Find the Rate of Change of Distance Between Two Cars Moving at Different Speeds

[AquaGlass]

Related Rates Help please!

Homework Statement

Two cars start moving from the same point. One travels south at 60 mi/h and the other travels west at 25 mi/h. At what rate is the distance between the cars increasing two hours later?"

Homework Equations

So far I have the equation..
(z^2) = (x^2) + (y^2)

I know dy/dt = 60 mi/h and dx/dt = 25 mi/h.

The Attempt at a Solution

I implicity differentiated the first equation and ended up with:
dz/dt = (50x +120y)*(1/2z)

Since I do not know what x or y is I am unable to find the rate the distance between the two cars is increasing.

Thank you so much for your help!

 

[rocomath]

x^2+y^2=z^2

taking the derivative

x\frac{dx}{dt}+y\frac{dy}{dt}=z\frac{dz}{dt}

You're told that one travels 60 miles per hour and the other, 25 miles per hour. How far would they each have gone after 2 hours?

 

[AquaGlass]

do I multiply the whole equation by 2 then?

 

[rocomath]

do I multiply the whole equation by 2 then?

Not the equation. Read my last sentence!

 

[AquaGlass]

After two hours, they would have gone 120 miles and 50 miles?

 

[rocomath]

Correct, now plug those values in. And the rate doesn't change, just the distance.

 

[PFStudent]

Hey,

Welcome to the forum AquaGlass.

After two hours, they would have gone 120 miles and 50 miles?

Exactly. But how did you get that? Once you figure out that, you will see how to re-express x and y.

Or in other words what are the following proportional to,

<br /> {\frac{dx}{dt}} = {\frac{?}{?}}<br />

<br /> {\frac{dy}{dt}} = {\frac{?}{?}}<br />



-PFStudent

 

[AquaGlass]

well then you get Zdz/dt = some number, but what about the Z then.. aren't we trying to find dz/dt? how do you cancel out the z?

 

[rocomath]

Use the Pythagorean theorem to find your missing length.

x^2+y^2=z^2

 

[AquaGlass]

ohhh ok i see! thank you so much!