# Thread: Find the second derivative

1. ## Find the second derivative

Hey, i'm having a bit of trouble after finding the first derivative of these two. After I find dx/dt and dy/dt I combine them into dx/dy and I'm stuck. Please help! Silly parameter problems

Find the second derivative of:

x= t+ sin(t) , y= t- cos(t)

2. Originally Posted by Inexcess
Hey, i'm having a bit of trouble after finding the first derivative of these two. After I find dx/dt and dy/dt I combine them into dx/dy and I'm stuck. Please help! Silly parameter problems

Find the second derivative of:

x= t+ sin(t) , y= t- cos(t)

$\displaystyle x= t+ \sin(t)$

$\displaystyle \frac{dx}{dt} = 1+\cos(t)$

$\displaystyle \frac{d^2x}{dt^2} = -\sin(t)$

$\displaystyle y= t- \cos(t)$

$\displaystyle \frac{dy}{dt} = 1+\sin(t)$

$\displaystyle \frac{d^2y}{dt^2} = -\cos(t)$

Does this help?

3. EDIT: tooooo sloooooooooow

Originally Posted by Inexcess
Hey, i'm having a bit of trouble after finding the first derivative of these two. After I find dx/dt and dy/dt I combine them into dx/dy and I'm stuck. Please help! Silly parameter problems

Find the second derivative of:

x= t+ sin(t) , y= t- cos(t)
I think it would be easier to find the second derivative of each and then 'combine' them.

$\displaystyle \frac{d^2y}{dx^2} = \frac{\frac{d^2y}{dt^2} }{\frac{d^2x}{dt^2}}$

Let me know if you're still stuck.

4. Well, in the notes it says to find the first derivative you do you the derivative of both. From there, you put dy/dt over dx/dt. Then, to get dx/dy you take the derivative of this (quotient rule) and then to find the second derivative you do d/dt (dy/dx) divided by dx/dt. Sorry, I don't know how to make the math symbols on here yet.

5. Scratch that, I think i'm just having a huge brain fade. I've come to 1+sin(t)/ 1+cos(t) . Is it possible to reduce this to just 1+tan(t). I really think i'm losing brain cells if I can't remember something this simple..

6. Originally Posted by Inexcess
Scratch that, I think i'm just having a huge brain fade. I've come to 1+sin(t)/ 1+cos(t) . Is it possible to reduce this to just 1+tan(t). I really think i'm losing brain cells if I can't remember something this simple..
I think you should refer to both my and pickslides posts. Firstly though, what exactly does the question ask you to do? Does it ask you to find $\displaystyle \frac{d^2y}{dx^2}$ ?

If so, then the easiest way to do so is find the individual second derivatives with respect to t, as per pickslide and then use the information in my post to
combine them.

$\displaystyle \frac{1 + \sin{t}}{1 + \cos{t}} \neq 1 + \tan{t}$ ! That is not how you cancel fractions.

7. Well, this section of homework is based on parametric equations. In class, we went over the method of find d2y/dx2 for these equations and it's not as simple as just taking the 2nd derivative of both and combining it. Or maybe it is. but yes, the directions just say take the second derivative.

8. Originally Posted by pomp
EDIT: tooooo sloooooooooow

I think it would be easier to find the second derivative of each and then 'combine' them.

$\displaystyle \frac{d^2y}{dx^2} = \frac{\frac{d^2y}{dt^2} }{\frac{d^2x}{dt^2}}$
This is NOT true!

Let me know if you're still stuck.
For example, if $\displaystyle x= t^2$, $\displaystyle y= t^3$, then $\displaystyle t= x^{\frac{1}{2}}$ so $\displaystyle y= x^{\frac{3}{2}}$, $\displaystyle \frac{dy}{dx}= \frac{3}{2}x^{\frac{1}{2}}$, and $\displaystyle \frac{d^2y}{dx^2}= \frac{3}{4}x^{-\frac{1}{2}}$.

But $\displaystyle \frac{dx}{dt}= 2t$, $\displaystyle \frac{dy}{dt}= 3t^2$, $\displaystyle \frac{d^2x}{dt^2}= 2$, and $\displaystyle \frac{d^2y}{dt^2}= 6t$ so that $\displaystyle \frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{6t}{2}= 3t= 3x^{\frac{1}{2}}$

You can treat first derivatives "as if they were fractions" but you cannot do that with higher derivatives!

9. Wow yes, you are completely right. When I went away and actually calculated both different ways I arrived at different results but was in a rush and assumed I'd made a bookkeeping error, not a school boy one! I do apologise!