Implicit/Explicit Differenciation, and Inflection point of a Graph

[clragon]

I have two questions, one about differentiating, and the other is about finding the inflection point on a function. Any help would be greatly appreciated.

Question 1

Homework Statement

If ay^3 = x^4, show that d^2y/dx^2 = 4x^2/9ay^2

now, this question can be done by both implicit or explicit differentiation. I can leave the ay^3 = x^4 as it is and start simplifying right away, or I could isolate y by making the equation y = (x^4a^-1)^(1/3)

Homework Equations

knowledge of the rules of differentiating.

The Attempt at a Solution

I am going to list two different attempts, one implicit and one explicit.

Implicit
ay^3 = x^4
a3y^2 dy/dx = 4x^3

dy/dx = 4x^3/ a3y^2

d^2y/dx^2 = (12x^2((a3y^2)-6ya(4x^3)(dy/dx)) / (3y^2a)^2
d^2y/dx^2 = (36x^2y^2a - 24yax^3(dy/dx) ) / (3y^2a)^2
*I then subbed the previous found value for dy/dx into dy/dx in this equation.
d^2y/dx^2 = (36x^2y^2a - 96yax^6/3y^2a) / (3y^2a)^2
d^2y/dx^2 = (36x^2y^2a - 32x^6/y) / 9y^4a^2

I stopped here because I could find no way to make the a^2 at the bottom become just a, similarily the y^4 could not be reduced down to y^2 like in the question.

Explicit

y = (x^4a^-1)^(1/3)
y' = (1/3)(x^4a^-1)^(-2/3)(4x^3a^-1)
y' = [(x^4a^-1)^(-2/3)(4x^3a^-1)]/3
y'' = [3(12x^2a^-1)^(-2/3)(x^4a^-1)^(-5/3)(4x^3a^-1)]/9
y'' = [-2(48x^5(a^-2)(x^4a^-1)^(-5/3)]/9
y'' = [-96xa^-1(x^4a^-1)(x^4a^-1)^(-5/3)]/9
y'' = -96x / 9a [(x^4/a^1)^(1/3)]^2
since (x^4/a^1)^(1/3) = y
y'' = -96x / 9ay^2

as you can see, I got a lot further by the final answer is still wrong... I have part of the answer but have a -96x instead of 4x...

could someone please be kind enough to solve this question, or tell me where I went wrong... if it is possible please also provide a solution to the implicit way of solving this question.

Question 2
1. Find the point of inflection of y = x^4 - 4x^3 + 6x^2 + 12x

at a glance, this question may seem very easy, but for some reason I can't get the right answer (-0.4 and 2.4)

2. to find inflection point, you find the 2nd derivative of the function and solve for x/b]

The Attempt at a Solution

y = x^4 - 4x^3 + 6x^2 + 12x
y' = 4x^3 - 12x^2 + 12x + 12
y'' = 12x^2 - 24x + 12
let y'' = 0
0 = 12(x^2 - 2x + 1)
0 = 12(x-1)(x-1)
x = 1

I put this function into a graphing calculator, it showed no apparent change in concavity at x = 1. I set the min y = -100 and max y = 100 while keeping min x = -20 and max x = 20 to get a better view. the graph seemed to be a parabola but like the answer suggested, the concavity is slightly different between -0.4 and 2.4.

anyone wishing to see the graph can just use http://www.coolmath.com/graphit/index.html" and paste in the function. remember the set the min y, max y, min x, max x values so that you can see the graph as described by me.

Thanks for you time.

 

[chanvincent]

Question 1:
For your first attempts, you are stucked in the last step:

$$\frac{d^2y}{dx^2} = \frac{(36x^2y^2a - 32x^6/y)}{ 9y^4a^2}$$

You could simplify it one step further to get

$$\frac{d^2y}{dx^2} = \frac{4x^2}{ay^2} - \frac{32x^6}{9a^2y^5}$$

The first term on the RHS is very similar to what you want to prove, but the second term mess up the whole thing, therefore you must find a way to get rid of the second term...
Try to expand the $$x^6$$ into $$x^2x^4$$ and substitude the $$x^4$$ in the second term by the original equation, $$x^4 = ay^3$$
This will solve your problem...

For the second attempt, try to do it in a simpler maner...
$$y = (\frac{x^4}{a})^{1/3} = a^{-1/3}x^{4/3}$$

$$\frac{dy}{dx} = \frac{4}{3}a^{-1/3} x^{1/3}$$

$$\frac{d^2y}{dx^2} = \frac{4}{3} \frac{1}{3} a^{-1/3}x^{-2/3} = \frac{4}{9} a^{-1/3}\frac{x^2}{x^{8/3}} = \frac{4}{9} a^{-1/3}\frac{x^2}{a^{2/3}y^2} = \frac{4x^2}{9ay^2}$$

For question 2, Your solution is abosulotely correct, maybe the answer is wrong or you misunderstand the question..

Good Luck

 

[clragon]

thank you so much :D

your second solution was so much easier than mine, but I'm curious of where I went wrong in my solution, did I make a arithmetic mistake? or does using the power rule in this equation eventually give me -96x at the top?