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course mth 277
9/4 6pm
120827Fill in your responses to the questions below sometime after you have submitted a couple of exercises using the Submit Work Form, per the preceding email. Then submit a copy of your responses using the Submit Work Form at
http://www.vhcc.edu/dsmith/submit_work.htm.
If you have questions use the Question Form at
http://vhcc2.vhcc.edu/dsmith/forms/question_form.htm
Your Assignments Page is located at
http://vhcc2.vhcc.edu/dsmith/geninfo/qa_query_etc/multivar/assignments.htm
If you scroll a little ways down the page you will see the following links:
Precalculus II open qa 09
Precalculus II open qa 10
Precalculus II open qa 12
You should work through the the last of the three (qa12). Work each qeustion out on paper. It's likely that you will understand some of the questions completely and after working them out will be able to verify that you have solved the problems correctly. If so you can simply indicate 'OK' in the self-critique, and there will be no reason for you to have to fill in your solutions to such problems. In some cases you might have missed a minor point or two; if this is the case you can again simply address the question in your self-critique, this time discussing what you missed and demonstrating your understanding. On problems you simply do not understand you should provide the entire solution where prompted.
You will then see a series of qa's reviewing some of your first-semester calculus. You may submit any exercises you wish. If you don't submit them, I'll assume that you understand these procedures.
For a review of first-semester calculus:
The page http://vhcc2.vhcc.edu/cal1fall/ is the homepage for the instructor's Calculus I course. The Open QA's for that course would provide a good review of the basic topics covered in most first-semester classes, and should at least give you a good start on some of the topics that might not have been covered in your first-semester course.
The coverage of topics is roughly as follows:
The open qa's for Assignments 11-12 cover the rules of differentiation.
The open qa's for Assignments 13-15 introduce applications of differentiation, tangent lines and the differential. ** Do this and submit it. **
The open qa for Assignment 16 addresses implicit differentiation.
The open qa's for Assignments 1-10 introduce you in a unified way to the concepts of differentiation and integration, beginning with the concept of rate of change.
The open qa for Assignment 10 is fairly challenging and introduces the central ideas of integration in the context of finding present and future values of an investment.
Most multivariable calculus students will benefit greatly from this refresher on standard first-semester calculus topics. All these ideas extend into three dimensions and therefore appear in Multivariable Calculus.
A listing of the Calculus I qa's, in the order mentioned above, follows:
qa11 on the rules of differentiation for the basic functions
qa12 on the product, quotient and chain rules
qa13 on applications of the chain rule
qa14 on tangent line approximations
qa15 on the differential and the tangent line
qa16 on implicit differentiation
qa01 on the idea of rate of change
qa02 more about the idea of rate of change
qa03 graphical representation of rate of change
qa05 average rate of change, changing rate of change
qa06 finding change in quantity given rate of change information
qa07 rate of change functions
qa08 approximate depth graph constructed from rate function
qa09 predictor-corrector
qa10 income streams
Problem Set
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Questions to be submitted:
In addition to Precalculus I qa_12 and Calculus qa14 and specified above, submit the following questions in the usual manner:
Explain your reasoning as you answer each question. If you have done the Precalculus II exercise specified above, you should find many of these questions to be straightforward. Some, however, are challenging. Do your best on each question.
`q001. Consider the equation x^2 + 4 y^2 + 9 z^2 = 36. The set of points is x-y-z space satisfying this equation forms an ellipsoid, sort of like a football with rounded ends.
If z = 0, the equation is simplified. Sketch the resulting curve in an x-y coordinate plane, and describe your sketch.
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You get x^2/36 + y^2/9 = 1. This is an ellipse with vertices at (+- 6,0) and (0, +- 3).
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If z = 1, the equation is again simplified. Sketch the resulting curve in an x-y coordinate plane, and describe your sketch. Where would this curve be located in x-y-z space?
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x^2 + 4 y^2 + 9 (1)^2 = 36. x^2 + 4y^2 = 36-9. x^2/27 + y^2/6.25 = 1. Another ellipse but with vertices at pts (+- sqrt(27), 0,1) and (0, +- sqrt(6.25),1). This ellipse would be flat in the x-y plane, but would exist 1 unit upwards 'in the air' so to speak. Were the x-y plane a table, the ellipse would lie on a plane parallel to the table but 1 unit above it.
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If z = 2 the equation is really simplified. What is the equation for z = 2, and what point or points of the x-y plane satisfy this equation? What are the corresponding points in x-y-z space?
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x^2 + 4 y^2 + 9 (2)^2 = 36. x^2 + 4 y^2 + 36 = 36. x^2 + 4 y^2 = 0. 4y^2 = -x^2.
Looks like you will end up with imaginary solutions.
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There are lots of imaginary solutions, but they aren't relevant to this course.
There is a single real solution: x = 0, y = 0.
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Now what do you get if x = 0? Describe the curve, and do your best to describe its location in x-y-z space.
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x^2 + 4 y^2 + 9 z^2 = 36. y^2/9 + z^2/4 = 1. Ellipse in the y-z plane with vertices at (0,+- 3, 0) and (0,0, +- 2). Its center is at the origin.
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What do you get if y = 0? Describe the curve, and do your best to describe its location in x-y-z space.
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x^2 + 9 z^2 = 36. x^2/36 + x^2/4 = 1. Ellipse in the x-z plane, center at the origin, vertices at (+- 6, 0, 0) and (0,0, +- 2).
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`q002. Consider the equation x^2 + 4 y^2 - 9 z^2 = 36.
If z = 0, the equation is simplified. Sketch the resulting curve in an x-y coordinate plane, and describe your sketch.
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x^2 + 4 y^2 = 36. x^2/36 + y^2/9= 1. Ellipse in x-y plane, center at origin, vertices (+- 6,0) and (0, +- 3).
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If z = 1, the equation is again simplified. Sketch the resulting curve in an x-y coordinate plane, and describe your sketch. Where would this curve be located in x-y-z space?
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x^2 + 4 y^2 - 9 z^2 = 36. x^2 + 4 y^2 - 9 +9 = 36+ 9. x^2 + 4 y^2 = 45. x^2/45 + y^2/11.25 = 1. Ellipse in x-y plane, center at origin, vertices at ( +- sqrt(45), 0,1) and (0, +- sqrt(11.25),1). The x-y plane in which this ellipse lies is shifted 1 unit upwards in the z direction.
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If z = 2 the equation is really simplified. What is the equation for z = 2, and what point or points of the x-y plane satisfy this equation? What are the corresponding points in x-y-z space?
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x^2 + 4 y^2 - 9 (2)^2 = 36. x^2 + 4 y^2 - 36 = 36. x^2 + 4y^2 = 72. x^2/72 + y^2/18 = 1.
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Now what do you get if x = 0? Describe the curve, and do your best to describe its location in x-y-z space.
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x^2 + 4 y^2 - 9 z^2 = 36. 4 y^2 - 9 z^2 = 36. y^2/9 - z^2/4 = 1. This is a hyperbola in the y-z plane with asymptotes z= +- 2/3y. The vertices are (0, +- 3, 0) with symmetry about the z axis.
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What do you get if y = 0? Describe the curve, and do your best to describe its location in x-y-z space.
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x^2 + 4 y^2 - 9 z^2 = 36. x^2 -9z^2 = 36. x^2/36 - z^2/4 = 1. Hyperbola in x-z plane, vertices at (+- 6, 0, 0), symmetry about the z axis. Asymptotes z= +- 2/6 x or +-1/3 x.
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Summarize the differences between your results for this question and the results you obtained for question `q001.
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If z = 0 or some number then you get an ellipse in both situations. However, if z is included as a variable, the equation becomes the general form for a hyperbola in q002 instead of an ellipse. This is because of the change of signs (-z as opposed to + z).
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`q003. What is the slope of the curve x^2 + 4 y^2 = 36 at the first-quadrant point where x = sqrt(20)?
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x^2 + 4 y^2 = 36. f(x) = sqrt( 9- x^2/4). f'(x)= -x/4 (9-x^2/4). f'(sqrt(20)) = -sqrt(20)/4 (9- (sqrt(20))^2/4 = -2sqrt(5). -4.472.
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What is a vector parallel to the line tangent to the curve at this point?
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m = -4.472.
x= sqrt(20)
y= 2.
2= -4.472 (sqrt(20)) + b.
b= 21.999... 22.
y= -4.472 x +22.
4.472 x i + yj = 22.
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`q004. The surface x^2 + 4 y^2 + 9 z^2 = 36 is an ellipsoid in 3-dimensional space.
The equation 2 x + y - z = 4 describes a plane in 3-dimensional space.
If you solve the second equation for z, and plug the result into the first equation, you will eliminate z and obtain the equation of a curve in the xy plane.
What is this equation?
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z= 2x + y -4. x^2 + 4y^2 + 9 (2x + y -4)^2 = 36.
37x^2 + 13y^2 + 36xy - 144x - 72y + 108.
Not sure what to do with the 36xy...
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Put this equation into the form
a x^2 + b x + c y^2 + d y = e.
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37x^2 -144x + 13y^2-72y +36xy = -108
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Complete the square on x, and complete the square on y, to get an equation of the form a ( x - h ) ^2 + b ( y - k ) ^ 2 = f. What is your equation?
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37(x^2 - 144x/37) + 13(y^2 - 72y/13) + 36xy = -108.
37(x-72/37)^2 + 13(y -36/13)^2 +36xy = -108 + 37*(-72/37)^2 + 13*(-36/13)^2.
37(x-72/37)^2 + 13(y -36/13)^2 +36xy = 131.8
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Your equation is that of an ellipse, centered at the point (h, k). Graph this ellipse and describe your graph.
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37(x-72/37)^2 + 13(y -36/13)^2 +36xy = 131.8
(x-72/37)^2/3.56 + (y-36/13)^2/10.138 + 36xy/131.8 = 1.
Ellipse with center at (72/37, 36/13). I admit I used technology to graph this one to save some time. It looks like the ellipse axes are on sloped lines themselves (not actually parallel to the x or y axis). So the ellipse is 'lop-sided'. Its tilted to the left.
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You probably need technology on this one.
Those xy terms rotate the ellipse, so just completing the square doesn't work.
We could talk about rotations, but we probably won't.
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`q005. Show that the point where x = 6 cos(theta) and y = 3 sin(theta) lies on the ellipse x^2 + 4 y^2 = 36.
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You can do this logically. When you get the equation in the proper ellipse form, it becomes x^2/36 +y^2/9 = 1. Ellipse with vertices (+- 6,0) and (0, +-3). In polar equation form, x= r cos(theta) and y = r sin(theta). r being the distance away from the origin and theta being the angle from 0 degrees. If x = 6 cos(theta) and y = 3 sin(theta), then r = 6 and 3 respectively, which are the distances of the vertices points from the origin.
It can further be proven that these points are on the ellipse by letting theta = 90 degrees or pi/2 radians. When this is the case, x = 6 cos(pi/2) = 6 and y = 3 sin(pi/2)= 0. This is the coordinate for one of the vertices.
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Show that as theta takes on all values between 0 and 2 pi, the points where x = 6 cos(theta) and y = 3 sin(theta) match the points of the ellipse x^2 + 4 y^2 = 36 in a one-to-one fashion.
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If theta = pi/6, x = 6 cos(pi/6)= 3*sqrt(3). y = 3 sin(pi/6)= 3/2.
(3*sqrt(3))^2 + 4 (3/2)^2 = 36. 36=36.
theta = pi/4, x = 6 cos(pi/4)= 3*sqrt(2). y = 3 sin(pi/4)=3/2* sqrt(2).
(3*sqrt(2))^2 + 4 (3/2* sqrt(2))^2 = 36. 36=36.
theta = pi/3, x = 6 cos(pi/3)= 3. y = 3 sin(pi/3)= 3/2* sqrt(3).
(3)^2 + 4 (3/2* sqrt(3))^2 = 36. 36=36.
This is for the 1st quandrant. Since the center of the ellipse is at the origin, there is symmetry in each remaining quadrant, so this pattern will hold true for all quadrants.
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Very good.
Check my notes.
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