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But if we take that approach, then we can try:
So which answer is a correct one???
No, you're integrating r with respect to theta. You instead need to integrate:
.Your upper limit of integration for 3cosθ should be pi/2, not pi. (For which θ does 3cosθ vanish? Which values does r = 3cosθ take for θ between pi/2 and pi?)
Your method should then be fine, provided you use the square of r, and double your result at the end, noting the symmetry of both shapes.
find the distance between the line r(t)= (1+t, 2-2t, -1+4t) and point (1,0,1)
With the above problem, I attempted it and thought I had a good idea of how to solve it, but the anwer I'm getting is different from the one that they've given.
Could someone help me find where i went wrong?so, I calculated the distance between these two as
(1+t-1, 2-2t-0, -1+4t-1) = (t, 2-2t, 2+4t)and then to find the value of t, from the equation of the line, i took d= (1, -2, 4)
and since these two are orthogonal, calculated the dot product of (t, 2-2t, 2+4t) and (1, -2, 4) and solved for t to get t=-4/21
and then from that length of t as (-4/21)^2 and squared it to get t=4/21
Check the bolded.
What does 'SVD' stand for?
So many banned posters in this thread... (or rather, poster)
Hi;
Thanks bobbym and zetafunc although I do not entirely understand that.You're welcome ganesh. Now, will we leave them as they are or a little editing will do?
You are welcome to create a thread in Help Me if you need any more help -- we'll do our best to respond to any troubles you might have.
Hi bobbym; I realize from experimentation that your answer T#76 is absolutely right. But I'm not able to simplify and get your answer. May I know the approach you used to arrive at your response? Or anyone else can help? Many thanks.
m9m
Ah, so \prime is the right one to use... and to think all this time I've been deliberately trying to avoid apostrophes in LaTeX!
If you are reflecting in the x-axis, then yes.
However, using a different axis of reflection will give you a different 'mirror image'. For instance, if you were reflecting in the y-axis instead, the 'mirror image' of (6,1) would be (-6, 1).
This is really just an application of diagonalisation. Let's consider the following system:
Suppose that
and that
.Then, if ' denotes differentation with respect to t, we write x' = Ax. We now make a change of variable. Put x = Py, where P is invertible, and:
and
Then, differentiating with respect to t, x' = Py' so that the system x' = Ax reduces to Py' = APy. That is:
Py' = APy => y' = P⁻¹APy.
Does P⁻¹AP look familiar to you?
What are U(16) and X16?
Not a problem. Sometimes the best way to become unstuck on a maths problem is to simply do something else.
Use the substitution
.Yes, all complex numbers can be represented in the form a + bi, where a, b are real numbers. With them, we can define addition and multiplication as follows:
Some examples:
We also have some other properties, such as conjugation:
There exist even more interesting algebraic structures, such as the quaternions (which are not commutative) and the octonions (which are not associative!).
For the purposes of your calculations, you most likely only need to know that i² = -1: that should suffice.
Do you know what a complex number is?
Yes, that's correct. In fact, you already completed the 'proof' under your 'solving' heading -- just do everything backwards.
See the quadratic formula.
Use the Euclidean algorithm.
I'm not sure what you're getting at. The error function and e^(x^2) are not the same function.
Yes, that's fine. When you square, you introduced the other solution, which does not satisfy the original equation -- which is why it's always important to check that your final result makes logical sense. (You sometimes also see this type of thing when doing kinematics problems -- you may end up getting a 'negative time' through solving a quadratic, which usually does have some physical meaning, but may be irrelevant to your problem.)
It would be best to consult a table of Laplace Transforms.
Your first can be split using partial fractions, and then using the linearity of the Laplace transform.
For the second, what is w? If w is a constant, then you can read this result off directly from the link above.
If you're not permitted to use a table, then you can compute the inverses directly using the formula:
where gamma = Re(s), constructed so that gamma is always greater than every singularity of F(s) (this requires some knowledge of complex analysis). However, the nature of your problem seems to indicate that they want you to use a table of Laplace transforms.
I don't understand what you're asking. Is this related to a problem you're working on? If so, it would help if you posted the problem in full detail, or provided a bit more information.
I can't do anything more for you without additional information about a,b. Is this what you were looking for?
No, you're not distributing the minus sign over the bracket.
Your problem is simply dealing with the brackets (and it seems you made a small error in the formula).
i.e.