Answers to Review Questions
1. The integrals are
> int(x/sqrt(1+x^2),x);
2 1/2
(1 + x )
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> int(x*exp(2*x),x);
1/2 x exp(2 x) - 1/4 exp(2 x)
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> int(1/(x^2-4),x=3..5);
1/4 ln(3) - 1/4 ln(7) + 1/4 ln(5)
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>
2. The first integral converges; the second one does not.
3. The integral that computes the desired area is
> Int((y+1) -(y^2/2-3),y=-2..4);
4
/
| 2
| y + 4 - 1/2 y dy
|
/
-2
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>
4. The volume of the region is given by the following
integral
> Int(sqrt(3)*(1-x^2), x=-1..1);
1
/
| 1/2 2
| 3 (1 - x ) dx
|
/
-1
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>
5. Work is 625 ft lbs.
6. The angle is arccos(11/15) or about 43 degrees.
7. parametric form of the line is (4,-1,2)+t(2,1,-3)
symmetric form is (x-4)/2 = (y+1) = (z-2)/(-3).
8. plane is 17x -6y -5z = 32
9. parametric form is (1,1,1)+t(1,2,3)
symmetric form is (x-1) = (y-1)/2 = (z-1)/3
10. Distance = 25/3
11. The first one is a saddle - - concave up in the x-direction
and concave down in y.
The second one is an ellipsoid (football shaped)
The third one is a hyperbola with one sheet which
does not intersect the x-axis.
12. The level curves of the first are hyperbolas;
the second are circles centered on the x-axis
and passing thru the origin.
13. The level surfaces of the first are hyperboloids with
two sheets that do not intersect the y,z - plane.
the second are planes parallel to the vector (3,-2,4).
14. The partial derivatives are the following
The x- derivative - - Maple code: diff(x*cos(x^2*y),x);
2 2 2
cos(x y) - 2 x sin(x y) y
The y-derivative - - diff(x*cos(x^2*y),y);
3 2
- x sin(x y)
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The x- derivative - - diff(exp(x*y)*sqrt(x^2+y^2),x);
2 2 1/2 exp(x y) x
y exp(x y) (x + y ) + ------------
2 2 1/2
(x + y )
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The y-derivative - - diff(exp(x*y)*sqrt(x^2+y^2),y);
2 2 1/2 exp(x y) y
x exp(x y) (x + y ) + ------------
2 2 1/2
(x + y )
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15. The unit vector for the first is (4/5, 3/5)
The unit vector for the second is
(-2,27,32)/sqrt(1757)
16. tangent plane for the first is z=-1-2(x-1)-3(y-1)
tangent plane for the second is (x-1)+4(y-2)-(z-1)=0
17. The gradient vector points perpendicular to the level curves.
If the level curves were identified (i.e. the value of the function
given on each level curves), then you could determine
the direction and estimate the magnitude of the gradient.
This is because the gradient points in the direction of
maximum increase.