Question

CALCULUS II--I need to find the second derivative of this polar equation:

x = 3t^2 - 6t
y = sqrt(t)
for t >= 0

Answer 1

This looks like a `parametric` equation :o not polar. Is that what you meant to type?
You appear to be still typing though, so I'll wait to see what ya got :3

Answer 2

I found \[\frac{ dx }{ dt } = 6t - 6\] and \[\frac{ dy }{ dt } = \frac{ 1 }{ 2 \sqrt{t} }\] Then, you put dy/dt over dx/dt find the derivative and put that over dx/dt.

Yes, I meant parametric, we are also learning polar so I got confused.

I am having trouble finding the derivative of dy/dx

Answer 3

\[\Large \color{royalblue}{\frac{dy}{dx}\qquad=\qquad \frac{1}{12\sqrt t(t-1)}}\]
Ok looks like you found dy/dx without any problems. (I simplified it down a tad bit).

So now we need to find,\[\Large \frac{d}{dx}\left(\color{royalblue}{\frac{dy}{dx}}\right) \qquad=\qquad \frac{\dfrac{d}{dt}\left(\color{royalblue}{\dfrac{dy}{dx}}\right)}{dx/dt}\]

So we already know that our denominator is \(\Large dx/dt=6(t-1)\).
So we need to find the derivative of the numerator with respect to t? Hmm

Answer 4

\[\Large \frac{\dfrac{d}{dt}\left(\color{royalblue}{\dfrac{dy}{dx}}\right)}{dx/dt}\qquad=\qquad \frac{\dfrac{d}{dt}\left(\color{royalblue}{\dfrac{1}{12\sqrt t(t-1)}}\right)}{dx/dt}\]So we just need to find the derivative of this messy thing right? :o

Answer 5

Pretty much. I'm lost as to how to do it..

Answer 6

Let's pull the constant out front. Distributing the sqrt(t) to each term in the brackets and applying rules of exponents gives us,
\[\Large \frac{1}{12}\left[(t^{3/2}-t^{1/2})^{-1}\right]'\]
This might be a little easier to work with, with can apply the power rule and chain rule from here.

Answer 7

as a double chk ...

since y = sqrt t, then y^2 = t

x = 3(y^2)^2 - 6y^2

x = 3(y^4 - 2y^2)

1 = 12y(y^2-1) y'

dy/dx = 1/12y(y^2-1)

which is what you worked it to to start with

Answer 8

\[\large \frac{1}{12}\left[(t^{3/2}-t^{1/2})^{-1}\right]'\qquad=\qquad -\frac{1}{12}(t^{3/2}-t^{1/2})^{-2}\color{green}{(t^{3/2}-t^{1/2})'}\]

Can you follow what I did here? Or did I jump ahead too quickly? :o

Answer 9

if we start from here:

1 = 12y (y^2-1) y' ; and product rule we get

0 = 12y' (y^2-1) y' + 12y (y^2-1)' y' + 12y (y^2-1) y''

0 = 12(y')^2 (y^2-1) + 12y (2y y') y' + 12y (y^2-1) y''

0 = 12(y')^2 (y^2-1) + 24y^2 (y')^2 + 12y (y^2-1) y''

0 = (y')^2 (y^2-1) + 2y^2 (y')^2 + y (y^2-1) y''

\[-\frac{(y')^2 (y^2-1) -+2y^2 (y')^2}{y(y^2-1)} = y''\]
and we know all those parts to define y'' :)

Answer 10

Yeah I got that and for the derivative of (t3/2−t1/2)′ I got \[\frac{ 3 }{ 2 } \sqrt{t}\ - \frac{ 1 }{ 2\sqrt{t} }\]

Answer 11

Yah that looks good!
Looks like should do some simplification though huh? :\ hmm

Answer 12

@amistre64 How did you get 1 = 12y (y^2-1) y'

Answer 13

\[-\frac{(y')^2 (3y^2-1) }{y(y^2-1)} = y''\]
\[-\frac{(\frac{1}{12y(y^2-1)})^2 (3t-1) }{\sqrt t(t-1)} = y''\]
\[-\frac{(\frac{1}{12\sqrt t(t-1)})^2 (3t-1) }{\sqrt t(t-1)} = y''\]
\[-\frac{\frac{1}{144t(t-1)^2} (3t-1) }{\sqrt t(t-1)} = y''\]
\[-\frac{ 3t-1 }{144t~(t-1)^3\sqrt t} = y''\]
and stuff

Answer 14

i eliminated the parameter, and implicit derived it

Answer 15

im thinking you might have to swap out t for x subs to have a dy^2/dx^2 tho

Answer 16

if method is your goal, then zepdrix has that covered

Answer 17

I know how to do it, I'm just having trouble doing it...also, our answer is supposed to be with respect to t

Answer 18

then hopefully we didnt mess up and we can use that as a dbl chk

Answer 19

by we mess up i meant I mess up lol

Answer 20

Okay, thanks guys! I'll try working it out and if I get stuck again I'll let you know lol

Answer 21

im thinking if i erred, it was in assuming my x' = 1 which is only true for dx/dx; but wrt.t, x' should have remained in tact to determine after the workings

Answer 22

I think you did mess up because I forgot that its an odd problem, so the answer is in the back. It says that the answer is \[\frac{ 1-3t }{ 144t ^{\frac{ 3 }{ 2 }} (t-1)^{3}}\]

Answer 23

But I still don't understand how to get it like that..I'm not great at implicit differentiation

Answer 24

nah, thats what i got

Answer 25

Whoops. lol The cubed looked like squared to me

Answer 26

\[-\frac{ 3t-1 }{144t~(t-1)^3\sqrt t} = y''\]
\[-\frac{ 3t-1 }{144t^{3/2}~(t-1)^3} = y''\]
\[\frac{ 1-3t }{144~t^{3/2}~(t-1)^3} = y''\]

Answer 27

Thanks!

Answer 28

youre welcome :)