http://calc3.askbot.com/questions/Open source question and answer forum written in Python and DjangoenCopyright Askbot, 2010-2011.Thu, 31 Jul 2014 11:05:02 -0500http://calc3.askbot.com/question/211/partial-derivatives/When showing that $\vec F$ is conservative, we take the partial derivatives, just wondering if that is the same as $\nabla f $? I know from there we find the function $f$ to make them equal? But I thought that the gradient was the partial derivatives?TiffanyThu, 31 Jul 2014 11:05:02 -0500http://calc3.askbot.com/question/211/http://calc3.askbot.com/question/210/volume-integrals/I was wondering if when we are setting up volume integrals, if we always integrate as 1? then $\delta \rho \delta \phi \delta \theta$ ? or do we use the function, and then if it ends up being 1, we go with that? For example number 5 on the review sheet for the final exam. It says "Let S denote the unit sphere. Evaluate $$\iiint_S (x^2 + y^2 + z^2) dV$$ " So I was wondering if we set it up as... $$ \int _0 ^{2\pi} \int _0 ^\pi \int _0^1 \rho^2* \rho^2 \sin \phi \delta \rho \delta \phi \delta \theta $$ or $$ \int _0 ^ {2\pi} \int _0 ^\pi \int _0^1 1 * \rho^2 \sin \phi \delta \rho \delta \phi \delta \theta $$TiffanyThu, 31 Jul 2014 11:01:05 -0500http://calc3.askbot.com/question/210/http://calc3.askbot.com/question/208/final-review/I feel like problem number 2 on the exam review should be easy, and I'm just making it hard for no reason. The problem states "Find the volume in the first octant (x, y, z all positive) and under the graph of the function $f(x,y)=4-(x^2 + y^2).$" With it being in the first octant, $\delta \theta$ should go from 0 to $2\pi$ since we need 1/4 of a circle. And r would go from 0 to 4? This part i'm not exactly sure of, I know that $(x^2 + y^2)$ is a circle, I'm not exactly sure what the 4- goes to, and graphing it on wolfram alpha didn't help much. It looks like it extends it up to 4 in the z direction? So this may be where my original mistake is. So i'm going to work this two ways. The first with $0\leq r \leq 4$ and the second being $ 0\leq r \leq 1$ With my reasoning behind one, because of the $x^2 + y^2$ denoting the unit circle. $$ \int _0 ^{\frac \pi 2} \int _0 ^4 4-r^2 r \delta r \delta \theta $$ $$ = \frac \pi 2 \int _0 ^4 4-r^3 \delta r$$ $$ = \frac \pi 2 * 4r- \frac 14 r^4 \bigg |_0 ^4 $$ $$ = \frac \pi 2 * 16-64$$ $$ = \frac \pi 2 * -48 $$ $$= \frac {-48\pi}2$$ $$ = -24\pi$$ and the second being $$ \int _0 ^{\frac \pi 2} \int _0 ^1 4-r^2 r \delta r \delta \theta $$ $$ = \frac \pi 2 \int _0 ^1 4-r^3 \delta r$$ $$ = \frac \pi 2 * 4r- \frac 14 r^4 \bigg |_0 ^1 $$ $$ = \frac \pi 2 * 4-\frac 14$$ $$ = \frac \pi 2 * \frac{15}4 $$ $$= \frac {15\pi}8$$ I'm not sure if either of these are right. So if not, if someone could explain the easiest way to think about this?TiffanyWed, 30 Jul 2014 20:48:12 -0500http://calc3.askbot.com/question/208/http://calc3.askbot.com/question/198/quiz-3-question-1/I'm having an issue on quiz 3, question 1. What I got was.. $$\int _0 ^{2\pi} \int _0^1 e^{-r^2} $$ $$= \int _0 ^{2\pi} e ^{-r^2} r \delta \theta \bigg |_0 ^1$$ $$= \int _0 ^{2\pi} e^{-1} \delta \theta $$ $$= 2\pi e^{-1} $$ Which I know isn't right, but I'm not sure where exactly I went wrong?TiffanyWed, 30 Jul 2014 09:17:18 -0500http://calc3.askbot.com/question/198/http://calc3.askbot.com/question/183/setting-up-spherical-integrals/I was wondering if anybody had an easily explainable way of how they choose the integration terms for spherical integrals. I know in class he was saying to use you as the point on z? So does that mean that the z term will always go from 0 to a certain number? For example in the last question on the quiz today, would the z integral have gone from 0 to the top function? or from the bottom of the function he gave, back to the top? ** Sorry I can't remember the question exactly** I feel like my thinking is wrong on how to come up with the terms, I thought your $\rho$ integral would go from the bottom function to the top, the $\phi$ integral would range from 0 to $\pi$ depending on how much of the sphere you want, and then the $\theta$ integral would range from 0 to $2\pi$ again ranging on how much of the sphere you are looking at. So I guess if anybody can remember how they went about question number 3 and could help me out that would be fantastic! :)TiffanyFri, 25 Jul 2014 13:45:28 -0500http://calc3.askbot.com/question/183/http://calc3.askbot.com/question/126/partial-derivative/So with the function $$f(x,y)=\frac{x^{2}-y}{y^{2}-x}$$ I don't like the quotient rule because it seems like it makes way more sense to just find$$\frac{\partial}{\partial x}(x^{2}-y)(y^{2}-x)^{-1}$$$$=2x(y^{2}-x)+(-1)(y^{2}-x)^{-2}(-1)$$ $$=2x(y^{2}-x)+(y^{2}-x)^{-2}$$ But wolfram alpha is getting $$\frac{x^{2} -2xy^{2}+y}{(x-y^{2})^{2}}$$ I've tried jumping through a bunch of algebra hoops to see if my answer can be simplified to its but I can't pull it off. Is my answer somehow wrong or is Wolfram doing some weird simplification that I can't do?SpaceManSpiffThu, 17 Jul 2014 12:21:25 -0500http://calc3.askbot.com/question/126/http://calc3.askbot.com/question/122/partial-derivatives/So I'm working on the exam review, and I'm looking at question 3 part a. I run into issues on my partial derivatives when they include e. I ended up with $\frac {\delta f}{\delta x} = e^{xy} + xye^{xy}$ and $\frac {\delta f}{\delta y} = x^2e^{xy}$ I don't feel like thats right though, and I can't get mathematica or wolfram alpha to give me answers. Any help would be greatly appreciated! TiffanyThu, 17 Jul 2014 11:15:53 -0500http://calc3.askbot.com/question/122/http://calc3.askbot.com/question/118/double-integrals/So with the problems he put on the board at the end of class today, I worked out the first double integral and was just hoping someone could verify if I'm doing it right. So he gave us the domain :  and the double integral as: $$\int\int x^2y\delta A$$ for which I came up with the integrals being: $$\int _0 ^2 \int _0 ^{4-x^2} x^2y \delta y \delta x$$ $$=\int _0 ^2 \frac 12x^2y^2\delta x \mid _0 ^{4-x^2}$$ $$=\int _0 ^2 \frac 12 x^2 (4-x^2)^2 \delta x$$ $$= \int _0 ^2 \frac 12 x^2(16-8x^2+x^4)\delta x$$ $$=\int _0 ^2 8x^2 -4x^4 + \frac 12 x^6 \delta x$$ $$=\frac 83 x^3 - \frac 45 x^5 + \frac {1}{14} x^7 \mid _0 ^2$$ $$= \frac 83 (8) - \frac 45(32) + \frac{1}{14}(128)$$ $$= \frac{64}{3} -\frac{128}{5} +\frac{128}{14}$$ This method seems to make sense to me(hoping that my terms of integration are correct), so I'm just hoping that someone else follows this too, or you could explain and easier way to do it. Thanks!TiffanyThu, 17 Jul 2014 10:21:11 -0500http://calc3.askbot.com/question/118/http://calc3.askbot.com/question/79/tangent-planes-level-curves-level-surfaces/I understand how to plug things into the tangent plane lines, and to get the equations, however in class he said that the answers to a and b from part two should be the same, but mine ended up different. We'll start with part a. $$f(x,y)=x^3y^2$$ $$f_x = 3x^2y^2 \quad f_y = 2yx^3$$ $$p_0 = \langle 1,2\rangle$$ $$ f(x_0,x_0)= 1^3(2^2) = 4$$ $$f_x(x_0,y_0)= 3(1^2)(2^2) = 12$$ $$ f_y(x_0,y_0)= 2(2)(1^3) = 4$$ $$ L(x,y) = f(x_0,y_0) + f_x(x_0,y_0)(x-x_0) + f_y(x_0,y_0)(y-y_0)$$ $$L(x,y) = 4 + 12(x-1) + 4(y-2)$$ for part b: $$F(x,y,z)=z-x^3y^2$$ $$p_0=\langle1,2,4\rangle$$ $$F_x= -3x^2y^2 \quad F_y= -2yx^3 \quad F_z=1$$ $$ f_x(x_0,y_0,z_0)(x-x_0)+f_y(x_0,y_0,z_0)(y-y_0)+f_z(x_0,y_0,z_0)(z-z_0)=0$$ $$f_x(x_0,y_0,z_0)= -3(1^2)(2^2)=-12$$ $$f_y(x_0,y_0,z_0)= -2(2)(1^3)= -4$$ So: $$-12(x-1)-4(y-2)+(z-4) =0$$ I know that if you set z=0 and then reverse the side of the equations you can end up with the same equation. Is that what he meant by it should be the same equation? Or did I screw up somewhere along the way. Also I know he explained the level surface and level curve. From what I got, a level curve is a 2D, and a level surface is a 3D? and basically to find those curves and surfaces you make sure all of the variables are on one side of the equation and set it equal to a constant? If someone could clarify that for me it would be greatly appreciated! :) *Comment*: Your answers *are* the same! To this totally clearly, you've just got to set $L(x,y)=z$ in your first (since that's how function values are interpreted geometrically) and then isolate the $z$ on one side in your second answer.TiffanyThu, 10 Jul 2014 09:55:49 -0500http://calc3.askbot.com/question/79/http://calc3.askbot.com/question/95/can-we-write-used-wolfram-alpha-as-work/I've been struggling to solve a system of 3 equations (for the optimization hw) for the past 2 hours, and I just can't seem to crack it. I've found every other critical point I need to, and this is essentially the last part I need. I have even used Wolfram Alpha to solve the equations, and have verified that the answers Wolfram Alpha gives are indeed the correct answers, and the solution to the system of equations that acts as a critical point, and my Absolute Minimum. If I write the before and after, and verify that it works, can I just put down "Used Wolfram Alpha" as my inbetween steps? The assignment sheet does say we can use a computer to solve equations...I'm just worried my solution won't be "clearly written and understandable by my fellow classmates." WesTue, 15 Jul 2014 11:50:07 -0500http://calc3.askbot.com/question/95/http://calc3.askbot.com/question/92/mathematica/So I'm trying to make a mathematica program that let's me manipulate the components and position of a directional vector and show the value of the directional derivative. Right now I'm still trying to get it to let me manipulate a directional vector on a plot. I tried to get the directional vector on the plot using Show[Plot3D[ x + y, {x, -10, 10}, {y, -10, 10}, ColorFunction -> "Rainbow", Mesh -> None], Manipulate[ Graphics3D[Arrow[{{u, v, u + v}, {a, b, c}}]], {u, -10, 10}, {v, -10, 10}, {a, -10, 10}, {b, -10, 10}, {c, -10, 10}], PlotRange -> All, AxesLabel -> {x, y, z}, ImageSize -> {400, 400}] Mathematica is fine with both of the graphics separately but when I use the Show command to put them in the same image I get an error message that says one of my options was not formatted as a rule and that skeleton is not a graphics3D primitive or derivative. Does anyone know what these messages are trying to tell me? **Update:Made it work. Everyone ignore this question.**SpaceManSpiffFri, 11 Jul 2014 15:29:17 -0500http://calc3.askbot.com/question/92/http://calc3.askbot.com/question/72/chain-rule/So I've just now been working on 14.4 the chain rule (late, I know). And I'm having some issues with 3 and 4. I understand the chain rule and how it works, and got 1 and 2 with no issues. My issue I'm running into now has to do with it being with respect to s and t now instead of x and y. So the problem states to use the chain rule to solve for $\frac {\partial z} { \partial s}$ and $\frac{\partial z}{\partial t}$ $$z = x^2y \quad x=\sin(st) \quad y= t^2+s^2$$ So I guess here is where I get confused. Do you replace x and y with their respective equations, then take the partial derivatives from there? Or is it some sort of combination of the original z equation plus the derivitavies of x and y separrately. The book gets the answer..$$ 2xyt\cos(st) + 2x2s, \quad 2xys\cos(st) +2x2t$$ So I'm lost, any help would be greatly appreciated! Thanks!TiffanyWed, 09 Jul 2014 20:54:50 -0500http://calc3.askbot.com/question/72/