I really need help w/ Related Rates

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nikchic5

Junior Member
Joined
Feb 16, 2006
Messages
106
Hello I was wondering if anyone could get me started on this problem. I am lost when it comes to Related Rates and I have a lot of questions about them. If anyone could help I would really appreciate it!

A plane flying horizontally at an altitude of 4 mi and10 a speed of 450 mi/h passes directly over a radar station. Find the rate at which the distance from the plane to the station is increasing when it is mi away from the station.

I don't even know where to start...

Thank you so so much!!!
 
You MUST have a drawing.
You MUST label appropriate parts, using clear and concise definitions.
You MUST establish are formulaic relationship between what is known and what is asked.
 
Re: OK?

nikchic5 said:
Ok well I drew the picture but I don't know where to go?
Click to expand...
Now label the picture. What sort of figure did you draw? A triangle? What does the "base" stand for? What does the "height" stand for? What variables have you used? What do they stand for?

And so forth.

Eliz.
 
so far...

A tringle. I got the base as 116 and the hyp as 232. Which would make the other side the suqare root of 40368. And one angle 90 degree. But I don't know where to go? Thanks
 
Is this what you drew?.

plane2ud.gif


You could call the hypoteneuse, say, D.

You know y, \(\displaystyle \frac{dx}{dt}\), and x

You need \(\displaystyle \frac{dD}{dt}\)

Differentiate with respect to time, enter in your known values, and solve for \(\displaystyle \frac{dD}{dt}\)

That should be m, not km, in the drawing.
 
Hello...

I just don't understand what the equation would be? I know I have all of those values but I don't know how to write the equation? And where did the 450 come from? Thanks!
 
Mistake..

OMG I am soo sorry! For some reason I was thinking of a TOTALLY different problem. I am so sorry! But hold on let me work on it...Thanks
 
Sorry so many questions...

What is the equation to differentiate with respect to time...sorry so many questions I just really want to understand this! Thank you so much!
 
Do the Pythagoras thing. You know that?

\(\displaystyle \L\\D^{2}=x^{2}+y^{2}\)

Let x be the planes horizontal path, y be its height. That makes sense, right?. y is vertical and x is horizontal.

Give it a shot and right back with your attempts.
 
What's Next?

Yea I know that...I already did that. D would be the swuare root of 116 or roughly 10.77. But I don't know the equation to put it all into. Thanks
Again sorry so many questions
 
The equation is set up. Now differentiate and solve for the variable you need.

\(\displaystyle \L\\D^{2}=x^{2}+y^{2}\)

Differentiate:

\(\displaystyle \L\\2D\frac{dD}{dt}=2x\frac{dx}{dt}+2y\)

Fill in your info and solve for \(\displaystyle \L\\\frac{dD}{dt}\)

I don't want to just give you the answer and you not understand what's going on. Do you see?. The rates that are changing are identified by the
differentiation notation. The constants will be D and y. They want to know how fast the distance between the station and plane are changing.
That is \(\displaystyle \frac{dD}{dt}\). See?. Since the plane is traveling at 450mi/hr, that rate is \(\displaystyle \frac{dx}{dt}\). it's at a constant height of 4 miles, so that doesn't change. D is your pythagoras thing, because they want to know how fast it changes when it is 10 miles away from the station. Is it a little clearer now?.
 
Thanks so much...

Ok thank you very much I FINALLY figured it out. Sorry I asked so many questions! But I ended up with 412. Thanks you again!!
 
Re: Thanks so much...

But I ended up with 412.
Click to expand...

Recheck. I get:

\(\displaystyle \L\\2(2\sqrt{29})\frac{dD}{dt}=2(10)(450)+2(4)=>\)

\(\displaystyle \L\\4\sqrt{29}\frac{dD}{dt}=9008\)

\(\displaystyle \L\\\frac{dD}{dt}=\frac{9008}{4\sqrt{29}}\)

\(\displaystyle \L\\\frac{dD}{dt}=418.2mi/hr.\)
 
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