# Instantaneous velocity and acceleration

1. Oct 7, 2004

### sb_4000

Hi,
This is probably an easy question for some of you, but I do not Undrestand it..

x=At^4 + (Bt + C)t^2 + D * sin(Et) where A, B, C, D, and E are constants. and I have to find instantaneous acceleration and velocity.

I know Instantaneous velocity is
v_x = dx/dt

Instantaneous acceleration dv_x/dt

But I do not know how to solve this equation..

Last edited: Oct 7, 2004
2. Oct 7, 2004

### arildno

You're absolutely correct in which formulas you are to use!
Now, let's find the velocity first:
You've been given x(t), could you state precisely why you find differentiating x(t) looks hard? (Refer to specific terms you find troublesome)

3. Oct 7, 2004

### sb_4000

do I just fond the derivative,
for example

4At^3+3Bt^2+2Ct+Dcos(Et), something like this, cause Im not sure how it should look.

4. Oct 7, 2004

### arildno

A bit too fast, there!
Let's take it slower:
1) You know that if h(t)=f(t)+g(t), then we have:
$$\frac{dh}{dt}=\frac{df}{dt}+\frac{dg}{dt}$$
that is, the derivative of a sum is the sum of the derivatives, right?
Let's use that to split up our task:
$$x(t)=At^{4}+Bt^{3}+Ct^{2}+D\sin(Et)$$
Hence, we have:
$$\frac{dx}{dt}=\frac{dF}{dt}+\frac{dG}{dt}+\frac{dH}{dt}+\frac{dI}{dt}$$
where I've introduced:
$$F(t)=At^{4},G(t)=Bt^{3},H(t)=Ct^{2},I(t)=D\sin(Et)$$
Now, as you can see, you have computed correctly the following terms:
$$\frac{dF}{dt},\frac{dG}{dt},\frac{dH}{dt}$$
But what about $$\frac{dI}{dt}$$?

5. Oct 7, 2004

### sb_4000

would it be sin(Et) only..

6. Oct 7, 2004

### arildno

Hint:
Have you learnt the CHAIN RULE yet?

7. Oct 7, 2004

### sb_4000

yeah I did in Cal1, but havent touched it in a while

8. Oct 7, 2004

### sb_4000

Sin(Et)+Dcos(Et)? something like this, Ill go back and read my Calculus book.

9. Oct 7, 2004

### arildno

OK, that's the rule you need to use, in order to differentiate Dsin(Et) correctly.
So, how would you proceed to do that?

10. Oct 7, 2004

### arildno

Your posts 8+5 have wrong suggestions.

11. Oct 7, 2004

### sb_4000

if I replace D, E with X, can I do that?

then I would get

xsin(xt)=sin(Xt)cos(Xt)?

12. Oct 7, 2004

### sb_4000

would the instantaneous velocity and acceleration be different in this question.

13. Oct 7, 2004

### arildno

NO!
It seems to me you find the chain rule kind of hard to understand;
that's fairly common, though. It IS rather difficult.
Let's suppose you've got two functions, f(t) and g(t).
You can now make yourself a new function, h(t), by substituting g(t) into all instances of "t" in f.
That is, we write:
h(t)=f(g(t))
Now, the chain rule says:
h'(t)=f'(g(t))g'(t).
Let's be absolutely clear about what this MEANS, in terms of how to compute the derivative of h(t) (that is, h'(t)) correctly:
1. Differentiate f(t) as usual, that is find $$\frac{df}{dt}$$
2. Substitute into all instances of "t" in your expression for $$\frac{df}{dt}$$ g(t)
3. The expression you've now found, is f'(g(t)), that is, one of the factors in h'(t).
4. Differentiate g(t) as usual, that is, find $$\frac{dg}{dt}$$ (or, which is the same, g'(t))
5. Finally, multiply together f'(g(t)) (found in 2&3) and g'(t).
You have now found h'(t)=f'(g(t))g'(t)

In order to use the chain rule, you must first find good choices for f(t) and g(t)!
We have: $$I(t)=D\sin(Et)$$
A good choice of f(t) is:
$$f(t)=D\sin(t)$$
(The reason why this is a GOOD choice, is that we know how to differentiate sine!)
We therefore set:
g(t)=Et
We do this, because we now see that we may write:
I(t)=f(g(t))
Agreed?
Hence, let's calculate $$\frac{dI}{dt}$$ according to the scheme above!
1. $$\frac{df}{dt}=D\cos(t)$$
Agreed?
2&3. $$f'(g(t))=D\cos(Et)$$
4. $$\frac{dg}{dt}=E$$
5. $$f'(g(t))g'(t)=DE\cos(Et)$$
Hence, we have:
$$\frac{dI}{dt}=DE\cos(Et)$$

Last edited: Oct 7, 2004
14. Oct 7, 2004

### arildno

OOPS!
Dreadfully sorry, wrote "cosine" at some places where I meant "sine".
This is now edited.

15. Oct 7, 2004

### sb_4000

If is the answer for the instantaneous velocity, what is the difference between that and instantaneous acceleration.

16. Oct 7, 2004

### arildno

In order to find the acceleration, you must differentiate your new-found expression for the velocity.

17. Oct 7, 2004

### sb_4000

This is what I get,

12At^2 + 6Bt + 2C - DEsin(Et)

is this correct?

18. Oct 7, 2004

### arildno

Again, you're right about the 3 first terms; you haven't done the last term correctly, though (it's that damn chain rule again..)

19. Oct 7, 2004

### sb_4000

-DE^2sin(Et)?

20. Oct 7, 2004

Precisely!