Sunday, September 25, 2016
Thursday, September 22, 2016
10.9#3
Shouldn't the 'j' component of the velocity for this be 3cos(3t)+0 since v(0)=i+0j+k?
*********************
No, because v(t)=∫_0^t a(s)ds+v(0), and ∫_0^t -9sin(3s)ds=3cos(3t)-3cos(3x0)
*********************
No, because v(t)=∫_0^t a(s)ds+v(0), and ∫_0^t -9sin(3s)ds=3cos(3t)-3cos(3x0)
Wednesday, September 21, 2016
Saturday, September 17, 2016
10.7#6 Hint
Dr. Taylor,
How do I go about finding this? The example in the book isn't very helpful here because they have a graph of the two surfaces and can tell what x and y are just by looking at the graph. What is a good
analytical way to arrive at the answer every time?
**********************************
Well the short answer has three steps 1) substituting one equation into the other to get a single equation in two variables, 2) remember what you learned in Calc2 about parameterizing plane curves, then 3) re-substitute your parameterization of a planar curve back into 3-D to get your intersection
Try that and let me know how it goes and what you try, I'll post your answer here.
How do I go about finding this? The example in the book isn't very helpful here because they have a graph of the two surfaces and can tell what x and y are just by looking at the graph. What is a good
analytical way to arrive at the answer every time?
**********************************
Well the short answer has three steps 1) substituting one equation into the other to get a single equation in two variables, 2) remember what you learned in Calc2 about parameterizing plane curves, then 3) re-substitute your parameterization of a planar curve back into 3-D to get your intersection
Try that and let me know how it goes and what you try, I'll post your answer here.
Friday, September 16, 2016
10.8#3 (edited)
I can not figure out how to solve this problem. I find the derivative of
the vector then set my limits from t=1 to t=2 and then use the formula:
Integral from 1 to 2 sqrt((4^2)+(4x)+(2/x)) and this does not work. What am
I doing wrong?
*************************
You have the limits of integration correct, but it looks like you forgot to square the derivatives of the y and z-components of r.
*********************************
Hello Professor. I've got the steps down to do this problem. I found the upper and lower T values needed, and I have set up the equation. The problem is, I have no clue how to integrate the mess after it is all set up. A clue or step in the right direction would be greatly appreciated!
************************
Well, a good thing to ask yourself might be something like: "Huh. There are very few functions that I've been taught how to integrate. The arc-length formula has the square root of some stuff for an integrand. What stuff could be inside that square root that I might reasonably be expected to come up with an anti-derivative for, after taking the square root?"
In fact there are very few things that could be inside the square root that would allow you to do that integral, and if you've done the sum of squares of the components of the derivative correctly when you look at that you see that it turns out to be one of those very few things.
Thursday, September 15, 2016
10.7#1
I am having a bit of an issue with problem 1 on
the 10.7 WebWork homework. I believe that WebWork is not grading my
work correctly (it is marking a correct answer as incorrect). The problem is to find the domain of the vector function r(t)=<e^(-4t),t/sqrt(t^2-49),
However, WebWork marks this answer as incorrect:
****************************
No, it's grading you correctly (this time, although it does make mistakes sometimes). It's interesting that Wolfram alpha is incorrect, isn't it?So, you made a combination of two mistakes: the domain of the second component is (-INF,-7) U (7,INF) and the domain of the third component is (-INF,INF)--the cube of a negative number is negative, so likewise for the cube root. Hence the domain is (-INF,-7) U (7,INF) .
10.7#16
I can't seem to solve question 16 on 10.7 and need help. I got part A, but my method doesn't work for parts B or C, so I think I just got lucky on part A. Thanks.
*******************
One parameterization of the line will be
r(t)=u+tv,
where u=<1,-1,2> and v=<4,1,-4>, but v can be replaced by α<4,1,-4> for any nonzero constant α and u can be replaced by any other vector with time on the line, i.e. u can be replace by
<1,-1,2> +β<4,1,-4> for any real number β. Since u+0v = <1,-1,2> and u+1v =<5,0,-2>, all we have to do to get the answer to (a) is replace v by v*=v/5, so now we have the equations
u+0v* = <1,-1,2> and u+5v* =<5,0,-2>, and your answer <1,-1,2>+t<4/5,1/5,-4/5> is the correct one. In more generality, you would need to look at something like
r(t)=(<1,-1,2> +β<4,1,-4>)+tα<4,1,-4>
and choose α,β to fit your parameters. For example in case (b) r(5)= (<1,-1,2> +β<4,1,-4>)+5α<4,1,-4>=<1,-1,2> and
r(7)= (<1,-1,2> +β<4,1,-4>)+7α<4,1,-4>=<5,0,2> so we get equations
5α+β=0 and 7α+β=1 from which 2α=1=>α=1/2, and β=-5/2,
r(t)=(<1,-1,2> -5/2<4,1,-4>)+t/2<4,1,-4>
= <-9,-7/2,12> +t<2,1/2,-2>
*******************
One parameterization of the line will be
r(t)=u+tv,
where u=<1,-1,2> and v=<4,1,-4>, but v can be replaced by α<4,1,-4> for any nonzero constant α and u can be replaced by any other vector with time on the line, i.e. u can be replace by
<1,-1,2> +β<4,1,-4> for any real number β. Since u+0v = <1,-1,2> and u+1v =<5,0,-2>, all we have to do to get the answer to (a) is replace v by v*=v/5, so now we have the equations
u+0v* = <1,-1,2> and u+5v* =<5,0,-2>, and your answer <1,-1,2>+t<4/5,1/5,-4/5> is the correct one. In more generality, you would need to look at something like
r(t)=(<1,-1,2> +β<4,1,-4>)+tα<4,1,-4>
and choose α,β to fit your parameters. For example in case (b) r(5)= (<1,-1,2> +β<4,1,-4>)+5α<4,1,-4>=<1,-1,2> and
r(7)= (<1,-1,2> +β<4,1,-4>)+7α<4,1,-4>=<5,0,2> so we get equations
5α+β=0 and 7α+β=1 from which 2α=1=>α=1/2, and β=-5/2,
r(t)=(<1,-1,2> -5/2<4,1,-4>)+t/2<4,1,-4>
= <-9,-7/2,12> +t<2,1/2,-2>
Wednesday, September 14, 2016
Exact Answers? (edited)
Hello, Professor, this is ********** in your MAT 267 MWF class. In question 1 of section 10.8 on the "Test 1 Review," it asks us to compute the length of a curve. To solve, I set up an integral and then plugged the integral into my calculator's integral solver to get a numerical answer (28.73). However, the answer key lists the answer as (1/27)(85^(3/2)-8), which computes to approximately 28.73. My question is, on the exam if I had skipped the step of doing a u-substitution to solve the integral by hand and instead set up the initial integral for arc length, plugged it in to a calculator, and given a numerical answer, would I be deducted points?
Sincerely,
***************
It depends on the question. If the question asks for an exact answer, you would not get full credit. If the question just asks for an answer you can use your calculator.
******************
Pardon me, reading your new post I found that under "arc length" you specify we need an exact answer (no calculator answer).
******************
******************
Pardon me, reading your new post I found that under "arc length" you specify we need an exact answer (no calculator answer).
******************
No, I said it depends on the way the question is worded.
Exam Q
Professor,
I was wondering if section 10.8 will be included on our exam this Friday?
***************
10.8 is fair game
Tuesday, September 13, 2016
Q on 10.8#2
Hello, Dr. Taylor.
I am not sure what integration technique I can use to integrate the magnitude of r'(t).
Should I be approaching this problem
differently?
Thanks.
******************
Well, r'(t)=<10√2, 10e^(10t), -10e^(-10t)>, so
thus
at which point you can look up the answer in your table of integrals.
Remark: I bet someone thought that you would never need to know hyperbolic trig functions
Monday, September 12, 2016
Test 1 Review Question
Hello Professor,
I was wondering if there was an answer key to the "Test 1 Review" so I can make sure I am doing my work correctly.
Thank you
******************
There is an answer key to Practice Test 1, how did that one go? For the review, what specific questions do you have, and what did you do?
Test 1 Question
Dr. Taylor,
While looking at the practice exam I noticed that there was a link to "Test 1" in addition to the review and practice. Is this how we will be taking the actual test on Friday, or are we doing it in class? Thanks.
*******************
The link to test 1 takes you to another practice exam. We'll be taking the exam in class.
Sunday, September 11, 2016
10.5#17 again
Hello, Dr. Taylor.
I am lost on this question - if I need to find the angle of intersection between the planes, I would assume I can just find the angle between their two normal vectors using the dot product formula
A dot B = ||A||||B||cos(theta), where A and B are the two normal vectors to the planes. I plotted both planes in Wolfram and visually it seems like that should give the right answer. I also found an answer on MathExchange that uses this method to solve a similar problem (and defines the angle of intersection between two planes as being the angle between the normal vectors). However, no matter the angle I find (obtuse or acute) with this method, the angle is reported as being incorrect
How should I go about doing this if this is not the correct method?
*****************
You aren't doing anything wrong, except you let webwork fool you into thinking that the normal vector of the first plane is <1,-4,1>. It's not. Notice that the coefficient of x is -4, while that of y is 1? The coefficient of x goes in the x-position and of y goes in the y-position. This means that the normal vector is <-4,1,1>.
I am lost on this question - if I need to find the angle of intersection between the planes, I would assume I can just find the angle between their two normal vectors using the dot product formula
A dot B = ||A||||B||cos(theta), where A and B are the two normal vectors to the planes. I plotted both planes in Wolfram and visually it seems like that should give the right answer. I also found an answer on MathExchange that uses this method to solve a similar problem (and defines the angle of intersection between two planes as being the angle between the normal vectors). However, no matter the angle I find (obtuse or acute) with this method, the angle is reported as being incorrect
How should I go about doing this if this is not the correct method?
*****************
You aren't doing anything wrong, except you let webwork fool you into thinking that the normal vector of the first plane is <1,-4,1>. It's not. Notice that the coefficient of x is -4, while that of y is 1? The coefficient of x goes in the x-position and of y goes in the y-position. This means that the normal vector is <-4,1,1>.
Saturday, September 10, 2016
10.5#8 hints
Hello! I have a question about this problem. I feel that I am missing
something here. I don't know where to begin, and I am unsure of the
process. Any help would be greatly appreciated!
***********
OK, first of all, go back to the lecture notes, we covered this. The line of intersection consists of points that are in both planes. The parametric form of a line is given by data that consist of 1) a vector (which we called u in the lecture notes) with endpoint at some single point in the line (which must consist a point in both planes) and 2) a vector parallel to the line (which we called v in the lectures notes and which must therefore of a vector in both planes, hence perpendicular to both normal vectors--hint, hint, hint). The problem helps you choose both of these by constraining u so that its z-component is zero and constraining v so that its x-component is 12.
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(You can click on the image below to enlarge)
10.5#10
On problem 10 of 10.5, two parametric equations are given. To solve for
the intersection we solve for two variables, usually t and s in the
equations. Unfortunately these equations have only t’s. The problem can
be solved by assuming the equations for L2 use a different variable,
independent from t.
******************
No typo, really, but you make a useful point. It's easy to make the mistake of thinking that the t in the parametric equation of a line refers to anything real. It doesn't, it's just a dummy variable. It's just a way of labeling points on the line. For example L1: x=2+12t, y=6+8t, z=4+2t is another labeling that would do just as well, and the dummy variable associated t associated with this labeling would be different.
It's true that if you expect t to be a real thing and look for an intersection by setting 14+6t=-18+7t, 14+4t=-10+6t, 6+t=-5+4t, there will almost always be no such t. And yes, the correct way to look for an intersection would be to allow give two different names to the parameters for the two different lines. The names t and s would work as well as any other two names.
(to misquote Wild Bill Shakespeare, a variable by any other name would be just as good, or smell as sweet. Wild Bill had much to say on this subject, btw:
JULIET
******************
No typo, really, but you make a useful point. It's easy to make the mistake of thinking that the t in the parametric equation of a line refers to anything real. It doesn't, it's just a dummy variable. It's just a way of labeling points on the line. For example L1: x=2+12t, y=6+8t, z=4+2t is another labeling that would do just as well, and the dummy variable associated t associated with this labeling would be different.
It's true that if you expect t to be a real thing and look for an intersection by setting 14+6t=-18+7t, 14+4t=-10+6t, 6+t=-5+4t, there will almost always be no such t. And yes, the correct way to look for an intersection would be to allow give two different names to the parameters for the two different lines. The names t and s would work as well as any other two names.
(to misquote Wild Bill Shakespeare, a variable by any other name would be just as good, or smell as sweet. Wild Bill had much to say on this subject, btw:
JULIET
'Tis but thy name that is my enemy;)
Thou art thyself, though not a Montague.
What's Montague? it is nor hand, nor foot,
Nor arm, nor face, nor any other part
Belonging to a man. O, be some other name!
What's in a name? that which we call a rose
By any other name would smell as sweet;
So Romeo would, were he not Romeo call'd,
Retain that dear perfection which he owes
Without that title. Romeo, doff thy name,
And for that name which is no part of thee
Take all myself.
10.5#17
Hi Dr. Taylor,
I've attempted this problem 8 times and my answers are
still wrong. I used the equation for the angle between planes. Am I doing
something wrong?
*****************************
Well, 142 radians is a lot of radians and 2.49 degrees is very few degrees. That should make you suspicious about which one is which. In fact, 2.49 radians divided by π x 180 degrees = 142 degrees, whereas 142 radians divided by π x 180 degrees amounts to 8136 degrees. So: at the very least you have degrees and radians reversed.
Now let's see, you didn't show me how you computed your angles but you should have had:
cos(θ)=u.v/(|u| |v|)
=<4,-4,-1>.<4,-4,-3>/(√(4^2+(-4)^2+(-1)^2)√(4^2+(-4)^2+(-3)^2))
=(16+16+3)/(√(16+16+1)√(16+16+9))
=39/(√37√45)
=0.956
which is a number very close to 1, hence we expect that θ is close to zero. In fact computing the arccos what we arrive at is θ=0.30 radians, which is about 17.1 degrees.
I've attempted this problem 8 times and my answers are
still wrong. I used the equation for the angle between planes. Am I doing
something wrong?
*****************************
Well, 142 radians is a lot of radians and 2.49 degrees is very few degrees. That should make you suspicious about which one is which. In fact, 2.49 radians divided by π x 180 degrees = 142 degrees, whereas 142 radians divided by π x 180 degrees amounts to 8136 degrees. So: at the very least you have degrees and radians reversed.
Now let's see, you didn't show me how you computed your angles but you should have had:
cos(θ)=u.v/(|u| |v|)
=<4,-4,-1>.<4,-4,-3>/(√(4^2+(-4)^2+(-1)^2)√(4^2+(-4)^2+(-3)^2))
=(16+16+3)/(√(16+16+1)√(16+16+9))
=39/(√37√45)
=0.956
which is a number very close to 1, hence we expect that θ is close to zero. In fact computing the arccos what we arrive at is θ=0.30 radians, which is about 17.1 degrees.
10.5#11
Following the book, I believe that I have correctly solved for the plane's
equation. The normal vector is expressed as <-2,4,1> and following the book
my specific points should lead to -2(x--4)+4(y--3)+1(z-2)=0. Am I correct
in this assumption or am I missing something?
*****************
Nope, not correct. A parametric equation for a line looks like r(t)=u+tv, where the vector u just gives a point on the line and the vector v determines the direction of the line: the direction of the line is read from the coefficients of t. Instead you've used the u.
Friday, September 9, 2016
Wednesday, September 7, 2016
question regarding section 10.5
Professor,
I’m wondering about an equation, equation 4, given on page 569 of our textbook. Do these statements show the right relationship?
__________________________________
The first equation is fine, it describes the line segment starting at the vector (point) r_0 and passing through to the endpoint vector (point) r_1, because r(0)=r_0 and r(1)=r_1. Note that this equation can be rewritten as
r(t) = (1-t)r_0 + t r_1
= r_0 - t r_0 + t r_1
= r_0 + t (r_1 - r_0)
= r_0 + t v
where v= (r_1 - r_0), which is just the form of a line as we described today in lecture, and allows all real values of t, i.e. does not require 0≤t≤1.
I'm not sure what the second equation means, since I'm not sure what you mean by r_a and r_b, but it would give some line segment not necessarily beginning or ending at r_a and r_b/
(Edit the morning after: after thinking a bit more about your question I think I understand where you were coming from: you wanted to know if 0 and 1 are special or if any numbers would do. The answer is no, any numbers will not do and 0 and 1 are special. The whole point is that
(Edit the morning after: after thinking a bit more about your question I think I understand where you were coming from: you wanted to know if 0 and 1 are special or if any numbers would do. The answer is no, any numbers will not do and 0 and 1 are special. The whole point is that
(1-t)u+t v
interpolates between u and v as t goes from 0 to 1, no matter what vectors u and v are.)Tuesday, September 6, 2016
10.4#5
I need help on this problem on how to find the orthogonal vectors
of 'a' and 'b'. I've tried it many times! I computed the cross product of
'a' and 'b' and then used the dot product using each vectors. Please
explain to me what I'm doing wrong.
(you can click on the image to enlarge it)
************
Well, you're doing a number of things wrong, starting with your interpretation of the problem.
First of all, your supposed to find two unit vectors u^ and v^, each of which is orthogonal to both of the vectors a and b. One thing you choose to do is to compute the cross product a x b, because this will give you a vector orthogonal to both a and b. It looks like you made a simple arithmetic error in this computation however, because for example
a x b = (-3i-5j+4k)x (-2i+4j+4k)
= -12( i x j)+10( j x i)-12( i x k)-8( k x i)-20( j x k)+16( k x j)
= -22( i x j) - 4( i x k) - 36( j x k)
= -22 k + 4 j - 36 i
= - 36 i + 4 j - 22 k
While this vector is orthogonal to both a and b, it is not a unit vector. Since the norm of this vector is ||a x b||=√(36^2+4^2+22^2)=√1796=2√449, the unit vector pointing in the direction of a x b is
a x b/(2√449). Of course another unit vector perpendicular to both a and b is b x a/(2√449), which points in the opposite direction
of 'a' and 'b'. I've tried it many times! I computed the cross product of
'a' and 'b' and then used the dot product using each vectors. Please
explain to me what I'm doing wrong.
(you can click on the image to enlarge it)
************
Well, you're doing a number of things wrong, starting with your interpretation of the problem.
First of all, your supposed to find two unit vectors u^ and v^, each of which is orthogonal to both of the vectors a and b. One thing you choose to do is to compute the cross product a x b, because this will give you a vector orthogonal to both a and b. It looks like you made a simple arithmetic error in this computation however, because for example
a x b = (-3i-5j+4k)x (-2i+4j+4k)
= -12( i x j)+10( j x i)-12( i x k)-8( k x i)-20( j x k)+16( k x j)
= -22( i x j) - 4( i x k) - 36( j x k)
= -22 k + 4 j - 36 i
= - 36 i + 4 j - 22 k
While this vector is orthogonal to both a and b, it is not a unit vector. Since the norm of this vector is ||a x b||=√(36^2+4^2+22^2)=√1796=2√449, the unit vector pointing in the direction of a x b is
a x b/(2√449). Of course another unit vector perpendicular to both a and b is b x a/(2√449), which points in the opposite direction
Monday, September 5, 2016
10.4#8
Hello, Dr. Taylor.
I solved this problem by projecting vector QP onto vector QR, finding the angle measure between vectors QP and QR with the arccos function, and then calculating sqrt(14)sin(theta) to find the
distance d. However, I am also aware that using cross products rather than projections is the focus of this lesson - how can this problem be solved using cross products? Using the formula ||U x V|| = ||U|| ||V|| sin(theta), I would end up with 14sin(theta) instead of sqrt(14), since both QP and QR
have magnitudes of sqrt(14).
Thanks for your assistance.
************
OK, some short answers are 1) you started on one possible right track using projections, but then you got off the trail and started bushwhacking through the weeds, and 2) that nice formula
||U x V|| = ||U|| ||V|| sin(theta)
you use comes in three pieces, it's up to you to figure out how to take the pieces you want.
But when doing these problems I think it's important to take an organized approach, and the beginning should be TO DRAW A PICTURE OF THE PROBLEM from a perspective that captures the essential features. This three-D plot captures EVERYTHING:
I solved this problem by projecting vector QP onto vector QR, finding the angle measure between vectors QP and QR with the arccos function, and then calculating sqrt(14)sin(theta) to find the
distance d. However, I am also aware that using cross products rather than projections is the focus of this lesson - how can this problem be solved using cross products? Using the formula ||U x V|| = ||U|| ||V|| sin(theta), I would end up with 14sin(theta) instead of sqrt(14), since both QP and QR
have magnitudes of sqrt(14).
Thanks for your assistance.
************
OK, some short answers are 1) you started on one possible right track using projections, but then you got off the trail and started bushwhacking through the weeds, and 2) that nice formula
||U x V|| = ||U|| ||V|| sin(theta)
you use comes in three pieces, it's up to you to figure out how to take the pieces you want.
But when doing these problems I think it's important to take an organized approach, and the beginning should be TO DRAW A PICTURE OF THE PROBLEM from a perspective that captures the essential features. This three-D plot captures EVERYTHING:
but maybe it's too much information? This simpler diagram says what we need to know:
See that right triangle? You know the size of the hypotenuse, and the piece || to QR is the projection so you know it's size. The other piece is the answer you need. You can get that in several ways. Hint: cos^2(x)+sin^2(x)=1.
Sunday, September 4, 2016
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