Spring 1997 Math 152 Exam 3 Solutions: Vectors, Quadric Surfaces, Motion, Functions, Deriv, Exams of Mathematics

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Spring 1997 Math 152 Exam 3 Solutions Mon, 21/Apr/

A-1b, B-2e [ 9.3–9.4 : The Dot and Cross Products ] We cannot cross a scalar (dot product) with a vector.

A-2d, B-1b [ 9.3: The Dot Product ] With and , the scalar projection of onto is

comp

A-3c, B-3a [ 9.6: Quadric Surfaces ] The graph is that of a hyperboloid of two sheets, having standard form , or a permutation thereof. (That is, with the two minus signs possibly in front of other terms.) The key is that there be two minus signs with the constant side of the equation set to 1. Accordingly, this eliminates two of the choices. Of the remaining choices, only one contains the point , which the picture shows is on the surface.

A-4a, B-5d [ 9.7: Vector Functions and Space Curves ] The parameter value which corresponds to is. Now plug and chug.

tangent line:

or

A-5d, B-4b [ 9.9: Motion in Space: Velocity and Acceleration ] Speed is the magnitude of velocity:. Constant speed thus implies , whence

(since ). That is, the acceleration vector points in the direction of the unit normal vector.

A-6c, B-7a [ 10.1: Functions of Several Variables ] Set , a nonzero constant; rearrange; then complete the square.

circles

(Positive sum of squares nondegenerate curves.)

A-7e, B-6c [ 10.3: Partial Derivatives ] Applying the Product Rule, we have

A-8e, B-9c [ 10.4: Tangent Planes and Differentials ] With and , we have

tangent plane:

i.e.,

A-9b, B-10d [ 9.5: Equations of Lines and Planes ] Equate the -, -, and -slots and simultaneously solve.

These values of and also satisfy the third equation. Therefore, the lines intersect at. Assume the lines are perpendicular. Then their direction vectors are orthogonal. Hence their dot product is zero. But , a contradiction. Accordingly, the lines intersect, but are not perpendicular.

A-10a, B-8e [ 10.3: Partial Derivatives ] We numerically approximate the partial derivatives of at the point. In this regard, is especially well-behaved with respect to at. The forward, backward, and central difference quotients all yield the same approximation. (“All roads lead to Rome...”)

Only one of the five choices matches this approximation for , so we’re done! But read on... For the curious, a numerical approximation to may be obtained by using the central difference quotient.

Of course, the textbook never spoke of central difference quotients! However, averaging the forward and backward dif- ference quotients also yields zero.

Alternatively, reason geometrically: as increases along the line , the function values increase, level off , then decrease. That is, they peak at , signifying a local max of the curve , at which the slope is zero.

A-11, B-13 [ 9.3: The Dot Product ] Place the cube in the first octant with one end of the diagonal at the origin. Then simply compute the angle between the vectors (diagonal along the bottom face) and (full diagonal).

or or

A-12, B-12 [ 9.5: Equations of Lines and Planes ] Let’s name our point:. To construct a vector normal to the plane, first pick any two points on the line ; say and. Then take a cross product.

Now use the vector equation of a plane through the point.

or

A-13, B-11 [ 9.3: The Dot Product ] Pick any point in the plane; say,. Form the vector from our point to :

. Read off the normal vector to the plane from the statement of the problem:. The object of our desire is either , the length of the vector projection of onto , or comp , the absolute value of the scalar projection of onto. Let’s compute the latter.

or

A-14, B-16 [ 9.8: Arc Length and Curvature ]

(a) The curvature formula stated on the exam is in terms of vector functions. So let’s embed our parabola (a curve in the -plane ) in 3-space:. The osculating circle touches the parabola at the given point . (b) Compute the radius of curvature ; i.e., the reciprocal of the curvature at. With and , we have

Thus our radius is.

(c) Starting at , “walk” units in the direction of the unit normal to arrive at the osculating circle’s center,. Via clever geometric visualization: As increases, the parabola is traversed from left to right. At , the unit tangent vector points due east; i.e.,. Accordingly, the unit normal vector points due north; i.e.,

. This is because the unit normal points toward the concave side of the curve. Start at and walk due north (in the direction of ) a radius of unit; you’re now at the center of the osculating circle, . Via brute force: Here we’ll use Phil Yasskin’s method of computing , , and , respectively. (“More is less.”) This is actually easier because we already computed the pieces we need during the determination of.

via the right-hand rule!

Now proceed as before to get the center at.

(d) Now that you know the center and radius, write down the equation of the osculating circle.

or

A-15, B-14 [ 9.9: Motion in Space: Velocity and Acceleration ]

A-16, B-15 [ 10.2: Limits and Continuity ] As along the line ,

As along the line ,

Inasmuch as these directional limits differ, the multivariable limit does NOT exist.