# Homework Help: 2nd derivative

1. Nov 10, 2014

### Klaz

1. The problem statement, all variables and given/known data
Show that for a projectile d^2 (v^2)/dt^2 = 2g^2

2. Relevant equations

3. The attempt at a solution
Since we are only dealing with the y component for this. 2nd derivative would be 2. But since gravity acts on it then d^2 (v^2)/dt^2 = 2g. I don't get how to get to 2g^2. Can someone help me please.

2. Nov 10, 2014

### Staff: Mentor

Start with a general expression for the vertical velocity of a projectile.

3. Nov 10, 2014

### Staff: Mentor

Step by step. What's the first derivative of v^2?

4. Nov 10, 2014

### Klaz

First derivative would be 2v.

5. Nov 10, 2014

### Klaz

Vertical velocity = v - at. Im still confuse on what to do with this equation. Since dv/dt = a.

6. Nov 10, 2014

### Klaz

If Vy=v^2-gt then the first would be dv/dt = 2v-g.

7. Nov 10, 2014

### Staff: Mentor

That's not complete. In that equation v is a function of time, so apply the chain rule.

that would be $v = v_o - g t$. Square both sides so that you have an expression for v2. Then do your derivatives. Remember that dv/dt = g.

8. Nov 10, 2014

### Staff: Mentor

I don't know where you got that expression for Vy. It's not correct.

9. Nov 10, 2014

### PeroK

It's your maths that's letting you down here. I don't think you understand what is being asked. You are asked to differentiate $v^2$. Not $v$.

I would first work out what $v^2$ is. That seems logical to me: you are asked to differentiate $v^2$, with respect to time (twice). So, let's first have $v^2$ as a function of $t$.

$v^2 = \dots$

Can you do that?

10. Nov 10, 2014

### Staff: Mentor

As gneill already pointed out, that is not complete.

What you found is the first derivative with respect to v. But what you need is the first derivative with respect to t. So, keep going. (Think chain rule.)

11. Nov 10, 2014

### Klaz

If I apply chain rule. Let u=(v-gt)^1/2. Then : dv/dt = (u)^1/2
dv/dt = du/dt × dv/du = -g/2 (v-gt)^1/2. Am I on the right track? Then my next step would be to differentiate my answer again.

12. Nov 10, 2014

### Staff: Mentor

Let u = v^2
du/dt = du/dv * dv/dt.

Hint: What is dv/dt for a projectile?

13. Nov 10, 2014

### Klaz

du/dt = 2v * -g = -2vg.
2nd derivative:
du/dt = -2g * - g = 2g^2.
Thank you. Can I ask one more question. What is wrong with my first attempt using the chain rule? Can i still get the same answer if I kept going with my solution?

14. Nov 10, 2014

### Staff: Mentor

I couldn't follow what you were trying to do.

Where did you get this?