# Calculus questions

• Sep 4th 2008, 10:51 AM
kezman
Calculus questions
1 I need to know the convergence of this series

$\sum_{k=1}^{\infty} \frac{2k+5}{\sqrt{3{k^3} - 1/2}}$

2 I have to obtain the expression of dy/dx for the following function

3xy - 2x^2y + y^3 =0

3 $5x^2 + 3y^2 = 16$
The x radius is $\sqrt{5}$ ?
Thanks
• Sep 4th 2008, 11:09 AM
11rdc11
Use the integral test for the 1st problem to see if it converges
• Sep 4th 2008, 11:16 AM
Moo
Quote:

Originally Posted by 11rdc11
Use the integral test for the 1st problem to see if it converges

It's better to compare :

As n approaches infinity, 2k+5 ~ k and sqrt(3k²-1/2) ~ k^(3/2)

Therefore, (2k+5)/sqrt(3k²-1/2) ~ k/k^(3/2)=1/k^(1/2)

By comparison with the Riemann series, since 1/2<1, the series is not convergent.

Quote:

2 I have to obtain the expression of dy/dx for the following function

3xy - 2x^2y + y^3 =0
Implicit differentiation, do you know it ?

Quote:

3 5x^2 + 3y^2 = 16
The x radius is \sqrt{5} ?
Thanks
This is the equation of an ellipse, which is x²/a² + y²/b² = 1

So transform it like it should be.
• Sep 4th 2008, 11:27 AM
kezman
on 2 ive got this

$dy/dx$ $= (2x^2 - 3x)/2y$
• Sep 4th 2008, 11:35 AM
Moo
Quote:

Originally Posted by kezman
on 2 ive got this

$dy/dx$ $= (2x^2 - 3x)/2y$

Hmmm I don't think you applied the chain rule properly ! o.O

Would you mind showing your working ?

I get something like dy/dx=(4xy-3y)/(3x-2x²+3y²) (I did it quickly)
• Sep 4th 2008, 11:41 AM
kezman
y^3 = y(2x^2 - 3x)
y^2 = x(2x -3) (is this legal?)
and then I derive the sqrt
• Sep 4th 2008, 11:46 AM
Moo
Quote:

Originally Posted by kezman
y^3 = y(2x^2 - 3x)
y^2 = x(2x -3) (is this legal?)
and then I derive the sqrt

This is the problem !

The derivative of y^3 wrt x is 3y'y², by the chain rule. Remember, y is a function of x ! The RHS is weird too.