When substituting for x I have used both r sint and r cost and both seem to work. However when using what would at first seem to be obviously correct x=rcost leads to a negative integrand and 0 as an upper bound of integration. My question is that is this the only reason why rsint is chosen as an arbitrary substitution.

-Aron ]]>

Anyway, any help would be appreciated! Thanks in advance for your time :)

I have been given this assignment to do question 9 and still haven't been able to complete it after hours of research and attempts to tackle this. Any form of help would be grateful, take your best shot to answer it.

Best Regards,

Lilshrekboye29

Hey there,

I have tried this question and I seem to be completely stuck.

Could somebody please help in solving this?

I am getting two different solutions. When integrating

4I(0 pi/2)I(0 acost)I(0 (a^2-r^2)^(1/2)) r dz dr dt = 1.205

(correct in the solutions manual)

I(-pi/2 pi/2)I(0 acost)I(-(a^2-r^2)^1/2 (a^2-r^2)1/2) r dz dr dt = 2.09

(incorrect in the solutions manual)

if if this is true. I would like to know why it is incorrect to integrate through the entire region. And correct to integrate through the sub region and multiply by 4. ]]>

Suppose that x thousand units of a particular commodity are sold each month when the price is p dollars per unit, where p(x) = 5(24-x).

(a) Let E be the total monthly consumer expenditure, that is, the total amount of money consumers spend on this commodity in one month. Find a formula expressing E as a function of the price per unit.

(b) Sketch the graph of the function E(p).

(c) What market price produces the greatest total monthly consumer expenditure? How many units will be sold each month at this optimal price? ]]>

[tex]

\lim_{(x,y) \to (0,0)}\frac{x^{2}+x|y|}{\sqrt{x^{2}+y^{2}}}=0

[/tex]

I have no idea about how to proceed, I was thinking about using the squeeze theorem but the non-absolute x in the numerator apparently makes it a void attempt:

[tex]

\lim_{(x,y) \to (0,0)}\frac{x^{2}}{\sqrt{x^{2}+y^{2}}} + \frac{x|y|}{\sqrt{x^{2}+y^{2}}}

[/tex]

1st limit:

[tex]

\frac{x^{2}}{\sqrt{x^{2}+y^{2}}} \leq \frac{x^{2} + y^{2}}{\sqrt{x^{2}+y^{2}}} =\sqrt{x^{2}+y^{2}}

[/tex]

So by the squeeze theorem, the 1st limit tends to 0; as for the 2nd, the same method doesn't apply:

[tex]

\frac{x|y|}{\sqrt{x^{2}+y^{2}}}

= \frac{x\sqrt{y^{2}}}{\sqrt{x^{2}+y^{2}}}

[/tex]

Thanks in advance to whoever can help ]]>

Would like a little help on this problem here if thats ok?

Your help is greatly appreciated!

20190115_202340.jpg

I am stuck on that :/ I know differentiate part but have problem with graphs :/ Thanks.

I always seem to have good luck coming here for help so here goes!

Here is a picture of the question and diagram set up first for clarity.

Problem 2-135.jpg

The basic solution involves finding the position vector of AB.

R_AB = (Bx-Ax)i + (By-Ay)j + (Bz-Az)k

R_AB = -1.5i + 3j +1k

Unit_Vector of R_AB is just the R_AB / Magnitude_R_AB

Magnitude of R_AB = sqrt(12.25) = 3.5

Unit_R_AB = (-1.5)/3.5 i + 3/3.5 j + 1/3.5 k

Now to solve for F_parallel I need to project the Force Vector onto Unit_R_AB

I'm having difficulty setting up the cartesian coordinates for the Force vector though.

I'm trying to conceptualize how the Force Vector takes the form given in the study material.

The force vector's cartesian form is [-90*cos(60)*cos(45)]i + [90*cos(60)*sin(45)]j + [90*sin(60)]k

I understand the k component because its 60 degrees from the z axis.

I'm trying to understand why the i and j components are multiplied by two trig angles though. I'm probably missing something very basic and I would appreciate help in understanding this better.

I don't want to be that guy who just memorizes formulas just for the test!

The solution after solving the parallel force is simple arithmetic to solve the perpendicular force and I understand that part of the problem.

Thank you again for the help!

And what is the result of b), I don't get it solved.

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Thanks.

I have a little problem that I could use a little help to figure out.

The assignment is to find possible extreme points and classify them according to the second order condition.

The function is:

f(x)=ln(x)-3x^2-6x+10

Thanks in advance! ]]>

Would like a little help on this problem here if thats ok?

Your help is greatly appreciated!

Screen Shot 2019-01-11 at 1.57.30 PM.png Screen Shot 2019-01-11 at 1.55.27 PM.jpg

I have worked out the equation for the parabola:

x^2 = -28(y-8.5)

However, I am puzzled on how to move on from that point onwards.

Again, it's another study I am stuck on!

Unfortunately I was only given one example on this particular equation (especially a negative quadratic equation), but it doesn't suit the problem I am trying to solve on this one.

Here is that example:

Screen Shot 2019-01-11 at 2.07.51 PM.jpg

Screen Shot 2019-01-11 at 2.08.07 PM.jpgScreen Shot 2019-01-11 at 2.08.20 PM.jpg

I am unsure as to what I should do next. I am familiar with factorising and squaring but only worked on normal parabolas that already have their equations given. (The canvas is blank on this round!)

I feel that I need to substitute 6 metres somehow to give me an answer. I was trying to find a way to calculate everything that I have learnt but it doesn't seem to give me the answer I am looking for.

Trying not to rely too much on the desmos calculator but maybe I might need to for an answer.

Who knows? Would love your input on this. Cheers! :smile: