Uni Maths Questions

[M-D]

find all the solutions for

here is what i have done:









, k is an element of z






these are the solutions in polar form (are they correct?).also to convert them to Cartesian form i use the Argand diagram, however, what i don't understand is that here the modulus is 2 and -2 but the modulus must always be positive. i really don't know what to do   

[brightsky]

not exactly sure what you did there, but here's how i would go about it:

z^8 = 256
z^8 = 256 cis(0)
z = (256 cis (2kpi))^(1/8)
= 2 cis (2kpi/8)
= 2 cis(kpi/4)
so you want your argument to be in (-pi,pi]
-pi<kpi/4 =< pi
-1 < k/4 =< 1
-4 < k/4 =< 4
so k = -3, -2, -1, 0, 1, 2, 3, 4
so z = 2cis(-3pi/4), 2cis(-pi/2), 2cis(-pi/4), 2cis(0), 2cis(pi/4), 2cis(pi/2), 2cis(3pi/4), 2cis(pi)

if this is uni maths, then just replace rcis(t) with r*e^(i*t). but yeah, modulus = distance from origin so it is strictly non-negative.

[kamil9876]

Thanks Mao!

Could someone help me with this question as well?

Without having to find the linear transformation matrix, how do I calculate the dimensions of the image and kernel for the linear transformation:

R: R^3->R^3 by reflection in the plane x+y+z=1 ?

So I know that dim(Ker) = nullity of R and dim(Im) = rank of R but I don't know whether these facts are of any use in this question?

But this transformation is not linear. Is it supposed to be x+y+z=0 perhaps?

[Deleted User]

Hmm... Not Sure. It's from an exam paper and it says that it's x+y+z=1.

[M-D]

could someone please help with integrating this. i have done as much as i could.



after applying long divison i get:

i don't know to continue from here with the second integral. the first is easy. In the answers their is    so i think the numerator and denominatos would have to be changed somehow. please help. thanks

[brightsky]

split the second integral up. (2x+1)/(x^2+10) = 2x/(x^2+10) + 1/(x^2+10). you can integrate the first fraction using substitution (you will get ln(something)). the arctan arises when you integrate the second fraction.

[M-D]

thanks brightsky.

[Deleted User]

Hey could someone help me out with this question?

Find the point closest to the origin (0,0,0) on the surface given by the graph of the function z=x^2 - y + 1.

No idea where to start...

[b^3]

I feel like I'm about to go about doing this the long way, but anyways.
The distance between the origin and any point on the surface is given by

Now we want to minimize that, which we can achieve by minimising whats under the square root, so lets let .
Now to find a minimum, we will take the partial derivatives with respect to and and set to zero. (For there to be a local minimum, the partial derivatives in both directions has to be zero).

So the only solution we have so far is , which makes our partial derivative with respect to zero, but we still need   to be zero as well, so we can substitute into [2].

We probably should verify that this distance is a minimum (just for the sake of it )

As and , is a local minimum.
That is the point on the surface that is closest to the origin is

[Deleted User]

Thanks!!!

[Deleted User]

Another question if you guys don't mind.

Consider C[-1,1] = space of continuous functions defined on [-1,1] with the inner product
<f(x),g(x)> = antidiff (f(x)*g(x) dx) from -1 to 1.

Use the Gram-Schmidt procedure to find an orthonormal basis for the subspace of C[1,1] spanned by 1 and x^2.

I'm used to dealing with vectors but not so much with functions so could you guys show me how to find the first element of the basis?

u1 = v1/magnitude of v1 so if v1 = 1, wouldn't u1 just be = 1? Why is the answer u1 = 1/sqrt(2)?

Thanks

[mark_alec]

Because <v1,v1> = |v1|^2 and its value is equal to 2.

[M-D]

i have a question which say: solve the following separable differential equations whose right hand side depends only on .

1)

2)

how would i go about solving this. it does not really look like a separable diff. equation because it is only with respect to . thanks

[Alwin]

i have a question which say: solve the following separable differential equations whose right hand side depends only on .

1)

2)

how would i go about solving this. it does not really look like a separable diff. equation because it is only with respect to . thanks

Hint:



So you can 'flip' both sides and then you can integrate in terms of y

[M-D]

thank you very much Alwin. I have another question relating to implicit differentiation and points where the tangent line to a curve is vertical. Here's the question:

Consider the curve given by the equation:

(a) Find the equation of the tangent line to the curve at the point

(b) Find all the points on the curve where the tangent line is vertical.

Solution to (a)



let and



equation of the line is

how should go about part (b)