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Your solution:
The cross product's coefficients come out to be:
-sqrt(t)*(t+2) - (t^2-6t+5) i +
sqrt(t)*t^2 - (t-1)/(t+2) j +
t^3 - 5t^2 + 1 k
Therefore, the domain of t is (-inf,2) u (2,inf).
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negative values of t are not in the domain because of the sqrt(t) factor
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Question: Describe the graph of G(t) = (sin t)i + (cos t)j + (4/3)k
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Your solution:
The graph of x = sin t, y = cos t results in an ellipse centered at the origin. Since z is constantly equal to 4/3, it is irrelevant.
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The ellipse is located in the plane z = 4/3, which is 4/3 of a unit above the xy plane.
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Question: Given F(t)= (t)i - 5(e^t)j +(t^3)k, G(t) = ti - (1/t)k and H(t) = (t*sin t)i + (e^-t)j, find H(t) dot [G(t) X F(t)]
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Your solution:
The cross product of G and F:
-5e^t/t i + (t^4+1) j - 5te^t k
The dot product of this and H:
-5 sin(t)*e^t + t^4*e^-t + e^-t j
e^-t(t^4 + 1) - 5e^t*sin(t)
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Question: Find a vector function F whose graph is the curve given by the equation x/5 = (y-3)/6 = (z+2)/4.
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Your solution:
Multiply through by 120:
24x = 20(y-3) = 30(z+2)
24x - 20y + 60 - 30z - 60 = 0
24x - 20y - 30z = 0
This equation describes a single plane in space.
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x/5 = (y-3)/6 = (z+2)/4
does not define a plane, but is rather the symmetric equation of a straight line through (0, 3, -2) and parallel to 5 i + 6 j + 4 k.
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The vector-valued function
G(t) = (2 + 3 t) i + (-5 - 2 t) j + (4 - t) k
is an example of a function which describes a straight line.
What is the function F(t) for the line described by the given equation?
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The normal vector is equal to - a, b, and c come from the above equation:
<24,-20,-30>
We then create a point using the equation:
(5,9,2)
Then construct an equation using the vector and point:
24(x-5) - 20(y-9) - 30(z-2) = 0
24x - 120 - 20y + 180 - 30z + 60 = 0
24x - 20y - 30z = 120
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Given Solution:
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Self-critique (if necessary):
I'm not exactly sure what you were looking for in this question. You wanted a vector equation, but I fail to see how you could get a vector to represent a plane (in this case). Am I missing something?
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Check my note above.
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Question: Find the limit as t -> 2 of ((t^4-2)/(t-2))i + ((t^2-4)/(t^2-2t))j + ((t^2 + 3)e^(t-2))k.
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Your solution:
Break it into three limits:
t^4-2/t-2 i + t^2-4/t^2-2t j + t^2+3*e^(t-2) k
@&
t^4-2/t-2 i + t^2-4/t^2-2t j + t^2+3*e^(t-2) k
means
t^4 - (2/t) - 2 i + t^2 - (4/t^2) - 2t j + t^2+ (3*e^(t-2)) k
not
(t^4-2)/(t-2) i + (t^2-4)/(t^2-2t) j + (t^2+3)*e^(t-2) k
which is what you probably intended.
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Apply L'Hopital's Rule to the first two:
4t^3/-2 + 2t/2t-2 + t^2+3*e^(t-2)
Substitute:
32/-2 + 4/2 + 7*e^0
-16 + 2 + 7 = -7
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Question: How many revolutions are made by the circular helix R(t) = (sin t)i + (cos t)j + (3/4)tk in a vertical distance of 12 units.
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Your solution:
Plot points for a range of 12 units.
0 | 0 1 0
1 | 0.84 0.5 0.75
2 | 0.909 -0.4 1.5
3 | 0.14 -0.98 2.25
4 | -0.75 -0.65 3
5 | -0.95 0.28 3.75
6 | -0.28 0.96 4.5
7 | 0.657 0.754 5.25
8 | 0.989 -0.15 6
9 | 0.41 -0.91 6.75
10| -0.5 -0.83 7.5
11| -0.999 0.004 8.25
12| -0.537 0.84 9
From the numbers, you can see there are approximately two revolutions in the space of 12 units.
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sin(t) and cos(t) both complete cycles when t changes by 2 pi.
A complete cycle of these functions will result in a revolution.
It is important to understand the circular definition of the sine and cosine functions.
You wouldn't use t values 0, 1, 2, ..., 12 to represent a cycle. You don't know what the values of the sine and cosine are for those values, and you shouldn't be using a calculator to understand these functions.
You know the exact values of these functions for t values which are multiples of pi/6 and pi/4, so if you were making a table you would use multiples of one or the other (or both) of these values.
For t = 0, pi/6, pi/3, pi/2, 2 pi/3, 5 pi/6, ... you get values of sin(t) equal to 0, 1/2, sqrt(3)/2, 1, sqrt(3)/2, 1/2, ....
You can approximate sqrt(3) / 2. It should be common knowledge that to 4 significant figures sqrt(2) = 1.414 and sqrt(3) = 1.732, so a good approximation is sqrt(3) / 2 = .866.
This would allow you to sketch a fairly good graph of the sine function, and a similar analysis would provide a good graph of the cosine function.
Once you understand the graphs of these functions, it's sufficient to know that as t goes from 0 to pi/2 to pi to 3 pi/2 to 2 pi the values of the sine go from 0 to 1 to 0 to -1 to 0, and the values of the cosine from 1 to 0 to -1 to 0 to 1.
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@&
A few brief notes:
Watch out for order of operations.
See my response on your question.
You need to thoroughly understand the circular model of the sine and cosine function. It's important again and again in this course.
See the more detailed notes I inserted into your document.
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