Huge problem with relation rates.

[cd246]

Homework Statement

I search all over my textbook and did not show any (simple) examples on relation rates(seriously). Well, The first problem is, x^2+y^2=25 find dy/dt when x=3. dx/dt =4 and it says find the indicated rate and assume x>0 and y>0

Homework Equations

I did my best considering the circumstances, I derived and got 2x(dx/dt)+2y(dy/dt)=0, applied 3 and got 6+2y=0 then 2y=-6 .

The Attempt at a Solution

Then divided and got -6/2y=-3y
The answer stated -3 so I guess I am close. With these, what is the difference between implicit diff. and relation rates? and what is the difference in steps?

 

[SpitfireAce]

ok, first just plug x=3 into x^2+y^2=25 and solve for y. Then implicitly differentiate x^2+y^2=25... you should end up with dy/dt=(-2x(dx/dt))/(2y) and you get dy/dt=-3. Frankly, I've only done related rates problems in the form of word problems, you just have to figure out from the wording, the equation you want to use and the givens.. then you get whatever else you need and finally implicitly differentiate
6+2y=0? why in the world did you get rid of dy/dt and dx/dt?

 

[olgranpappy]

Homework Statement

I search all over my textbook and did not show any (simple) examples on relation rates(seriously). Well, The first problem is, x^2+y^2=25 find dy/dt when x=3. dx/dt =4 and it says find the indicated rate and assume x>0 and y>0

Homework Equations

I did my best considering the circumstances, I derived and got 2x(dx/dt)+2y(dy/dt)=0, applied 3 and got 6+2y=0 then 2y=-6 .

that's wrong. you set the rates to one... why? at x=3 you have

6(dx/dt)+2y(dy/dt)=0

*not* 6+2y=0

so then, if (dx/dt) is 4 we have

6*(4)+2y*(dy/dt)=0

now all you have to do is to use the equation (x^2+y^2=25) to find y at x=3 and plug in above to find (dy/dt).