Mark's Calc III - Individual question feed

How can I include typeset mathematics in my posts?

Mark — Sun, 22 Jun 2014 18:56:58 -0500

Many questions I see have beautifully typeset mathematics. For example, instead of seeing something like > integrate from -oo to oo e^(-x^2) w/resp to x some users input a beautifully typeset version of this. How can I include such typeset mathematics in my own posts? $$\lim_{\stackrel{h\rightarrow0}{h\in\mathbb R}}$$

Final Review Number one

asmith14 — Thu, 31 Jul 2014 17:38:09 -0500

Let D denote the solid pyramid with vertices located at (4,0,0), (0,1,0), (0,0,2), and the origin. Set up $\iiint_{D} f(x,y,z)dV$. So when we were talking about this in class we drew the graph first and then set up the equation ax+by+cz=d and then d was substituted for 4 because it was the least common multiple of the points. My question is how were the other numbers found in the equation which made it x+4y+2z=4 ?

Partial derivatives

Tiffany — Thu, 31 Jul 2014 11:05:02 -0500

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?

Volume integrals

Tiffany — Thu, 31 Jul 2014 11:01:05 -0500

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 $$ \int _0 ^ {2\pi} \int _0 ^\pi \int _0^1 1 * \rho^2 \sin \phi \delta \rho \delta \phi \delta \theta $$

Final Review

Tiffany — Wed, 30 Jul 2014 20:48:12 -0500

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?

Exam III review sheet

Anonymous — Wed, 30 Jul 2014 10:08:58 -0500

So I am completely confused on how to solve #1 on the final exam review sheet: Let $D$ denote the solid pyramid with vertices located at $(4,0,0),(0,1,0),(0,0,2)$ and the origin. Set up $$\iiint_D f(x,y,z)dV$$ in the order $dzdydx$. We did one similar to this in class but we were given an equation for the plane. I guess the real question is, how would we go about finding an equation of a plane that we could work with?

Section 16.3

Anonymous — Wed, 30 Jul 2014 07:15:50 -0500

Number 8 in section 16.3 asks, Evaluate $\int (10x^4=2xy^3)dx-3x^2y^2dy$ where $C$ is the part of the curve $x^5-5x^2y^2-7x^2=0$ from $(0,0)$ to $(3,2)$. Can someone explain how to solve something like this?

Number 6 on Exam 2

asmith14 — Mon, 28 Jul 2014 14:47:43 -0500

I have a question on number six from the last test because I am unable to figure out what I did wrong. The question was to use Lagrange multipliers to find the extremes of $f(x,y)=x+y^2$ subject to the constraint $x^2+y^2=1$. Can someone please offer any help as to where I went wrong or any tips that might help for the final. I do not know how to draw the graph on here but this is my work for part C. $$\nabla f=<1,2y>$$ $$\nabla g=<2x,2y>$$ then from those I got $$1=2\lambda x$$ $$2y=2\lambda y$$ $$x^2+y^2=1$$ and solved and found $$x=1/2\lambda$$ $$y=\pm \sqrt {3/4}$$ $$\lambda =1$$ then I put those together to find my critical points to be $$(1/2,\sqrt {3/4)}:{max}$$ $$(1/2,-\sqrt {3/4)}:{min}$$. Please let me know if you have any ideas I to what I should of done and how to fix it.

quiz 3 question 1

Tiffany — Wed, 30 Jul 2014 09:17:18 -0500

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?

Review for final Exam

asmith14 — Tue, 29 Jul 2014 18:56:25 -0500

I was looking back over Exam 1 and question number ten is still giving me a problem. Can someone please explain the right path that I should have taken to get the right answer? The question was Let $p(t)=<1+3t, 2+4t>$ and let $q(t)=<3-t, -4+3t>$. Find the point of intersection between the paths parametrized by p and q. What I attempted was to solve for u and t $$1=3t+3-u \rightarrow -2+3t=-u \rightarrow (2-3t)=u$$ $$2+4t=-4-3u$$ $$2+4t=-4-3(2-3t)$$ $$6+4t=-6+9t$$ $$12=5t$$ $$t=12/5$$ but then I don't know what to do complete the problem after that.

Section 16.2

Anonymous — Tue, 29 Jul 2014 10:06:31 -0500

I'm having trouble remembering how to parametrize a line when given 3 points. In section 16.2, #10, it asks to compute $\int <1/xy,1/(x+y)>\cdot dr$ along the path from (1,1) to (3,1) to (3,6) using straight line segments. I'm really lost on what to do here.

Line Integrals

Anonymous — Tue, 29 Jul 2014 06:58:55 -0500

I'm hoping someone can show me what I am doing wrong with this. In section 16.2, question #3, it asks you to compute $\int (z\cos(xy))ds$ along the line segment from $(1,0,1)$ to $(2,2,3).$ Here is my calculations: $$P(t)=<1,0,1>+$$ $$\int_1^2 (1+2t)\cos(2t+2t^2)\sqrt{1^2+2^2+2^2}dt$$ $$u=2t+2t^2$$ $$\frac{du}{2(1+2t)}=dt$$ $$3\int_1^2 (\cos(u)(1+2t)\frac{1}{2(1+2t)})du$$ $$\frac{3}{2}\int_1^2 (\cos(u))du$$ $$\frac{3}{2}\sin(u)\biggr|_1^2$$ $$\frac{3}{2}\sin(2)-\frac{3}{2}\sin(1)$$ where did I go wrong?

How do I get the bounds of integration for line integrals?

Christina — Mon, 28 Jul 2014 12:06:32 -0500

I am currently working on the homework for 16.2. Number one, for example, states: **"Compute $\int\limits_Cxy^2ds$ along the line segment from (1,2,0) to (2,1,3)"**. I am setting it up as follows: For my parameterized line I get: $$\vec{p}(t) = \langle1,2,0\rangle + t\langle1,-1,3\rangle$$ Giving me: $$x=t+1$$ $$ x' = 1$$ $$y=2-t$$ $$ y'= -1$$ $$z=3t$$ $$ z'= 3$$ Then setting up an integral to compute: $$\int(t+1)(2-t)^2\sqrt{(1)^2+(-1)^2+(3)^2}dt$$ What I am not sure of is how to know what my bounds of integration are. Do I plug my point values in for x, y, and z? Or in this case, for x and y and then solve for t? Or just for x? I would love any help on this or to know if what I have done so far is correct or not correct...Thanks!

Question #2 on quiz

Anonymous — Fri, 25 Jul 2014 09:05:21 -0500

I don't remember the exact wording of question #2 on the quiz but I think it asked what the area above the curve $z=-\sqrt{x^2+y^2}$ and below $z=\sqrt{x^2+y^2}$ inside the cylinder $x^2+y^2=1$; evaluate $\int\int\int (z) dV$. My answer came out to be zero but I don't think that is correct. If anybody understands this problem or has any ideas, please post them.

Setting up spherical integrals

Tiffany — Fri, 25 Jul 2014 13:45:28 -0500

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! :)

Setting up exponential function in Cartesian coordinates

Christina — Thu, 24 Jul 2014 14:52:49 -0500

I am hoping for some help looking at the exponential function in Cartesian land. If we are given $e^{-(x^2+y^2)}$ and asked to set this up over the domain of a disk of radius R both in polar and Cartesian coordinates, is this what polar would look like? $$\int_0^{2\pi}\int_0^R e^{-(r^2)}rdrd\theta$$ In turn, is this what the Cartesian set-up would look like? $$\int_{-R}^R\int_0^\sqrt{R-x^2} e^{-(x^2+y^2)} dydx$$ Any help or comments would be great. Thanks! **COMMENT** - Thanks Gear Junky, that makes sense. I am struggling to remember to visualize the projection onto the xy plane. As far as evaluating it, I would definitely evaluate the polar coordinates. I was just practicing because he said we may have a problem like this on the quiz to set up in both but evaluate one.

Can Anyone Validate Problem #3

Gear Junky — Thu, 24 Jul 2014 21:30:49 -0500

So for problem #3 on the In-Class worksheet I worked it through, but didn't have enough time to validate my answer with the rest of my group. I approached it as: $$\int_0^{2\pi} \int_0^{\sqrt{\pi/2}} \int_0^{\cos(r^2)} Z \delta z r \delta r \delta \theta$$ $$=2\pi \int_0^{\sqrt{\pi/2}} \int_0^{\cos(r^2)} Z \delta z r \delta r$$ $$=2\pi \int_0^{\sqrt{\pi/2}} \frac{\cos^2(r^2)}{2} r \delta r$$ And by the Double angle formula I got: $$=\frac{\pi}{2}\int_0^{\sqrt{\pi/2}}r(1+\cos(2r^2)\delta r$$ Using U-sub: $$u=2{r^2}$$ $$\delta u=4r \delta r$$ So I ended up with: $$=\frac{\pi}{8} \int_0^\pi 1+\cos(u) \delta u$$ $$=\frac{\pi}{8}(u+\sin(u)) |_0^\pi$$ Getting an answer of: $$\frac{\pi^2}{8}$$ If anyone got a different answer or can see an error in my math I would greatly appreciate your input.

Quiz question

asmith14 — Thu, 24 Jul 2014 18:08:34 -0500

This maybe a simple question but can someone please help me with one of the problems that was said to be similar on the quiz tomorrow? Set up the Cartesian plane for $\iint\limits_{Disk} e^{-(x^2+y^2)} dA$

Graphical explanation

Anonymous — Thu, 24 Jul 2014 10:40:31 -0500

Can someone give a graphical explanation (pictures please) as to how you can find the answer to question 1b of todays worksheet? $$\int_0^1\int_0^{\sqrt{1-x^2}}\int_0^{\sqrt{1-x^2-y^2}}1dzdydx$$ No one in my group could really grasp what McClure was saying in class.

Setting up an integral for the volume under a plane

Anonymous — Thu, 24 Jul 2014 10:14:06 -0500

I just want to know if I am doing this correctly. On the worksheet from today, number 6 asks: Set up an integral representing the volume under the plane $x+2y+z=2$ and in the first octant. I came up with the following: $$\int_0^2\int_0^\frac{2-x-z}{2}\int_0^{2-2y-z} dxdydz$$ am I doing this right or am I completely lost?

Quiz 1 Review

Dylan — Wed, 23 Jul 2014 20:24:27 -0500

I would also be posting quiz 2, but sadly it was lost by an anonymous individual named SpacemanSpiff. **1:** Prove that scalar multiplication of vectors is distributive over the addition of 2-D vectors. We can start by defining two arbitrary vectors, $\vec{u}=\langle u_1, u_2 \rangle$, and $\vec{v} = \langle v_1, v_2 \rangle$. Now let $a$ be a real number. Then we just add our arbitrary vectors together and multiply the resulting vector by $a$, and break apart the vectors into their components, then put them back together: $a(\vec{u} + \vec{v}) = a(\langle u_1, u_2 \rangle + \langle v_1, v_2 \rangle $ $= a\langle u_1+v_1, u_2 + v_2 \rangle $ $= \langle a(u_1+v_1), a(u_2+v_2) \rangle $ $= \langle au_1 + av_1, au_2+av_2 \rangle $ $= \langle au_1,au_2 \rangle + \langle av_1,av_2 \rangle $ $= a\langle u_1, u_2 \rangle + a \langle v_1, v_2 \rangle $ $= a\vec{u} + a\vec{v}$. And our proof is done. **2:** The intersection between the $xy$ plane and the plane $x+2y -3z = 6$ is a line. Find an $xy$ equation for that line. The key here is to see that the intersection occurs at $z=0$, we just take the equation for the plane and set $z=0$: $x+2y = 6$. This is our $xy$ equation. **3:** Find a parametrization of the line containing the point $(1,2,3)$ and perpendicular to the plane $x+2y-3z = 6$. The coefficients on $x$, $y$, and $z$ form a vector perpendicular to the plane, as we showed in class. So we just take this vector $\langle 1, 2, -3 \rangle $ as the direction vector for our line, and then use the point we want to be on the line as our $\vec{p_0}$: $\vec{l}(t) = \langle 1, 2, 3 \rangle + \langle 1, 2, -3 \rangle t $ **4:** Find all values of $t$ such that the vector $\vec{p}(t) = \langle t^2, t \rangle $ is perpendicular to the vector $\vec{q}(t) = \langle t, -1 \rangle $. Two vectors are perpendicular if and only if their dot product is equal to 0, so we can just set the dot product of these two vectors equal to 0: $\langle t^2, t \rangle \cdot \langle t, -1 \rangle = t^3 - t$ $ = t(t^2-1) = 0$ This gives us solutions where $t = 0$, $t= 1$ and $t=-1$. So for those values of $t$, the vectors are perpendicular.

In class problem #5

Anonymous — Thu, 24 Jul 2014 09:57:59 -0500

I think that I understand how to set up the domain of this integral except for the inner integral. The question asks: "Let $D$ denote the set in $\mathbb{R^3}$ lying above the cone $z=\sqrt{x^2+y^2}$ and inside the sphere $x^2+y^2+z^2=4$. Set up the following integrals over $D$ as iterated integrals in spherical coordinates. You should think about which ones you can evaluate." would this be correct?: $$\int_0^\pi \int_0^\frac{\pi}{4} \int_0^2 (\rho^2)\rho^2\sin(\phi)d\rho d\phi d\Theta$$ comment: sorry, this is for part a.

In class problem

Anonymous — Thu, 24 Jul 2014 09:42:04 -0500

I am having trouble understanding #4e from todays worksheet: "Let $D$ denote the set in $\mathbb{R^3}$ lying above the cone $z=\sqrt{x^2+y^2}$ and below the plane $z=4$. Set up the following integrals over $D$ as iterated integrals in cylindrical coordinates. You should think about which ones you can evaluate. (e) An integral representing the volume of $D$." What is the significance of setting up an integral representing the volume of $D$?

Spherical and cylindrical problems

Anonymous — Wed, 23 Jul 2014 12:03:06 -0500

I am having a lot of trouble visualizing and understanding how to set up spherical/cylindrical integrals. In particular, I am having trouble with #4 on the homework sheet. "Let $D$ denote the three-dimensional domain above the cone $z=\sqrt{x^2+y^2}$ and inside the sphere $x^2+y^2+z^2<=4$. Evaluate $\int\int\int(x^2+y^2+z^2)dV$." By looking at this, I can guess that spherical coordinates would work well because $x^2+y^2+z^2=P^2$ but past that I am lost. Please help!

Mass of region between two graphs

Anonymous — Wed, 23 Jul 2014 12:18:56 -0500

I'm having problems with #5 on the homework for Spherical and Cylindrical Problems: "Figure 1 shows a 3D domain stuck between $z=x^2+y^2$ and $z=8-(x^2+y^2)$. Find the mass of the corresponding object." so far all I have come up with is $$\int\int\int_{r^2}^{8-r^2} (r^2) dzrdrd\Theta$$ Am I on the right track at least? If so, how do I find the domain of r and $\Theta$?

In class Problem

Anonymous — Wed, 23 Jul 2014 09:41:37 -0500

In class today my group and I had a little problem solving the second problem written on the board: Evaluate $\int\int\int sin((x^2+y^2+z^2)^{3/2}) dV$ where $D$ is defined as the top half of the solid unit sphere. This is how far we got $$\int_0^\pi\int_0^\pi\int_0^1 (sin((p^2)^{3/2})p^2 sin\phi dP d\phi d\Theta$$ $$\pi\int_0^\pi\int_0^1 sin(p^3)p^2 sin\phi dP d\phi$$ $$u=p^3$$ $$1/3 du=p^2 dP$$ $$\frac{\pi}{3}\int_0^\pi\int_0^1 sin(u) sin\phi du d\phi$$ $$\frac{\pi}{3}\int_0^\pi sin\phi - cos(u) \biggr|_0^1 d\phi$$ $$\frac{\pi}{3}\int_0^\pi 1-cos(1) d\phi$$ $$\frac{\pi}{3} (\phi-\phi cos(1)) \biggr|_0^\pi$$ $$\frac{\pi}{3} (\pi - \pi cos(1) $$ Did we start with the correct domain of integration or did we make a mistake in the calculations?

Improper double integrals

Anonymous — Mon, 21 Jul 2014 10:17:19 -0500

In section 15.2, question #19a, it asks: Consider the integral $\int\int 1/\sqrt{x^2+y^2} dA$, where $D$ is the unit disk centered at the origin. a) Why might this integral be considered improper? The link to see this graph did not work for me, but I used wolfram alpha and the graph looked kind of like a piece of paper being folded upwards. I don't know why this could be considered improper? Any thoughts?

Post-test analysis

Dylan — Fri, 18 Jul 2014 15:26:28 -0500

For those of you who want to join me in trying to figure out what happened on the test, did anyone manage to solve number 5? I got a monstrous set of equations after factoring out the $\mathrm{e}^{-(x^2+y^2)}$ that went up to sixth powers of $y$ and couldn't figure out what went wrong. I'm not stressing too much about it but I'm curious how we can solve it without getting those high rank polynomial equations. (I can't quite remember what the function was).

Final exam review: Test 1

SpaceManSpiff — Sun, 20 Jul 2014 18:14:44 -0500

Hey everyone, I think it's a good time for us to start reviewing things for the final. It is a short class after all and it'll be here before we know it. I think it's a good idea for us to use ask bot as a data base for important assignments, tests and quizzes. Creating these posts is a great way for the reviewer to review an assignment and if we all chip in and post explanations of these things on the website than we'll have a very complete resource for anything we could want to go over. I'll start by covering test 1. **Note**: Completing these takes a serious amount of time. It's conceivable that two people could be working on the same assignment at the same time. So before completing one of these reviews leave an answer to this question explaining what you will cover so we can get through it in the most efficient way possible. When your post goes up use the tag I've used on this post, "DataBase" so we can have them all in one location. 1) "An object moves according to the parametrization $p(t)= \langle - \cos (5t), \sin(5t) +2t \rangle$." A) Describe the motion determined by $p(t)$. You can make the motion described by this path more apparent by breaking it into the sum of two vectors. $$p(t)= \langle - \cos(5t), \sin(5t) \rangle+ \langle0, 2t \rangle$$ The first vector in this sum describes clockwise circular motion. We know it is circular motion because the $\cos(5t)$ term and the $ \sin(5t)$ term both have the same coefficient, $1$. We know it is clockwise because the $\cos(5t)$ term is negative, which tells us that it moves from negative $-1$ to $1$ over the interval $[0, \pi]$. The second vector describes linear motion along the positive $y$ axis. So we can conclude that this is a clockwise circular path that travels "up" along the positive $y$ axis. Here's a groovy picture. ![image description](/upfiles/calc3.askbot.com/14058793798597336.png) B) Write down a parametrization of the line tangent to the path at the point when $t= \frac{\pi}{2}$ To parametrize this line, we use the formula $l(t)= p_{0}+dt$ where $p_{0}$ is a point along the line and $d$ is a direction vector for the line. First we find $p_{0}$, which is the point on the circular path where $t= \frac{\pi}{2}$. To find it all we have to do is plug this value into $p(t)$. $$p_{0}= \langle -\cos(\frac{5 \pi}{2}),\sin(\frac{5 \pi}{2})+\pi \rangle$$ $$= \langle 0,1+\pi \rangle$$ Next we find $d$. This direction vector can be found by plugging in our value of $t$ into $p'(t)$. $$p'(t)= \langle 5 \sin(5t),5 \cos(5t)+2 \rangle$$ $$d= p'(\frac{\pi}{2})= \langle 5 \sin(\frac{5 \pi}{2}),5 \cos(\frac{5 \pi}{2}+2 \rangle$$ $$= \langle 5, 2 \rangle$$ Now we have everything we need to create our parametrization. $$l(t)= \langle 0,1+ \pi \rangle + \langle 5,2 \rangle t$$ $$= \langle 5t,2t+1+\pi \rangle$$ Here's another groovy picture. ![image description](/upfiles/calc3.askbot.com/14058810325082741.png) C) "Write down an integral representing the distance traveled by the object over the time interval $0$ to $3 \pi$." We use the formula $D= \int_{a}^{b} ||p'(t)||dt$ to answer this question and recall from part B that $$p'(t)= \langle 5 \sin(5t),5 \cos(5t)+2 \rangle$$ The magnitude of $p'(t)$ is the square root of the sum of the squares of the components. Or $$||p'(t)||=[25 \sin^{2}(5t)+(25 \cos(5t)+2)^{2}]^{\frac{1}{2}}$$ So our integral is $$D= \int_{0}^{3 \pi}[25 \sin^{2}(5t)+(25 \cos(5t)+2)^{2}]^{\frac{1}{2}}dt$$ 2) "Let $v= \langle 1,-2,3 \rangle$ and let $w= \langle -1,1,3 \rangle$. Find $proj_{v}w$, the vector projection of $w$ onto $v$". So remember that the formula for the projection of $w$ onto $v$ is $$proj_{v}w= \frac{v \cdot w}{|v|^{2}}v$$ $$=\frac{\langle 1,-2,3 \rangle \cdot \langle -1,1,3 \rangle}{1^{2}+2^{2}+3^{2}} \langle 1,-2,3 \rangle$$ $$= \frac{3}{7} \langle 1,-2,3 \rangle$$ 3) "Find all values of $t$ so that $ \langle 1,-t,-t^{2} \rangle$ is perpendicular to $ \langle 1,1,1 \rangle$." The given vector will be perpendicular to $ \langle 1,1,1 \rangle$ at all values of $t$ such that the dot product of the two vectors is equal to zero. $$\langle 1,-t,-t^{2}\rangle \cdot \langle 1,1,1 \rangle =1-t-t^{2}$$ The quadratic formula is then used to find the values of t such that this expression is zero. $$\frac{1 \pm [1-(4)(-1)(1)]^{\frac{1}{2}}}{-2}=- \frac{1}{2} \pm \frac{5^{\frac{1}{2}}}{2}$$ 4) "Prove that the two dimensional dot product is commutative." This means to prove $a \cdot b=b \cdot a$ This will be a component wise proof. Let $a= \langle a_{1},a_{2} \rangle$ and $b= \langle b_{1},b_{2} \rangle$ Then $a \cdot b= \langle a_{1},a_{2} \rangle \cdot \langle b_{1},b_{2} \rangle$ $$=a_{1}b_{1}+a_{2}b_{2}$$ $$=b_{1}a_{1}+b_{2}a_{2}$$ $$= \langle b_{1},b_{2} \rangle \cdot \langle a_{1},a_{2} \rangle$$ $$= b \cdot a$$ 5) "Describe the motions determined by the following two parametrized paths. What is the relationship between the two motions?" A) $p(t)= \langle 1,2,3 \rangle + \langle -2,2,-1 \rangle t$ B) $p(t)= \langle 1,2,3 \rangle + \langle -2,2,-1 \rangle \sin(t)$ A fits the formula for the parametrization of a line through the point $(1,2,3)$. B is a little more complicated. It also fits the formula for the parametrization of a line through the point $(1,2,3)$ but it has its direction vector multiplied by $sin(t)$ instead of $t$. From an intuitive perspective this may make it seem like this gives the line some sort of oscillating behavior, however, this simply means the speed that an object travels along this path oscillates. So this path is a line segment through the point $(1,2,3)$. 6) "Find the equation of a plane that contains the points $(2,2,-1),(0,1,-1)$ and $(-2,3,-1)$. Or explain why no such plane exists." This problem can be solved with no math whatsoever. All the points contain a $z$ coordinate of $-1$. So the plane containing them is $z=-1$. 7) "Find the parametrization of the line through the point $(1,2,3)$ and perpendicular to the plane $2x-y+3z=6$." To solve this problem we use the formula for the parametrization of a line given earlier $$l(t)= p_{0}+dt$$ We have been given the point $p_{0}$ so all we need is a normal vector to plug in everything to the formula. Recall that the components of the normal vector are the respective coefficients of the equation of the plane. This make the normal vector $d= \langle 2,-1,3 \rangle $. So our line is $$l(t)= \langle 1,2,3 \rangle + \langle 2,-1,3 \rangle t$$ 8) "Parametrize the intersection between the sphere of radius $2$ centered at the point $(1,1,1)$ with the plane y=2" First things first. Define your sphere. $$S=(x-1)^{2}+(y-1)^{2}+(z-1)^{2}$$ Next set $y$ equal to $2$ to find your intersection with the plane $y=2$. $$(x-1)^{2}+(z-1)^{2}=3$$ This equation describes a circle centered at $(1,0,1)$ with radius $3^{ \frac{1}{2}}$ Now we plug these values into the parametric equation of a circle. $$p(t)=c+rv \cos(t) + ru \sin(t)$$ Where $c$ is the position vector of the center of the circle $r$ is the radius of the circle and $v$ and $u$ are two perpendicular unit vectors in the plane of the circle. In this case they can be the unit vectors $ \langle 1,0,0 \rangle$ and $ \langle 0,0,1 \rangle$. The parametrized circle is $$ \langle 1,0,1 \rangle +3^{\frac{1}{2}} \langle 1,0,0 \rangle \cos(t) + 3^{\frac{1}{2}} \langle 0,0,1 \rangle \sin(t)$$ Another Groovy Picture!! ![image description](/upfiles/calc3.askbot.com/14058907899496069.png) 9) "Let $f(x,y)=x^{2}-3y^{2}$. A) Compute $f(1,2)$ B) Compute $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$. I think we've all gotten really familiar with problems like these since this test so I'll just leave the correct answers. A) $$f(1,2)=1-3(4)=-11$$ B) $$\frac{\partial f}{\partial x}=2x$$ $$\frac{\partial f}{\partial y}=-6y$$ 10) "Let $p(t)= \langle 1+3t,2+4t \rangle$ and let $q(t)= \langle 3-t,-4+3t \rangle$. A) Find the point of intersection between the paths parametrized by $p$ and $q$. B) Do the motions parametrized by $p$ and $q$ collide?" To determine if the lines intersect set the components of each equal to each other. To make things easier use $t$ for the variable in one vector and $s$ for the other. $$1+3t=3-s$$ $$2+4t=3s-4$$ Solving the system shows us that $t=0$ and $s=2$. By plugging these values into their appropriate equations we see they both give the point $(1,2)$ as the location of intersection. We know the paths do not collide even though they intersect because $s \neq t$.

Directional Derivative equal to 10

SpaceManSpiff — Thu, 17 Jul 2014 15:43:34 -0500

So with # 4 on the review sheet. $$\nabla f=\langle y^{3},3xy^{2} \rangle$$ So at $(1,2)$ $\nabla f=\langle1,6 \rangle.$ So to determine if $D_{u}f$ is equal to $10$ we solve for $a$ and $b$ in the equation $$\langle1,6 \rangle \cdot \langle a,b \rangle=10$$ Here's where I feel confused. You end up with $a+6b=10.$ Now $a=4, b=1$ is an obvious solution to this equation. However, does $\textbf{u}$ have to be a unit vector to correctly answer this question? This leads to two other questions. If it does need to be unit vector do we find it by normalizing the vector $\langle4,1 \rangle$? If we do normalize this vector the components of the unit vector we construct will not solve the equation $a+6b=10.$ But I feel that since it will still have the same direction this may not matter. As an alternative method we could solve the system of equations below? $$ a+6b=10$$ $$a^{2}+b^{2}=1$$ I haven't bothered trying to solve that system yet because I'm not sure it's necessary. Thoughts?