Finding Concavity: Solve with x's and y's | Rutgers

[ACLerok]

http://www.eden.rutgers.edu/~cjjacob/images/calc.gif

I am given this problem to do but I'm not absolutely sure how to get through it. The way I would do is to find y' in terms of x's and y's and then plugin the point (1,2) and then take the derivative again and see if y'' is positve (curve up) or negative (curve down). Is this the best way to go about this problem or is it completely wrong or what?

 

[cookiemonster]

Yes. Concavity is determined by the second derivative.

So implicitly differentiate twice and get an expression for d^2y/dx^2 in terms of x and y. Then plug in (1,2).

And whatever you do, ignore that ugly-looking expression for y(x)! That will make your life very, very, very miserable.

cookiemonster

 

[ACLerok]

Yes. Concavity is determined by the second derivative.

So implicitly differentiate twice and get an expression for d^2y/dx^2 in terms of x and y. Then plug in (1,2).

And whatever you do, ignore that ugly-looking expression for y(x)! That will make your life very, very, very miserable.

cookiemonster

I calculated the first derivative and got 15x^(2)y+5x^(3)y'-3y^(2)-6xyy'+3y^(2)y'=0. is this correct and if so, what do i do after this? express in terms of y'?

 

[cookiemonster]

Yes. Now solve for y' and then differentiate again. Then solve for y'' and substitute away any y' you see. Then just plug in values.

cookiemonster

 

[ACLerok]

Thanks! I got it.