# Homework Help: Differentiation Problem

1. Aug 10, 2010

### eddysd

1. The problem statement, all variables and given/known data

a=(t^2 + 1)i - j + tk

Find da/dt

I multiplied out the brackets so a= it^2 + i - j + tk

then got as far as da/dt= 2it + 0 + 0 + ?

a) is this right so far?
b) how do i integrate tk with respect to t?

Any help would be greatly appreciated.

2. Aug 10, 2010

### Char. Limit

I would assume a constant k, in which it's simple.

3. Aug 10, 2010

### eddysd

Thinking about it, the question says "Given a=(t^2)i -j + tk find:"

da/dt

Do you think that the bold is relevant? There are another two parts one for b and db/dt then the third part is d/dt(a.b). The whole question is worth 7. Any ideas?

4. Aug 10, 2010

### Staff: Mentor

Yes, the bolded things are very relevant. i, j, and k on the right side are unit vectors. da/dt will also be a vector quantity.

5. Aug 10, 2010

### Staff: Mentor

For both a and b, differentiation and integration of vector functions is done component-wise.

As for the integration, here's an example:
$$\int 2t\bold{i} dt = t^2\bold{i} + \bold{C}$$

6. Aug 10, 2010

### eddysd

so would da/dt= 2ti + 0j + 1k ?

7. Aug 10, 2010

### Staff: Mentor

Yes. This could also be written as da/dt = 2ti + k

8. Aug 10, 2010

### eddysd

Thank you for the help! But just to check I understand, I'm going to do the second and third parts.

b=2tj - k

so db/dt=2j ?

and the third part, d/dt(a.b) how would I times the two together?

Would a.b= t^2i - 2tj -tk?

So d/dt(a.b)=2ti - 2j -k ?

9. Aug 10, 2010

### eddysd

Another question whilst I'm at it, if f=5x^5.y

is df/dx= 25x^4.y + 5x^5.ydx?

10. Aug 10, 2010

### Staff: Mentor

What does 5x^5.y mean?
Is it 5x5* y or 5x5y?
Is f a function of two independent variables, x and y?

11. Aug 10, 2010

### eddysd

Sorry for being unlclear, it's (5x^5)* y. And yes, it is f(x,y).

12. Aug 10, 2010

### Staff: Mentor

A better way to write it would be f = 5yx5.

So you're trying to find fx, or $$\frac{\partial f}{\partial x}$$.

You don't need to use the product rule. Since x and y are independent variables, y is considered a constant when differentiating with respect to x.

Why do you have dx in one part of your answer? It should not be there.

13. Aug 10, 2010

### eddysd

I just remembered something from school, I thought you had to put something like that in when differentiating y with respect with x. So df/dx would just be 25yx^4?

14. Aug 10, 2010

### eddysd

Implicit differentiation? Or do you not do that when the function is f(x,y)?

15. Aug 10, 2010

### Staff: Mentor

No.
That's the answer, but you should indicate that you're doing a partial differentiation. There are two (at least) kinds of notation: one with subscripts, which in this case would be fx, and another is similar to df/dx but with different letters -- $$\frac{\partial f}{\partial x}$$

No.
No, not then either.

If you have z = f(x, y) and you want the total differential, it will involve differentials of x and y.

dz = fx dx + fy dy

or
$$dz = \frac{\partial f}{\partial x}~dx + \frac{\partial f}{\partial y}~dy$$