- 14 | Differentiating Functions of Several Variables
- 15 | Optimization: Local and Global Extrema
- 16 | Integrating Functions of Several Variables
- MA 124: Homework
- MA 124: Workshop
- Review Problems

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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?

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.

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

$$

\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)}

$$

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?

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

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 \(f_y\).

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 \(R2\).

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 \((1, 2, 3)\).

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 \((x,y,z)\).

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?

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.

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

\( c = \sqrt{a^2+b^2}\)

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

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?

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

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.

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.

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

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.

Sorry I thought I had typed the third term.

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}

$$