Mark's Calc III - Individual question feed

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?

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!

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 $\Theta$?

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?

Setting up double integrals

Anonymous — Thu, 17 Jul 2014 12:15:19 -0500

I am working on #16 from 15.8 and I'm wondering if I have set up my integrals correctly. The question asks: Evaluate $\int\int x^2 \hspace{2 mm} dA$ over the region in the first quadrant bounded by the hyperbola $xy=16$ and the lines $y=x, y=0$, and $x=8$. I came up with $$\int_4^8\int_{16/x}^x x^2\hspace{2 mm} dydx$$ is this correct?

Double Integrals From 15.8

Anonymous — Thu, 17 Jul 2014 11:07:00 -0500

So I am having trouble with #9 from 15.8: Compute $$\int_0^1\int_\sqrt{y}^1 \sqrt{x^3+1} \hspace{2 mm} dxdy$$ here is my work $$\int_0^1\int_0^{x^2} \sqrt{x^3+1} \hspace{2 mm} dydx$$ $$\int_0^1 y\sqrt{x^3+1} \biggr|_0^{x^2} dx$$ $$\int_0^1 x^2\sqrt{x^3+1} \hspace{2 mm} dx$$ I used $u$ substitution here $\hspace{14 pc}$ Let $ u=x^3+1$ $$du=3x^2 dx$$ $$1/3 du=x^2 dx$$ $$1/3 \int_0^1 \sqrt{u} \hspace{2 mm} du$$ $$1/3(2/3 \sqrt{u^3}) \biggr|_0^1$$ $$=2/9$$ what am I missing here?

Evaluating the max/min using Lagrange method on the turn-in homework

Kyouko — Wed, 16 Jul 2014 01:42:44 -0500

In class our usual problem might state that we are looking to see whether we have a max or min at and intersection between two functions, and to solve this we use the lagrange method by setting both gradients equal to each other times a constant $\nabla f = \nabla \lambda g$, where $g(x,y)$ is equal to a constant. In this problem, we are given two constraint functions $g(x,y)$ and $h(x,y)$ that are both equal to $x$. Do we simply bring $x$ over to either side of the inequality equation and set that equal to 0 to create the "third" equation to solve our system? Or is there something else that I'm missing? $$ f(x,y) = x^3 - 2xy +y^2 -3x $$ constrained by: $$ y^2-4 \leq x \leq 1-y $$ As a follow up, I solved my system of equations such that the third equation looked like so: $$ 1-y=x $$ But as I solved this system for \lambda , $x$ and $y$, I ended up getting an imaginary root. looking at the pictures people have posted, I don't think that this should be the case. Graphing this function in matlab yielded this image (not sure how to get the constraints on it too) ![image description](http://puu.sh/adwJ3/8fa87f4a2f.png) And the corresponding contour plot to go with it ![image description](http://puu.sh/adwU2/a6087db4ad.png)

Can we write "used Wolfram Alpha" as work?

Wes — Tue, 15 Jul 2014 11:50:07 -0500

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."

Using Lagrange multipliers for the homework

Dylan — Mon, 14 Jul 2014 22:12:00 -0500

I'm slightly confused by a certain step I'm at in the homework. I have looked for extrema in the whole unrestricted function, and found them without much difficulty, but for finding the extrema of the "edges" of the constrained area, I'm not sure how to apply our technique. By setting $\nabla f = \lambda \nabla g$, I get the two equations we need, but the third "equation" is actually the inequality we were provided with as a constraint. Can we use this inequality to solve for $x$ and $y$, or is there another method that I'm missing?

Rate of change of potential

Anonymous — Fri, 11 Jul 2014 07:14:49 -0500

In section 14.5, question #6, it asks: Suppose the electric potential at $(x,y)$ is $ln\sqrt {(x^2+y^2)}$. Find the rate of change of the potential at $(3,4)$ toward the origin and also in a direction at a right angle to the direction toward the origin. I know how to calculate $D_uf$ but I am lost on what $u$ would be in this problem.

Find the direction

Anonymous — Fri, 11 Jul 2014 07:23:36 -0500

I am stumped on how to begin question #16 in section 14.5: Find the directions in which the directional derivative of $f(x,y)=x^2+sin(xy)$ at the point $(1,0)$ has the value 1. I'm sure $<0,1>$ would work but I don't see how to find all other directions.

Chain rule

Tiffany — Wed, 09 Jul 2014 20:54:50 -0500

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!

Does anyone know any tips or ideas on visualizing the functions?

asmith14 — Tue, 08 Jul 2014 16:57:54 -0500

I am having trouble matching the functions from the problem sheet that we did today (July 8) and I was wondering if someone could explain how they are matching the functions with their pictures or if they have any ideas of how they can be matched. I do know know how to create the pictures online but it is number 7 on the problem sheet.

How do I accurately visualize and describe level curves?

Justin — Wed, 02 Jul 2014 14:16:17 -0500

For the 14.1 homework tonight, many of the problems ask you to describe the level curves of the function. I understand how to do the rest of the problem, but how does one visualize what the level curves will look like? Is a 3D graphing program necessary, or is it possible to look at the function and tell? For example (Exercise 14.1.1), let $ f(x,y) = (x−y)^2$. The level curves of this function looks like lines of slope $1$... how could one tell that from the function alone? *Comment*: We haven't covered level curves in class, but they will be **major** when we get back after the fourth of July!!

Finding the cosine of the angle between two curves?

Tiffany — Mon, 30 Jun 2014 14:42:31 -0500

So, I've been trying to work on the homework 13.2, and I'm stuck on both questions 6 and 7. But we'll start with question 6. "6. Find the cosine of the angle between the curves $\langle0,t^2,t\rangle$ and $\langle\cos(\frac{\pi t}{2}),\sin(\frac{\pi t}{2}),t\rangle$ where they intersect." Well I can tell that they intersect at t=1 And from the book on page 336 it states... " $ \cos(\theta) $ = $\mathbf {\frac{\vec r' * \vec s'}{|\vec r'||\vec s'|}}$ = $ \mathbf {\frac {\vec r'}{|\vec r'|} \cdot \frac{\vec s'}{|\vec s'|}} $ " So letting $\vec r$ = $\langle0,t^2,t\rangle$ and $\vec s$ = $\langle \cos(\frac{\pi t}{2}), \sin(\frac{\pi t}{2}),t\rangle$ I end up with $\vec r'$ = $\langle 0,2t,1\rangle$ and $\vec s' $ = $\langle (\frac{-1}{2} \pi \sin(\frac{\pi t}{2})), (\frac 12 \pi \cos(\frac{\pi t}{2})), 1\rangle $ using this to come up with | $ \vec r'$| I get |$\vec r'$| = $\sqrt{0^2 + (2t)^2 + 1^2} $ = $\sqrt{4t^2 +1} $ and |$\vec s'$|= $\sqrt{(\frac{-1}{2}\pi \sin(\frac {\pi t}{2}))^2 + (\frac 12 \pi \cos(\frac{\pi t}{2}))^2 + 1^2) } $ Now when we plug this into the $\cos(\theta)$ equation, do we substitute t=1 for the t in all of the equations to find the answer, or how exactly are we supposed to approach this problem? I end up with this ridiculous answer with a whole bunch of randomness in it that doesn't match up with the back of the book, so I'm beyond confused on this problem, and feel like there is a much easier way than how i'm trying to figure it out. HELLLPPPP! :) *Comment*: When you take the norm of $\vec{s}'$, note that you get a $\sin^2+\cos^2$. That yields a significant simplificatoin that might help.

How do you find a parametric equation for a hypercycloid?

Anonymous — Mon, 30 Jun 2014 13:26:15 -0500

In section 10.4, problem number 7: A wheel of radius 1 rolls around the outside of a circle of radius 3. A point P on the rim of the wheel traces out a curve called a hypercycloid. Assuming P starts at the point (3,0), find the parametric equations for the curve. This isn't an assigned problem, I am just curious and I am not sure how to geometrically break this problem apart to

How do I visualize space curves?

Wes — Sun, 29 Jun 2014 19:02:51 -0500

I'm having a bit of trouble visualizing some of the space curves on the homework (section 13.1, #1-4). I can draw sketches of what they look like in the x-y and the x-z planes, but when it comes to "combining" the two, I'm struggling. Do any of you have any suggestions? Or is this not even necessary for the homework as it just says to "investigate?"

Finding area of a parallelogram

Tiffany — Wed, 25 Jun 2014 12:27:36 -0500

So I've been working on problem 7 in 12.4. Following the book on page 315 about parallelograms. > "Given two vectors, we can put them tail to tail and form a parallelogram...The height of the parallelogram, h, is |**A**|$ \sin(\theta)$ , and the base is |**B**|, so the area of the parallelogram is |**A**||**B**|$ \sin(\theta)$, exactly the magnitude of |**A** x **B**|." The problem states "7. Find the area of the parallelogram with vertices (0, 0), (1, 2), (3, 7), and (2, 5)." When I sketched out the parallelogram with the given points, it looked like the base of the parallelogram would be the vector from the origin to (1,2) (**B**), and the top vector would be (2,5)(**A**). (I'm not sure if this is where my mistake was, or if it was in trying to find $\theta$.) Using those two points as the vectors, I ended up with |**A**| = $\sqrt{29}$ and |**B**| =$ \sqrt{5}. $ I then used the dot product |**A**| * |**B**| = |**A**||**B**|$\cos$($\theta$) to attempt to find theta. Plugging everything in I ended up with $$ \theta = \cos^{-1}(\frac {12} {\sqrt{29} \sqrt{5}}) $$ Plugging that into the area equation from the book, I ended up with.. $$ area = \sqrt{29}\sqrt{5}(\sin(\cos^{-1}(\frac {12} {\sqrt{29}\sqrt{5}})))$$ Which unless I did wrong in my calculator gave me 0. But the back of the book showed that the answer should have been 1. So I'm not exactly sure where I went wrong, or if I did the whole problem wrong. So if anybody could give some input that would be fantastic! Thanks!