 PM?

let ship A begin at the origin. its position as a function of time is ...

$x = 0$

$y = -25t$

ship B starts at (70,0). its position as a function of time is ...

$x = 70$

$y = 15t$

using the distance formula, the distance between the two ships at any time t is ...

$D = \sqrt{(70-0)^2 + [15t - (-25t)]^2}$

$D = \sqrt{4900 + 1600t^2}$

$\frac{dD}{dt} = \frac{1600t}{\sqrt{4900+1600t^2}}$

now ... evaluate $\frac{dD}{dt}$ for $t = 6$ hrs
December 6th 2008, 08:38 PM
fattydq

Quote:

Originally Posted by skeeter

let ship A begin at the origin. its position as a function of time is ...

$x = 0$

$y = -25t$

ship B starts at (70,0). its position as a function of time is ...

$x = 70$

$y = 15t$

using the distance formula, the distance between the two ships at any time t is ...

$D = \sqrt{(70-0)^2 + [15t - (-25t)]^2}$

$D = \sqrt{4900 + 1600t^2}$

$\frac{dD}{dt} = \frac{1600t}{\sqrt{4900+1600t^2}}$

now ... evaluate $\frac{dD}{dt}$ for $t = 6$ hrs

Again I understand the whole thing until you involve the final step, the fraction.
December 7th 2008, 03:18 AM
skeeter

take the derivative of D w/r to t using the chain rule ...

$D = (4900 + 1600t^2)^{\frac{1}{2}}$

$\frac{dD}{dt} = \frac{1}{2}(4900 + 1600t^2)^{-\frac{1}{2}} \cdot 3200t$

$\frac{dD}{dt} = 1600t(4900 + 1600t^2)^{-\frac{1}{2}}$

$\frac{dD}{dt} = \frac{1600t}{\sqrt{4900 + 1600t^2}}$

clear enough?