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APPM 2350
Spring 2012
Exam 1 Solutions
February 15, 2012
Problem 1
a
In order to determine the standard equation of the plane of intersection of two surfaces, we must find a relationship between x, y, and z when both surface equations are true. The easiest way to do this is to multiply the second equation by − 12 and then add the two equations:
x^2 + 2y^2 − z^2 +
x = 1
−x^2 − 2 y^2 + z^2 + y + z = 0
Which, when added yields
y + z +
x = 1
b
A normal vector for the plane can be made from the coefficients on x, y, and z, so
~n = 〈
or ~n = 〈 1 , 2 , 2 〉
c
The principle axes x, y, and z are given respectively by setting y = z = 0, x = z = 0, and x = y = 0 respectively. We plug into the plane to get the three points:
x axis: (2, 0 , 0)
y axis: (0, 1 , 0) z axis: (0, 0 , 1)
d
The three points from part c, above, form a triangle in the first octant. We can take 3 points and turn them into 2 vectors, and take the magnitude of the cross product of the 2 vectors to get the area of the parallelogram formed by the 2 vectors. The area of the triangle is half of that. Let’s call the three points from part c, x, y, and z for the intersections of the x, y, z axes.
~v = xy~ = 〈− 2 , 1 , 0 〉
w ~ = xz~ = 〈− 2 , 0 , 1 〉 v × w =... = 〈 1 , 2 , 2 〉
and so 12 |v × w| = 32.
Problem 2
a
To find B, we use the formula B = T × N , and so
T × N = 〈
, 0 〉 × 〈 0 , 0 , 1 〉 =... = 〈
b
We need to find ~v, but we don’t have a position vector ~r(t) that we can take a derivative of! Recall though that T = (^) |vv|. We’re given T in the problem statement, and we are told that particle moves at a constant speed of 2, so |v| = 2. A little algebra:
v = |v|T = 〈
c
Acceleration seems as hopeless as velocity, but again, we can use constant speed to our advantage. If speed is constant, then aT , tangential acceleration, is 0. Thus, acceleration is just
a = aN N = κ|v|^2 N = 3(2^2 )〈 0 , 0 , 1 〉 = 〈 0 , 0 , 12 〉
d
Not enough information.
e
We don’t know anything about time t = 5t∗^ except that the particle moves at a constant speed! Use the acceleration information from up above:
v · a = v · aN N = |v|T · aN N = |v|aN (T · N ) = 0
since T and N are always orthogonal to each other.
c
Hyperboloid of 2 sheets, centered on the y axis:
−x^2 + y^2 − z^2 = 1
d
Hyperboloid of 1 sheet, centered on the x axis:
−x^2 + y^2 + z^2 = 1
e
A plane parallel to the x axis that goes through the point (0, 1 / 2 , 1 /2) but not the origin. What does parallel to the x axis mean? Well, the plane goes on forever is some direction, and we want to make sure that it never crosses the x axis...so we can’t involve x in the equation for the plane. Think about what would happen if x were in the plane equation: then, we’d be able to set y = z = 0 and find some value for x where the plane crosses the axis. So, our equation looks like:
(something) y + (something else) z = (something else else)
Now, back to the problem statement where we remember that the coefficients have to be 1, 0, or -1. We therefore get y + z = 1
or, if you’re more of the negative, glass-half-empty type,
−y − z = − 1
