Frequently Asked Questions

Question from Webwork, Sections 13.2-13.3 Problem 7:
I have VECTOR-valued function r'(t)=(sin a, sin b, ct). Can you look at this problem and give me a little explaination?
From what other students have told me, it looks like there is a glitch in this problem, and you should instead work the problem as follows: VECTOR-valued function r'(t)=(sin(at),sin(bt),ct).
Editor's note: the vector-valued function notation is not working on this page due to HMTL, but consider the outer parentheses above as pointed brackets.

Question from Webwork, Sections 12.4-12.5 Problem 14:
I got t = 18/17, and then I tried plugging in 18/17 for t in x, y, and z but webwork says its wrong. HELP!
Assuming you got the first part correct, we're going to use that to find the second part. The first question I have is how do you know if a point is in the yz-plane? It should have x-coordinate equal to 0. Thus we set the answer from the first part for x equal to 0 and solve for t. Then we use that t in both the y and the z to get the point.
This is because the point Q of intersection must be on the line AND on the plane, so there needs to be an (x,y,z) that satisfies the plane equation and there needs to be a t that gives you (x(t),y(t),z(t)) in the parametrization of the line.

Question from Webwork, Sections 12.4-12.5 Problem 8:
How do you get the magnitude of the cross product of two constants which are named vectors?
For this problem, there is a formula for the magnitude of the cross product that uses magnitudes of the two vectors as well as the angle between them. See if you can find it in the notes or in the book.

Question from Webwork, Sections 12.4-12.5 Problem 5:
Do I use the parallelepiped equation and just divide the complete answer by 2 to get a triangle? Is there a different way to do this problem?
Recall from Section 12.4 that the norm of the cross product of two vectors v and w is equal to the area of the parallelogram spanned by these vectors (think back to the Parallelogram Law).
You are given a triangle OPQ, and you can divide a parallelogram into two triangles of equal area.
So use one side of the triangle as your vector v and use the other side (with same base point) as vector w to get the area of the parallelogram and just divide by two.

Question from Webwork, Sections 12.4-12.5 Problem 4:
For this problem I have vector <2, -3, -1>, and it wants me to calculate vXi vXj vXk .
There are two different ways to think of this problem: either entirely in i,j,k notation or entirely in < , , > notation.
In the first case, we rewrite <2,-3,-1> = 2i -3j -k and cross this with, for example, i. Then by distribution, we get:
(2i -3j -k)x(i)=2(ixi)-3(jxi)-(kxi), noting that order is very important here since the Cross Product is NOT commutative (in fact it is ANTI-commutative, meaning that switching the order gives a minus sign).
Also note that a vector crossed with itself is 0, and this fits with one of the equations we have: ||v x w|| = ||v|| ||w|| sin (theta).
Thus we get (0)-3(jxi)-(kxi)= (-3)(-k)-(j) = -j+3k.
Or, alternately, if we drop the i,j,k notation, we can think of, for example, i=<1,0,0> and cross <2,-3,-1> by <1,0,0> using the determinant form, through which we arrive at the same answer.

Question from Webwork, Sections 12.2-12.3 Problem 12:
This problem tests that you understand the properties of the dot product, including the Distributive Property, the Relation with Length Property, and Pulling out Scalars (parts of Thm 1 from 12.3).
For constants a,b,c and vectors u,v:
au . (bu + cv) = (au . bu) + (au . cv) by Distribution
= (ab)(u . u) + (ac)(u . v) by Pulling out Scalars
= (ab)||u||^2 + (ac)(u . v) by the Relation with Length Property
Luckily this problem gives us ||u|| and (u . v), otherwise we wouldn't be able to solve it!

Question from Webwork, Sections 11.5-12.1 Problem 1:
When the parabola is centered at the origin, the value "c" is the distance between the vertex and the focus. Since the [distance between the vertex and the focus] is the same as the [distance between the vertex and the directrix], the (endpoint of the perpendicular from the vertex to the) directrix is going to be a distance "c" from the vertex.
So if your vertex is (0,0) and your focus is (0,1), then your directrix is going to be y = - 1 . However, if your center is ( alpha , beta ) and your focus is ( alpha , beta + c ), then your directrix is going to be y = ( beta - c ) .

Question from WebWork, Sections 11.3-11.4 Problem 7:
Let me see if I can re-state the problem: there are two circles that intersect, and we need to find out the POLAR coordinates of one point where they intersect. The best way to think about this problem is to find the equations of both circles and then try to solve them together. Circle C is centered at the origin and has radius 5. This is an easy equation in polar coordinates. Circle K intersects the origin and lies along one of the axes, so it is like one of the problems we've looked at in class, and using the other end of the diameter, you should be able to find its polar equation, too.
Remember that the point has to lie on BOTH circles, so it has to work in BOTH polar equations.
To find the limits of integration, your best best is to draw both graphs to give you the correct intuition (e.g. which quadrants), but you must solve for them algebraically. Say you have two POLAR equations r_1(theta) and r_2(theta). To find their points of intersection, set r_1 = r_2 (since they have the same radius) and then use that to solve for theta. Note that you should obtain two angles, so you might need to check quadrants besides the first one.

Question from WebWork, Sections 7.6-7.7 Problem 7:
My answer is the sum of constants that are ln's and arctan's of different numbers, but I keep getting the answer wrong. Is this the correct way to enter them?
Yes, it is. You may have the correct terms with the wrong signs in front of them. Be careful during your u-substitutions:
for example, for the integral from -2 to 1,
(x=-2 to x=1) might be replaced with (u=8 to u=5).
Even though these u's are in the wrong order, you have to leave them that way because that's how the problem is stated.

Question about tutoring:
Where can I get free tutoring?
Free math tutoring is available in 39 Allen, 9:30 - 5:30 daily, M-Thurs and 9:30 - 3:00 Friday.
However, as some students have learned already, these tutors may have a different perspective than the one I am teaching, and you might be confused if they solve a problem a different way. My suggestion: FORM A STUDY GROUP WITH YOUR CLASSMATES!

Question from the WebWork, general:
How do you enter theta on WebWork?
You are not allowed to enter theta as a variable in WebWork unless the problem gives you theta to begin with (and that is unlikely). You must remember to end the problem with whatever variable it starts with by substituting back.

Question from WebWork, Sections 7.3-7.4 Problem 5:
I used Equation 25 for this problem. I don't know what went wrong.
It's easy to make mistakes when you rely on remembering a complicated formula like Equation 25. Alternatively, and perhaps to better help you prepare for this material, you can use the trig identity:
sin(2x) = 2 (sin x) (cos x)

Question from WebWork, Sections 7.3-7.4 Problem 11:
"It seems that when WebWork was updated they changed Section_7.3_7.4 problem #11 for the worse. It now has errors in it. We are writing to get them to correct it for subsequent semesters.
Thanks, Wayne Britt" (Course Coordinator)

Question from WebWork, Sections 7.3-7.4 Problems 11 and 13:
The answers I have entered for the integrals are displayed as incorrect, but I have differentiated to verify my answers, and they should be correct.
It looks like WebWork wants your answer without extra constants in it. For example, you can rewrite:
ln(|(x/2)+(.../2)|)+C =
ln(|(x+...)|(1/2))+C =
ln(|x+...|)-ln2+C =
ln(|x+...|)+C for some DIFFERENT constant C.
Rather than your answer being incorrect, I suspect it's just not in the form WebWork wants to see it in.

Question from WebWork, Sections 7.1-7.2 Problem 5:
It wants the error bound for T_20 of the integral from 0 to 4 of f(x)= 3e^(-x/4). I find that the second derivative of f(x)=(3/16)e^(-x/4). I use your technique and get as far as (3/16) >= (3/16)e^(x/4) >= (3/16)e^(-1). Can you tell me if I've made any mistake up till then? Im not quite sure what to do from here to find K2.
I'm not sure if you mistyped or if you didn't get all the way. If you need to get f''(x)=(3/16)e^(-x/4), then you'll want to proceed by:
0 <= x <= 4
0 >= -x >= -4
0 >= (-x/4) >= -1
1 >= e^(-x/4) >= e^(-1)
(3/16) >= (3/16)3^(-x/4) >= (3/16)e^(-1)

Now, K_2 = | f''(x) |, so we need to take the absolute value of (3/16)e^(-x/4). Since f''(x) is between two positive numbers, the absolute value doesn't matter and we choose the larger number, which is (3/16), and set K_2=(3/16).

Then we plug this into the Error(T_20) formula along with N=20, b=4 and a=0.

Question about WebWork due dates post-Gustav:
Is the Section 7.1-7.2 homework still due on Sept 4th?
I've postponed the due dates. I hope all is well with you, your loved ones, and your places of shelter during these trying times.
In case you feel overwhelmed, you may contact LSU’s Mental Health Services at 225-578-8774 or THE PHONE at 225-924-5781, a 24-hour crisis intervention and referral service of the Baton Rouge Crisis Intervention Center, provided especially for LSU students. All calls are confidential.

Question on Rogawski section 7.1 #1:
I keep getting the wrong answer for the integral from 0 to 4 of x^2 dx. I found (b-a)/N = 1 and the equation
T_4= 1/2(1)*(1(1^2)+2(2^2)+2(3^2)+1(4^2)) = 21.5.
It looks like you are missing x_0 = 0 in your expansion above. Then you should have:
T_4 = 1/2(1)*(1(0^2)+2(1^2)+2(2^2)+2(3^2)+1(4^2))
Remember that a = x_0 and b = x_N should ALWAYS be in T_N and S_N (and that they are the only ones that carry a coefficient of 1.