NAME: (print!) Dr. Z. Section: ALL E-Mail address: DrZcalc3@gmail.com MATH 251 (22,23,24 ) [Fall 2020], Dr. Z. , Makeup for Exam 1, Sunday, Oct. 25, 2020, 9:00-11:00am WRITE YOUR FINAL ANSWERS BELOW

1. (t, 0, 0), −∞ < t < ∞ ; x-axis .

2. 272

3. (−3, 3) .

4. z = 2x + 12y − 17

5. \frac{1}{2}

6. \frac{24999}{5000} or 4.9998

7. DNE

8. (−1, −1, 0)

9. 6 ; ⟨\frac{1}{3}, −\frac{2}{3}, −\frac{2}{3}⟩

10. 10 Types: Number, Function of variable(s), 2D vector of numbers, 3D vector of numbers, 2D vector of functions, 3D vector of functions, equation of a plane, parametric equation of a line, equation of a line, equation of a surface, equation of a line, DNE (does not exist).

Sign the following declaration: I Hereby declare that all the work was done by myself. I was allowed to use Maple, calculators, the book, and all the material in the web-page of this class but not other resources on the internet. I only spent (at most) 90 minutes on doing the exam. The last 30 minutes were spent in checking and double-checking the answers. I also understand that I may be subject to a random short chat to verify that I actually did it all by myself. Signed:

1. (10 pts.) Find a parametric representation, in terms of the parameter, t, of the line of intersection of the planes y = 0 and z = 0. What is the usual name for that line?

The type of the answers is: parametric equation of a line in 3D space

ans. (t, 0, 0), −∞ < t < ∞ ; The x-axis

The short way (perfectly OK!) is to say that obviusly a typical point on the line of intersection of y = 0 and z = 0 looks like (x, 0, 0) for some x, but the traditional letter for a parameter is t so it is the set of all points (t, 0, 0) where t goes from −∞ to ∞ 2. (10 points) Use any method to find \frac{∂^{2}h}{∂r^{2}} at (q, r) = (1, 1) where h(u, v) = u^{5} v^{6}, u = q^{2} r, v = q r^{2}

The type of the answer is: number

ans. 272

I told you use any method. Superfically it looks like that you need the chain rule, and some people did it that way, and got the right answer and full credit (but it took them a long time!). Other people messed up. In this simple problem, it is much more efficient to express the function directly in terms of q and r and then it is a piece of cake. The function is u^{5} v^{6} = (q^{2}r)^{5}(qr^{2})^{6} = q^{10}r^{5} · q^{6}r^{12} = q^{16}r^{17} So h_{r} = 17q^{16}r^{16} and h_{r}r = 17 · 16 · q^{15}r^{16} = 272 Finally, don’t forget!, plug-in q = 1 r = 1, and the type of the final answer is number.

3. (10 points) Find the point on the line y = x + 6 (in the xy-plane) where the rate of change of the function f(x, y) = x^{2} + y^{2} in the direction < 1, 1 > is zero.

The type of the answer is: point

ans. (−3, 3

The gradient of f, < f_{x}, f_{y} > equals < 2x, 2y >. The directional derivative in direction < 1, 1 > is < 2x, 2y > . < 1, 1 > /sqrt2. Setting this equal to 0 we get that x + y = 0. But out point also lies on the line y = x + 6 . Solving the simple system of two equations and two unknowns x, y, x + y = 0, y = x + 6 we get −x = x + 6 so −2x = 6 so x = −3 and y = −(−3) = 3. So the desired point is (−3, 3).

4. (10 points) Find an equation of the tangent plane to the following surface at the given point

z = x^{2} + y^{3} , (1, 2, 9) .

The type of the answer is: Equation of a plane

ans. z= 2x+12y-17

First way: Use the fact that for a surface given explicitly in the format z = f(x, y) the equation of the tangent plane through (x_{0}, y_{0}, z_{0}) (provided of course the point is a good one, i.e. f(x_{0}, y_{0}) = z_{0} is

z − z_{0} = f_{x}(x_{0}, y_{0}) ∗ (x − x_{0}) + f_{y}(x_{0}, y_{0}) ∗ (y − x_{0}) .

Step 0: Verify that f(1, 2) = 1^{2} + 2^{3} = 9 (yea!) f_{x} = 2x, f_{y} = 3y^{2} si f_{x}(1, 2) = 2, f_{y}(1, 2) = 3 · 2^{2} = 3 · 4 = 12 so

z − 9 = 2(x − 1) + 12(y − 2)

Simplifying

z − 9 = 2x − 2 + 12y − 24

Simplifyiny more

z = 2x + 12y − 17

Second way: Convert it to implicit format z − x^{2} − y^{3} = 0 and use the the formual grad(f)(x_{0}, y_{0}, z_{0}). < x − x_{0}, y − y_{0}, z − z_{0} >= 0.

5. (10 points) Compute f_{xyy}(1, 1) (aka \frac{∂^{3}f}{∂x∂^{2}y} (1, 1), if f(x, y) = ln(x^{2} + y). The type of the answer is: Number ans. \frac{1}{4} . f_{x} = \frac{2x}{x^{2}+y} (by the chain rule of calc1). This can be written f_{x} = 2x · (x^{2} + y)^{−1} Moving right along f_{xy} = 2x · (−1)(x^{2} + y)^{−2} and f_{xyy} = 2x · (−1)(−2)(x^{2} + y)^{−3}, so

f_{xyy}(x, y) = \frac{2}{(x^{2} + y)^{3}} .

Finally (don’t forget!) f_{xyy}(1, 1) = \frac{2}{(1^{2}+1)^{3}} = \frac{2}{8} = \frac{1}{4}.

6. (10 points) Use Linearization to apprximate \sqrt{(3.001)^{2} + (3.999)^{2}}. Part of the challenge is to find the appropriate function f(x, y) and the appropriate ‘nice’ point (x_{0}, y_{0}).

The type of the answer is: Number

ans. \frac{24999}{5000} OR 4.9998 (both are acceptable)

The underlying function is

f(x, y) = \sqrt{x^{2} + y^{2}} = (x^{2} + y^{2})^{1/2} .

The point of interest is (3.001, 3.999). The nice point nearby is (x_{0}, y_{0}) = (3, 4). The Linearization of f(x, y) at (x_{0}, y_{0}) is

L(x, y) = f_{x}(x_{0}, y_{0}) (x − x_{0}) + f_{y}(x_{0}, y_{0}) (y − y_{0}) .

f_{x} = \frac{1}{2}(x^{2} + y^{2})^{−1/2} · 2x = \frac{x}{x^{2}+y^{2}} . f_{y} = \frac{1}{2}(x^{2} + y^{2})^{−1/2} · 2y = \frac{y}{x^{2}+y^{2}} . So f_{x}(3, 4) = \frac{3}{\sqrt{3^{2}+4^{2}}} = \frac{3}{5}, f_{y}(3, 4) = \frac{4}{\sqrt{3^{2}+4^{2}}} = \frac{4}{5}. Getting

L(x, y) = 5 + \frac{3}{5} (x − 3) + \frac{4}{5} (y − 4) .

Warning: do not simplify. Leave it like that. So

L(3.001, 3.999) = 5+\frac{3}{5} (3.001−3) + \frac{4}{5} (3.999−4) = 5+\frac{3}{5} (0.001) + \frac{4}{5} (−0.001) = 5−\frac{1/1000}{5} = \frac{24999}{5000} = 4.9998

Remark: The exact value (up to 10 decial points) is 4.999800196 (not bad!)

7. (10 points) Decide whether the following limit exists. If it does find it. If it does not Explain why not?

lim_{(x,y,z)→(1,1,1)} \frac{x + 2y + 3z − 6}{3x + 2y + z − 6}

Hint: To prove that a limit exists at a designated point you first try to plug it in. If there are no issues you are done, and the value is the limit. Failing this, you try to use algebra to simplify and then plug it in, if possible. Failing this you try to prove that it does not exist. One way of doing it is to pick any two different lines through the point in question and show that if you approach the designated point via these lines you get different limits (in the sense of single-variable calculus). Remark: This was the most challenging problem, and most people did not get it. Please study it carefully. Plugging in gives 0/0. Trying to simplify also fails. So we are suspecting that the limit does not exist. We can do some preliminary investigations plugging into the function

f(x, y, z) = \frac{x + 2y + 3z − 6}{3x + 2y + z − 6}

Random values near the designated point (1, 1, 1). For example

f(1, 1, 1.0001) = 3 , f(1, 1.0001, 1) = 1

1 and 3 are very far apart! So this is an indication that the limit probably does not exist. In order to prove it conclusively we neet to find any two different lines passing through the designated point (1, 1, 1) and plug-it in and take the limits in the sense of calc1 (that DO EXIST!, some people claimed that the calc 1 limits do not exist!). A convenient line is (1 + t, 1, 1 getting that on this line

f(1 + t, 1, 1) = \frac{(1 + t) + 2 + 3 − 6}{3(1 + t) + 2 + 1 − 6} = \frac{t}{3(1 + t) + 2 + 1 − 6} = \frac{t}{3t} = \frac{1}{3}

So along this line the value of the function t is always \frac{1}{3}, and obviously the limit in the sense of calc1 exists and equals \frac{1}{3}. Another convenient line is (1, 1 + t, 1 getting that on this line

f(1, 1 + t, 1) = \frac{1 + 2(1 + t) + 3 − 6}{3 + 2(1 + t) + 1 − 6} = \frac{2t}{2t} = \frac{t}{t} = 1.

So along this line the value of the function t is always 1, and obviously the limit in the sense of calc1 exists and equals 1. But since these two calc1 limits do not agree, this is proof that the limit (in the sense of calc3) does not exist.

Comments:

1. You can take any two lines . Of course it is good to take convenient line. But the lines must pass throgh the designated point, not the origin (unless the point in question happened to be, as often is the case in the homework, and this confused many students) is the origin (0, 0, 0.

2. Another, more general approach is to consider an ‘abstract’ (general) line passing through (1, 1, 1)

(1, 1 + at, 1 + bt)

For a and b arbitrary numbers, find the limit along that line as an expression in a, b, and since it depends on a, b the limit does not exist.

3. Yet another (legal!) approach , for all those who love the origin, is to convert the problem to a limit about the origin doing a change of variable

x′ = x − 1 , y′ = y − 1 , z′ = z − 1 ,

Getting that we have to study

lim_{(x′,y′,z′)→(0,0,0)} \frac{x + 2y + 3z}{3x + 2y + z}

and then take lines through the origin! Warning: I was disappointed that quite a few people ‘faked’ it (unintenionally) writing gibberish. If you are not sure, please exercise intellectual honesty and admit that you don’t know how to do it.

8. (10 points) Find the minimum location(s) (i.e the point(s) where the function is the smallest) of the function

f(x, y, z) = x + 2y − z

in the closed pyramid whose vertices are

(1, 1, 0) , (−1, 1, 0), (1, −1, 0), (−1, −1, 0), (0, 0, 1)

The type of the answer(s) is: point

ans. (−1, −1, 0)

The function is linear so there are no cricial points (not even along the edges). All we have to do is plug-in the function at the five vertices and see who gives the mimimal value.

f(1, 1, 0) = 3 , f(−1, 1, 0) = 1 , f(1, −1, 0) = −1 , f(−1, −1, 0) = −3 , f(0, 0, 1) = −1 .

The minimum value is −3 at it happens at the location (−1, −1, 0).

9. (10 points) A certain particle of mass 2 kilograms has position function, expressed in meters, where t is time, expressed in seconds,

r(t) = ⟨ e^{t}, 2 cos t, 2 cos t ⟩ ,

(i) Find the magnitude of the force acting on it at time t = 0. Explain!

(ii) Find the unit direction of that force.

Reminder from physics: F = ma, i.e. the force equals the mass times the accelaration.

The types of the answers are (i) non-negative number (it is OK to say number) (ii) vector (or even better unit vector)

ans. (i) magnitude : 6 Newtons ; (ii) unit direction:⟨\frac{1}{3}, −\frac{2}{3}, −\frac{2}{3}⟩

r′(t) = ⟨ e^{t}, −2 sin t, −2 sin t ⟩ ,
r′′(t) = ⟨ e^{t}, −2 cos t, −2 cos t ⟩ ,

r′′(0) = ⟨ 1, −2, −2 ⟩ ,
F = 2⟨ 1, −2, −2 ⟩ = ⟨ 2, −4, −4 ⟩ ,
|F| = |⟨ 2, −4, −4 ⟩| = \sqrt{2^{2} + (−4)^{2} + (−4)^{2}} = \sqrt{36} = 6 .

direction: ⟨ 2, −4, −4 ⟩/6 = ⟨ \frac{1}{3}, −\frac{2}{3}, −\frac{2}{3}⟩ 10. (10 points) A certain function f(x, y, z) depends on three variables called x, y and z. At a certain time the rate of change of this function with respect to time happens to be 140. It is also known, that at that same time

• The rate of change of f(x, y, z) with respect to x is 1

• The rate of change of f(x, y, z) with respect to y is 2

• The rate of change of f(x, y, z) with respect to z is 3

• The rate of change of y with respect to time is two times The rate of change of x with respect to time.

• The rate of change of z with respect to time is three times The rate of change of x with respect to time. What is the rate of change of x with respect to time at that time?

The type of the answer is: Number

ans. 10 We use the Chain rule in its abstaract form

\frac{df}{dt} = \frac{df}{dx} · \frac{dx}{dt} + \frac{df}{dy} · \frac{dy}{dt} + \frac{df}{dz} · \frac{dz}{dt} .

Incorporationg the data

140 = 1 · \frac{dx}{dt} + 2 · \frac{dy}{dt} + 3 · \frac{dz}{dt} .

But

\frac{dy}{dt} = 2\frac{dx}{dt} , \frac{dz}{dt} = 3\frac{dx}{dt} ,

So going back

140 = 1 · \frac{dx}{dt} + 2 · (2\frac{dx}{dt}) + 3 · (3\frac{dx}{dt}) ,

giving

140 = \frac{dx}{dt}(1 + 4 + 9) = \frac{dx}{dt} · 14

Giving

\frac{dx}{dt} = 10 .

Going back to English: The rate of change of x with respect to time, at that above instant is 10.