# Math Help - Calculus questions

1. ## 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

2. Use the integral test for the 1st problem to see if it converges

3. 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.

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 ?

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.

4. on 2 ive got this

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

5. 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)

6. y^3 = y(2x^2 - 3x)
y^2 = x(2x -3) (is this legal?)
and then I derive the sqrt

7. 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.