Calc Test Tomorrow. Help me please

[MrCodeDude]

53. A right triangle in the first quadrant had the coordinate axes as sides, and the hypotenuse passes through the point (1,8). Find the vertices of the triangle such that the length of the hypotenuse is minimum. (I drew a picture)

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c = +/- sqrt [ x^2 + y^2 ]
(but only use positive one, right?)

What is the other equation? It isn't y = 8x, cause then you get x = 0, which can't happen.

 

[Ionizer86]

Take the c equation and derive it to find dc/dt. Then, set that value to zero, and when there is no change of c over change in time, you have the minimum length of the hypotenuse

 

[Zorba]

Only postitive because distance can only be postitive. Been a lot time since I have done optimization (sp?) so I am not sure about the second part

 

[Zorba]

Yeah what he said ^ is right

 

[MrCodeDude]

Originally posted by: Ionizer86
Take the c equation and derive it to find dc/dt. Then, set that value to zero, and when there is no change of c over change in time, you have the minimum length of the hypotenuse

But then you get dx/dt and dy/dt by doing dc/dt?

 

[crt1530]

Vertices are (0,0), (0,9), (9,0).

 

[Trevelyan]

Originally posted by: MrCodeDude

Originally posted by: Ionizer86
Take the c equation and derive it to find dc/dt. Then, set that value to zero, and when there is no change of c over change in time, you have the minimum length of the hypotenuse

But then you get dx/dt and dy/dt by doing dc/dt?

you don't take dc/dt... there's no time involved. firstly you gotta relate the variables x and y to each other so you only have one variable (other than c) in the equation for c. Then take the derivative with respect to that variable and set it to 0 and solve to get the minimum.

 

[crt1530]

OK...I'll help...incrementally. Try to solve it yourself, though. You have three points on the triangle. (0,0) (0,y) (x,0). And you know that y=mx+b. And that (1,8) must fit into that equation.

 

[crt1530]

You want to minimize the distance between (0,y) and (x,0).

 

[MrCodeDude]

You do know your points are wrong, right?

Because the slope through those points has to be -8. (take slope of the points (0,y) and (x,0), you get m = -y/x. Plug in (1,8), you get m = -8)

 

[crt1530]

That distance is....sqrt(x^2 + y^2). But we need to be able to do this in one variable because you have no idea what partial differential equations are.

 

[crt1530]

Um, you don't know what the hell you're talking about. Generally, when somebody is helping you, it's a good idea to keep your mouth shut unless you have a question. Telling the helper that what they are doing is wrong when you have no clue is somewhat irritating and leads to a disinclination to help.

 

[Shooters]

Originally posted by: MrCodeDude
You do know your points are wrong, right?

Because the slope through those points has to be -8. (take slope of the points (0,y) and (x,0), you get m = -y/x. Plug in (1,8), you get m = -8)

Actually, that's not quite right. If your two points are (x,0) and (0,y) then yes, the slope is m = -y/x, but you can't plug the point (1,8) into that equation and expect to get a correct answer.

Just remember that slope is change in y divided by change in x for any two points.

 

[crt1530]

The points I gave you, (0,9) and (9,0) yield y= -x + 9, which passes through point (1,8) and has slope -1. Good luck on your test.

 

[MrCodeDude]

I was informing you, your answer is incorrect. So if you were simply showing how you got those points, I don't care.

I have the answer of (0,0), (5,0), (0,10). Which is correct, you don't need to get off your pedestal though, just a bigger fall next time.

 

[Justin218]

Haha, you posted here and at genmay. Sorry I can't help you out though. Maybe next semester. I haven't seen anything like that in calc1?