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Use this page to ask anything related to MA 124. If you are registered and logged in, your posts will display your user name. If you would like to post anonymously, make sure you are logged out. Anyone is allowed to post.
Aori,
On webassign 11, question four asks us to find the volume of the solid bounded by the cylinders $$ x^{2}+y^{2}=9r^{2} $$ and $$ y^{2}+z^{2}=9r^{2} $$ . I am having trouble picturing the graph and I don’t really know how to approach the problem. Could you help?
The equation \(x^2 + y^2 = 9r^2\) represents the cylinder with the \(z\)-axis as its center line and radius 3r. The equation \(y^2+z^2 = 9r^2\) is the cylinder with the \(x\)-axis as its center line and radius 3r.
Recall that the volume of a solid above the region \(D\) and below the surface \(z = f(x,y)\) is given by
$$
\iint_D f(x,y) \; dA
$$
provided that \(f(x,y) \geq 0\) on \(D\).
In this problem, \(D\) corresponds to the projection of the cylinder \(x^2 + y^2 = 9r^2\) onto the \(xy\)-plane. The solid we would like to find the volume of lies above and below the \(xy\)-plane. Therefore we cannot make use of the formula above. We can use the generalized version of that formula: The volume of a solid above \(z= g(x,y)\) and below \(z = f(x,y)\) over a region \(D\) is given by
$$
\iint_D f(x,y) – g(x,y) \; dA.
$$
provided that \(f(x,y) \geq g(x,y)\) on \(D.\)
The cylinder along the \(x\)-axis can be split up into two functions \(f(x,y)\) and \(g(x,y)\). Use these functions in the above formula and integrate over the region \(D\) as previously defined.
Hi,
This comes from the homework problem #3, wherein I have to find the coordinates of the center of mass. Where
$$ D=\{(x,y)| \ |x|\le y, 0\le y\le4 \} $$
And
$$ \rho (x,y) = xy+20 $$
I found the mass by
$$ \int^2_0\int^4_x \rho(x,y) \; dy \; dx + \int^0_{-2}\int^4_{-x} \rho(x,y) \; dy \; dx $$
Now when I try to find the x and y-corrdinate of the center of mass, what would be the upper and lower bounds of my integration? I’m unsure since I had to separate the region and integrate them separately. Do I do the same for the coordinates?
Recall that the center of mass is given by \((\bar{x}, \bar{y})\) where
\(\bar{x} = \frac{M_y}{m}\), \(\bar{y} = \frac{M_x}{m}\),
$$
\begin{aligned}
M_x & = \iint_D y \rho(x,y) \; dA,
\end{aligned}
$$
$$
\begin{aligned}
M_y & = \iint_D x \rho(x,y) \; dA,
\end{aligned}
$$
and
$$
\begin{aligned}
m & = \iint_D \rho(x,y) \; dA
\end{aligned}
$$
The graph of the region \(D\) is given below:
If you split the region \(D\) at \(x = 0\) and consider each sub-region as a type I region then you would have two double integrals, as in the calculation of the mass \(m.\)
Aori,
On the webassign, this iterated integral is given and after I find the first integral, the second one comes out to zero and webassign says that is incorrect. What am I doing wrong?
$$
\int_{0}^{\pi}\int_{0}^{2 }7r(sin(\theta ))^2 drd\theta
$$
Excellent type setting! Just remember to wrap LaTeX code in either
$$ math stuff here $$or\( math stuff here \)Here is is how you do this integral:
\begin{aligned}
\int_0^{\pi}\int_0^2 7 r \sin^2 \theta \; dr \; d\theta
& = \int_0^{\pi} \sin^2 \theta \left[\int_0^2 7 r \; dr \right] \; d\theta \\
& = \int_0^{\pi} \sin^2 \theta \left[\left. \frac{7}{2}r^2 \right|_0^2 \right] \; d\theta \\
& = \int_0^{\pi} 14 \sin^2 \theta \; d\theta \\
& = 14 \int_0^{\pi} \frac{1-\cos 2 \theta}{2} \; d\theta \\
& = 14 \left(\frac{\theta }{2}-\frac{1}{4} \sin (2 \theta )\right) \Big|_0^{\pi}\\
& = 7 \pi.
\end{aligned}
$$
\begin{aligned}
\int_0^{\pi}\int_0^2 7 r \sin^2 \theta \; dr \; d\theta
& = \int_0^{\pi} \sin^2 \theta \left[\int_0^2 7 r \; dr \right] \; d\theta \\
& = \int_0^{\pi} \sin^2 \theta \left[\left. \frac{7}{2}r^2 \right|_0^2 \right] \; d\theta \\
& = \int_0^{\pi} 14 \sin^2 \theta \; d\theta \\
& = 14 \int_0^{\pi} \frac{1-\cos 2 \theta}{2} \; d\theta \\
& = 14 \left(\frac{\theta }{2}-\frac{1}{4} \sin (2 \theta )\right) \Big|_0^{\pi}\\
& = 7 \pi.
\end{aligned}
$$
\begin{aligned}
& \int_0^{\pi}\int_0^2 7 r \sin^2 \theta \; dr \; d\theta \\
& = \int_0^{\pi} \sin^2 \theta \left[\int_0^2 7 r \; dr \right] \; d\theta \\
& = \int_0^{\pi} \sin^2 \theta \left[\left. \frac{7}{2}r^2 \right|_0^2 \right] \; d\theta \\
& = \int_0^{\pi} 14 \sin^2 \theta \; d\theta \\
& = 14 \int_0^{\pi} \frac{1-\cos 2 \theta}{2} \; d\theta \\
& = 14 \left(\frac{\theta }{2}-\frac{1}{4} \sin (2 \theta )\right) \Big|_0^{\pi}\\
& = 7 \pi.
\end{aligned}
$$
I’m having trouble with number two on the written homework. It say to find the absolute minimums and maximums of the function
$$
f(x,y)=\frac{1}{x^2+y^2+1}
$$
on the disk \((x-1)^2+y^2 \leq 4\)
I tried using Lagrange multipliers but that only covers the perimeter of the disk. How do I find the critical points?
You are correct in that Lagrange multipliers finds the maximum and minimum values on the perimeter. To find the critical points on the interior of the disk, you need to solve the system
$$
\begin{cases}
f_x(x,y) = 0 \\
f_y(x,y) = 0
\end{cases}
$$
Any solution to this system that lies in the interior of the disk is a critical point.
Dear Aori,
I am struggling on problem 2 part b of the written homework. I cannot get all of my sines and cosines to cancel… any advice?
Problem 2 Part B:
Assume \(u = f(x, y)\) is a function all whose partial derivatives exist. Let \(x=e^s \cos(t)\) and \(y=e^s \sin(t)\)
Prove that:
$$
\left ( \frac{\partial u}{\partial x} \right )^2+\left ( \frac{\partial u}{\partial y} \right )^2=e^{-2s}\left [ \left ( \frac{\partial u}{\partial s} \right )^2+\left ( \frac{\partial u}{\partial t} \right )^2\right ]
$$
Some common errors to avoid:
(\cos \theta + \sin \theta)^2 \not = \cos^2 \theta + \sin^2 \theta
$$
See if any of the above errors apply. If so, fix them and try again. If not, show me your final answer and we’ll figure out where to go from there.
Remark: When typesetting equations on here, you’ll need to wrap them in either
\( equation \)
or
$$ equation $$
where the first wrapper is for equations used inline and the second wrapper is for centered equations on their own line.
In the second derivative test when minimizing/maximizing values for functions of two variables, why do we only check the second partial derivative of F with respect to x to determine if the point a local maximum or minimum, and not the second partial derivative of F with respect to y? Does it matter which one I test with?
Excellent question! It turns out that it doesn’t matter. Here is why: If \(D>0\) then
$$
f_{xx}f_{yy} – (f_{xy})^2 >0
$$
or, equivalently,
$$
f_{xx}f_{yy} > (f_{xy})^2
$$
which implies \(f_{xx}\) and \(f_{yy}\) have the same sign, as opposite signs would contradict our original assumption that \(D>0\). So it doesn’t matter which one we check.
I think it’s tradition to formulate the test in terms of \(f_{xx}\).
Hi, I’m trying to do a problem in WebEx, but I’m stuck. Can I have some help?
Find tangent plane and normal line equations of the following at the point (6,8,6)
$$ 2(x-5)^2 + (y-6)^2 + (z-4)^2 = 10 $$
Here’s my attempt at it:
For the plane:
$$ \vec{n} = \nabla f(6,8,6) \\ x+y+z = 20 \\ $$
which was right, but for the normal line:
$$ x = 4t +6 = z \\ y = 4t + 8 \\ $$
Which was wrong, but I’m not sure why.
Recall that an equation for the tangent plane to the surface \(z = f(x,y)\) at \(P(x_0,y_0,z_0)\) is given by
$$
\nabla F(x_0,y_0,z_0) \cdot \langle x-x_0, y-y_0, z-z_0 \rangle = 0
$$
where \(F(x,y,z) = f(x,y)-z\). That is, \(F\) is the function obtained by moving everything in the given equation to one side.
In your problem, \(F(x,y,z) = 2(x-5)^2 +(y-6)^2+(z-4)^2-10 = 0\). So that,
$$
\nabla F(x,y,z) = \langle 4(x-5), 2(y-6), 2(z-4) \rangle
$$
and
$$
\nabla F(6,8,6) = \langle 4,4,4 \rangle
$$
Applying the formula, we obtain the following:
$$
\begin{aligned}
\nabla F(x_0,y_0,z_0) \cdot \langle x-x_0, y-y_0, z-z_0 \rangle & = 0 \\
\langle 4,4,4 \rangle \cdot \langle x-6, y-8, z-6 \rangle & = 0 \\
4(x-6) + 4(y-8) + 4(z-6) & = 0 \\
x-6 + y-8 + z-6 & = 0 \\
x + y + z & = 20
\end{aligned}
$$
Recall that a vector equation for the line going through \(P(x_0,y_0,z_0)\) with directional vector \(\mathbf{v}\) is given by
$$
\mathbf{r}(t) = \langle x_0,y_0,z_0 \rangle + \mathbf{v} t
$$
In your problem, \(\mathbf{v} = \mathbf{n}\) and \(P(x_0,y_0,z_0)=(6,8,6)\).
That is,
$$
\mathbf{r}(t) = \langle 4t + 6, 4t + 8, 4t+6 \rangle
$$
which is what you have. Perhaps the form of the line doesn’t match the form that WebAssign requires. Remember a line can be expressed in vector form (as above), parametric form, or symmetric form.
Hello Aori,
I have a question from one of the problems in WebAssign.
Use the following equation
$$ \frac{\partial z}{\partial x} = – \frac{\partial F/\partial x}{\partial F/\partial z} $$
To find
$$ \frac{\partial z}{\partial x} $$
And
$$ xyz = \tan(x+y+z) $$
I’m quite confused, what’s the difference between partial of F and partial of z (both with respect to x). How do I approach this? Is this implicit?
Also, the answer didn’t have any \(\sec^2\) involved, which is weird since derivative of tan is \(\sec^2\).
Thanks ahead
By \(F\), we mean the function obtained by moving everything to one-side of the equation. For this example, that would mean \(F(x,y,z) = \tan(x+y+z) -xyz=0\). Now that you know what \(F\) is find its partial with respect to \(x\) and \(z\) and make use of the first formula to find \(\partial z / \partial x\).
Alternatively, you can use implicit differentiation without defining this function \(F\), as follows:
$$
\begin{aligned}
\frac{\partial }{\partial x} \left[ xyz \right] &= \frac{\partial}{\partial x}\left[\tan(x+y+z)\right] \\
\Rightarrow y \frac{\partial }{\partial x} \left[ xz \right] &= \frac{\partial}{\partial x}\left[\tan(x+y+z)\right] \\
\Rightarrow y \left(z + x \frac{\partial z}{\partial x} \right)&= \sec^2(x+y+z)\frac{\partial}{\partial x}\left[(x+y+z)\right] \\
\Rightarrow \cdots
\end{aligned}
$$
When you’re done differentiating both sides, you’ll have terms involving \(\partial z/ \partial x\) on both sides of the equation. Isolate these terms on one side of the equation. Then factor out \(\partial z/ \partial x\), so that you have \(\partial z/\partial x\) times some function of \(x\) and \(y\). Divide by that function to get \(\partial z/ \partial x\) by itself.
Thanks,
But when you were solving by implicit differentiation, why did the tan(x+y+z) turn into arctan(x+y+z)?
That’s an error. It should be arctan. So you were right to expect secant squared in your final answer. However, the final form of the answer can differ by using identities. Of course identities should be used to simplify your solution, and here I’m not so sure replacing secant squared with anything will simplify the final form.
Hi, I am working on the first problem of homework 1. I was wondering if I should use the fact that the rate of change in each row should be constant to prove that the graph is linear?
Yes, but of course you’ll need to explain why. Have a look at example 3.
Hi, I’m working on number 2 from homework 1 and have a question about how to approach it.
T(x,y) = 1/(1 + x^2 + y^2)
Find the rate of change of temperature with respect to distance at the point (2,1) in the (a) positive x and (b) positive y directions.
I’m thinking I should take the partial derivatives with respect to x and y, then plug in (2,1) for each. Or is there more to it?
Yes, that’s correct.