1. ## Vector Calculus.

If A = t2i - tj + (2t + 1)k and B = (2t 3)i + j - tk, evaluate at t = 1,

(i)
d/dt(A x B)

(ii)
d/dt(A xdB/dt)

i, j, k are vectors

It's the working I need help with. I get confused when it comes to differentiating the vectors when there's no scalar.

Thanks.

2. Am I right in assuming this is a cross product? if so, you need to use the formula:

and then differentiate the remaining function with respect to t. so for example, when multiplied the i vector becomes 2ti and hence, when differentiated becomes 2i. then repeat using the formula for j and k above and differentiate each term (remember that e1 is i, e2 is j ad e3 is k). this should give you the answer to the first part of the question.

For the second part, differentiate B : so B= 6t^2i - k

Then multiply this by A using the formula above. And then differentiate each term as in the first part.

Does this make sense? I have deliberately left the answer out so that you can have a go but if it gets difficult I can give you more detailed working

3. We use matrices for cross product , but that way seems a lot easier. I've done them both out, but I've got answers that don't look right.

For the first one I've got -i+(9t^2-2t-6)j+(3t^2+6t-3)k

What really gets me is, say for example I have 2ti+j-k (which is what i got for dB/dt) and i have to differentiate it, I don't know do I get 2i as the answer or 2i+j-k?

4. Originally Posted by Daithi
We use matrices for cross product , but that way seems a lot easier. I've done them both out, but I've got answers that don't look right.

For the first one I've got -i+(9t^2-2t-6)j+(3t^2+6t-3)k

What really gets me is, say for example I have 2ti+j-k (which is what i got for dB/dt) and i have to differentiate it, I don't know do I get 2i as the answer or 2i+j-k?
You're expected to know that the derivative with respect to t of $\displaystyle a_1(t) i + a_2(t) j + a_3(t) k$ is $\displaystyle \frac{d a_1}{dt} i + \frac{d a_2}{dt} j + \frac{d a_3}{dt} k$ ...... And you know the derivative of a constant is zero, right?

5. So for example

d/dx(2xi+3x^2j+2k)
=
2i+6xj+0k
=
2i+6xj

Is that correct?

6. Originally Posted by Daithi
So for example

d/dx(2xi+3x^2j+2k)
=
2i+6xj+0k
=
2i+6xj

Is that correct?
Yes.

7. Originally Posted by Daithi
We use matrices for cross product , but that way seems a lot easier.
Hi, that formula is what you get from calculating the determinant of the matrix containing unit vectors, i,j,k, and components of your vectors, that is:

$\displaystyle \left| \begin{array}{ccc} \bf{i} & \bf{j} & \bf{k} \\ a_x & a_y & a_z \\ b_x & b_y & b_z \end{array} \right| = \bf{i} \left| \begin{array}{cc} a_y & a_z \\ b_y & b_z \end{array} \right| - \bf{j} \left| \begin{array}{cc} a_x & a_z \\ b_x & b_z \end{array} \right| + \bf{k} \left| \begin{array} {cl} a_x & a_y \\ b_x & b_y \end{array} \right|$

$\displaystyle =\bf{i}\ (a_yb_z-a_zb_y)-\bf{j}\ (a_xb_z-a_zb_x)+\bf{k}\ (a_xb_y-a_yb_x)$

(except only i,j,k should display in bold on that last part, instead of all of it, but I can't get the latex to work properly )