# MA 124: The SandBox

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## 44 thoughts on “MA 124: The SandBox”

1. Anonymous says:

Aori,

For this question:
Let $$f(x, y) = 9 \cos x \sin y$$ and let $$S$$ be the surface $$z = f(x, y)$$. Find a unit vector \bar{u} that is
normal to the surface $$S$$ at the point $$(0, \pi/2, 9)$$.

I got the gradient vector at $$(0, \pi/2)$$ to be equal to 0, am I doing something wrong is there a concept I am missing here?

1. Let $$F(x,y,z) = f(x,y) – z = 0$$, then $$\nabla F(x,y,z) = \rangle -9 \sin x \sin y, 9 \cos x \cos y, -1 \langle$$. So, $$\nabla F(0, \pi/2, 9) = \langle 0, 0, -1 \rangle$$ is normal to the surface $$z = f(x,y)$$. This vector is already a unit vector.

2. Consider the equation $$2x^2yz=3xy+xz-yz$$. Assume that it determines a function $$z= f(x,y)$$. Find $$\partial z/ \partial x$$.

1. $$\frac{\partial}{\partial x} \left[ 2x^2y z\right] = \frac{\partial}{\partial x} \left[ 3xy + xz -yz\right]$$
$$2y\frac{\partial}{\partial x} \left[ x^2 z\right] = 3y\frac{d}{d x} \left[ x \right] + \frac{\partial}{\partial x} \left[ xz \right] – y\frac{\partial}{\partial x} \left[z\right]$$
$$2y\left(\frac{d}{d x} \left[ x^2\right] \cdot z + x^2\frac{\partial z}{\partial x}\right) = 3y + z+\frac{\partial z}{\partial x} \cdot x – y\frac{\partial z}{\partial x}$$
$$4yxz + 2yx^2\frac{\partial z}{\partial x} = 3y + z+x\frac{\partial z}{\partial x} – y\frac{\partial z}{\partial x}$$
$$4yxz-3y -z = x\frac{\partial z}{\partial x} – y\frac{\partial z}{\partial x} – 2yx^2\frac{\partial z}{\partial x}$$
$$4yxz-3y -z = \frac{\partial z}{\partial x}\left(x – y – 2yx^2 \right)$$
$$\frac{\partial z}{\partial x}= \frac{4yxz-3y -z}{\left(x – y – 2yx^2 \right)}$$

3. Anonymous says:

Aori,

Find the three numbers x, y, z that satisfy
$$x^2+y^2+z^2=1$$
and whose sum is a large as possible. How do you solve this.
Is the constraint x+y+z?

1. You want to maximize $$f(x,y,z) = x+y+z$$ subject to the constraint $$x^2+y^2+z^2 =1$$.

4. Anonymous says:

Aori,

\begin{aligned} R1 & = \lbrace (x, y) \; | \; 0 \leq x \leq 2, \; \; 0 \leq y \leq  1 \rbrace \\ R2 & = \lbrace (x, y) \; | \; 0 \leq x \leq 2, \; \; 0 \leq y \leq 3 \rbrace \end{aligned}

The function is $$f(x,y) = x^2-2x+y^2-4y+5$$

How do you find the maximum and minimum on the region, and the points at
which they occur. I already found the $$f_x$$ and $$f_y$$.

1. Find the critical points of $$f$$ on $$R1$$. These are the ordered pairs which make $$f_x$$ and $$f_y$$ equal to zero. If there are no critical points in $$R1$$, check the boundaries. Do the same for $$R2$$.

5. Anonymous says:

Hi Aori,
I have a question:
$$T(x,y,z) =100e^{-x^2+2x-y^2+4y-z^2+6z-14}$$

How to show that at every point $$(x, y, z)$$, the temperature increases most rapidly in the direction towards the point $$(1, 2, 3)$$.

1. Find the gradient of $$T$$. You should be able to express the gradient of $$T$$ in the form $$ke^{r(x,y,z)}\langle 1-x,2-x,3-x \rangle$$, for some constant $$k$$ and function $$r(x,y,z)$$. Now observe that this is a scaled vector pointing in the direction of $$(1,2,3)$$ from the point $$(x,y,z)$$.

6. Anonymous says:

Hi Aori,
I need help with this:
Show that every tangent plane to the cone $$z = \sqrt{x^2+y^2}$$ passes through the origin.Start by fixing a point $$(a, b, c)$$ on the cone and writing down the equation.
Can you eliminate c?

1. The formula for the tangent plane to the graph of $$z=\sqrt{x^2+y^2}$$ at $$(a,b,c)$$ is given by
$$z = f(a,b) + f_x(a,b)(x-a) + f_y(a,b)(y-b)$$
where $$f(x,y) = \sqrt{x^2+y^2}.$$

Find the partial derivative of $$f$$ with respect to $$x$$ and the partial derivative of $$f$$ with respect to $$y$$. Evaluate both at $$(a,b)$$. You’ll also need to evaluate $$f$$ at $$(a,b)$$. Use these results in the formula of the tangent plane.

Now that you have a formula for the tangent plane to the graph of $$z = \sqrt{x^2+y^2}$$, verify that the point $$(0,0,0)$$ satisfies the equation of the tangent plane. Since we found an equation for the tangent plane at an arbitrary point $$(a,b,c)$$ (not equal to the origin, since the tangent plane does not exist there) on the graph of $$z = \sqrt{x^2+y^2}$$, then any tangent plane to the graph of $$z = \sqrt{x^2+y^2}$$ passes through the origin.

2. Anonymous says:

How to answer the third part,can you eliminate c? I got a $$\sqrt{a^2+b^2}$$ and a mixed variable tangent equation.

3. $$c = \sqrt{a^2+b^2}$$

What did you get as the equation for the tangent plane?

4. Anonymous says:

Aori,

the equation is:

$$z = \sqrt{a^2+b^2}+\frac{ax-a^2}{\sqrt{a^2+b^2}} + \frac{by-b^2}{\sqrt{a^2+b^2}}.$$
I simplified it down to :
$$z = \frac{ax+by}{\sqrt{a^2+b^2}}$$
and I still have the c. Can you even eliminate it?

5. You need to carry out the last step in my first comment: Verify that the point $$(0,0,0)$$ satisfies the equation you obtained for the tangent plane. You do this by substituting the point into the equation and observing that the left- and right-hand sides are the same.

I don’t see how you still have a $$c$$ in the equation since everything is written in $$a$$’s and $$b$$’s

6. Anonymous says:

In the question it says $$(a,b,c)$$ and it asks can you eliminate $$c$$? I also thought that there is no $$c.$$ I don’t get that part of the question.

7. This part is relevant if you substitute $$c$$ for $$f(a,b)$$ in the equation of the tangent line provided in the first comment. Then you’ll want to eliminate $$c$$ be observing that $$c = \sqrt{a^2+b^2}$$.

In your case, you substituted $$\sqrt{a^2+b^2}$$ for $$f(a,b)$$, thereby skipping this part.

7. Anonymous says:

I am having trouble with this webassign problem. I have tried using the chain rule…
$$\frac{dV}{dt}=\frac{\partial V}{\partial l}\frac{dl}{dt}+\frac{\partial V}{\partial w}\frac{dw}{dt}$$
but it doesn’t seem to give me the right answer. It doesn’t work when tring to find the surface area either.

The length l, width w, and height h of a box change with time. At a certain instant the dimensions are l = 1 m and w = h = 4 m, and l and w are increasing at a rate of 3 m/s while h is decreasing at a rate of 2 m/s. At that instant find the rates at which the following quantities are changing.
A) Volume
B) Surface Area
C) The Length of the Diagonal

1. Aori Nevo says:

The chain rule gives you
$$\frac{dV}{dt} = \frac{\partial V}{\partial l} \frac{dl}{dt} + \frac{\partial V}{\partial w} \frac{dw}{dt} + \frac{\partial V}{\partial h} \frac{dh}{dt}$$

You’re missing the third term.

2. Anonymous says:

Sorry I thought I had typed the third term.

3. Aori Nevo says:

Then
\begin{aligned} \frac{dV}{dt} & =\frac{\partial V}{\partial l}\frac{dl}{dt}+\frac{\partial V}{\partial w}\frac{dw}{dt} + \frac{\partial V}{\partial h} \frac{dh}{dt} \\ & = wh \frac{dl}{dt} + lh \frac{dw}{dt} + lw \frac{dh}{dt} \end{aligned}
At the instant the dimensions of $$l = 1$$ m and $$w = h = 4$$ m, and $$l$$ and $$w$$ are increasing at a rate of 3 m/s while $$h$$ is decreasing at a rate of 2 m/s, the rate of change of the volume is
$$\frac{dV}{dt} = 4(4)(3) + 1(4)(3) + 1(4)(-2) = 52 \text{m/s}$$