how fast is distance between 2 cars changing half-hour later

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tsh44

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Two cars start at the same place and time. One travels west at constant velocity of 50 miles per hour and a second one travels south at a constant velocity of 60 mph. Approximately how fast is the distance between them changing one-half hour later?

I wrote the velocities as 25x and 30x for half an hour later. I think the distances each way would be the antiderivatives (12.5x^2)^2 and (15x^2)^2. I tried doing the pythag theorme and got d = 19.25x^2 as a guess and things didn't work out. I would love some help
 
Use Pythagoras. \(\displaystyle \L\\D^{2}=x^{2}+y^{2}\)

Differentiate:

\(\displaystyle \L\\D\frac{dD}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}\)

You want dD/dt. You are given dx/dt(west) and dy/dt(south). You can find D, x, and y by using Pythagoras and t=1/2 (1/2 an hour). For instance, after a 1/2 hour, 60(1/2)=30 miles. Can you finish?.
 
sorry Im a bit confused. so x would =25, y=30 and D would equal SQRT(1525)? or should i write them as 25x, 30x, etc.? I'm also not sure where you got that equation to derive. Thanks
 
tsh44 said:
I'm also not sure where you got that equation to derive.
Click to expand...
The Pythagorean Theorem should have been covered extensively in geometry and/or trigonometry, and should have come up frequently in algebra and again now.

To learn how it relates to right triangles, try reading some of the information available online. You will want to understand this, because you'll be using it a lot in calculus.

Eliz.
 
Draw a diagram. It can help.

carsed1.gif
 
Re: how fast is distance between 2 cars changing half-hour l

Hello, tsh44!

Did you make a sketch?

Two cars start at the same place and time.
One travels west at constant velocity of 50 miles per hour
and a second one travels south at a constant velocity of 60 mph.
Approximately how fast is the distance between them changing one-half hour later?
Click to expand...
Code: 
           50t     O
    A * ← - - - - *
        *         |
          *       |
         x  *     | 60t
              *   |
                * ↓
                  * B

Car #1 starts at \(\displaystyle O\) and drives west at 50 mph.
In \(\displaystyle t\) hours, it has driven \(\displaystyle 50t\) miles to point \(\displaystyle A.\)

Car #2 starts at \(\displaystyle O\) and drives south at 60 mph.
In \(\displaystyle t\) hours, it has driven \(\displaystyle 60t\) miles to point \(\displaystyle B.\)

Let \(\displaystyle x \,=\,AB\)
Then: \(\displaystyle \:x^2\:=\:(50t)^2 + (60t)^2 \:=\:6100t^2\)

Differentiate: \(\displaystyle \:2x\cdot\frac{dx}{dt} \:=\:12200t\;\;\Rightarrow\;\;\L\frac{dx}{dt}\:=\:\frac{6100t}{x}\;\) [1]

When \(\displaystyle t\,=\,\frac{1}{2}:\;x^2 \:=\:6100\left(\frac{1}{2}\right)^2 \:=\:1525\;\;\Rightarrow\;\;x\:=\:5\sqrt{61}\)

Substitute into [1]: \(\displaystyle \L\:\frac{dx}{dt}\;=\;\frac{6100\left(\frac{1}{2}\right)}{5\sqrt{61}} \;\approx\;78.1\text{ mph.}\)

 
The method I was getting at was like so:

Using Pythagoras and differentiaitng, we get:

\(\displaystyle \L\\D\frac{dD}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}\)

From the given data, if the westbound car had driven for 1/2 hour, then it has travelled x=50(1/2)=25 miles. The southbound car, y=60(1/2)=30.

The \(\displaystyle \L\\D=\sqrt{25^{2}+30^{2}}=5\sqrt{61}\)

So, using the derivative from above:

\(\displaystyle \L\\5\sqrt{61}\frac{dD}{dt}=(25)(50)+(30)(60)=3050\)

\(\displaystyle \L\\\frac{dD}{dt}=\frac{3050}{5\sqrt{61}}=10\sqrt{61}\approx{78.1}\)

Same as Soroban. We're either both right or both wrong. I hope it's the former.
I think it is. :wink:
 
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