1.(6pts) Which of the following vectors has the same direction asv=h−1,2,2i but has length 6?
(a) h−2,4,4i (b) D−√2,2√2,2√2E (c) h4,2,4i
(d) h2,4,4i (e) h0,6,0i
Solution.
Solution. The length of v is given by
p
(−1)2 + 22+ 22 =√9 = 3
We would like to make the length 6. The only vectors in the same direction as v are those of the form λv where λ is a positive number. Now |λv|=|λ| |v|=λ|v|. So we let λ= 2. 2.(6pts) Which line below is an equation of the line of intersection of the two planes 2x+ 3y−
4z = 10 and 3x−4y+ 2z =−2. Note (−1,0,2) is a point on both planes.
(a) th10,16,17i+h−1,0,2i (b) th2,3,−4i+h−1,0,2i (c) th10,16,16i+h2,3,−4i (d) th3,−4,2i+h−1,0,2i (e) th10,16,17i
Solution.
Solution. Actually (−1,0,2) is a point on neither plane since 2(−1) + 3(0)−4(2) = −10 and 3(−1)−4(0) + 2(2) = 1. However if you do what you should you should never notice.
The two normal vectors are h2,3,−4i andh3,−4,2i. Hence a vector parallel to the line is
i j k
2 3 −4
3 −4 2
=h−10,−16,−17i. Hence
th10,16,17i+h−1,0,2i is an equation of the line.
3.(6pts) Given two lines th1,3,2i+h1,1,1i and th3,−2,1i+h7,8,8i, find the point of inter-section, if any?
(a) (4,10,7) (b) (3,7,5) (c) (4,10,1)
(d) (4,7,10) (e) There are no points of intersection. Solution.
Solution. The equations are
t+ 1 = 3s+ 7 3t+ 1 =−2s+ 8 2t+ 1 =s+ 8
Then t = 3s+ 6 so 2(3s+ 6) + 1 =s+ 8 and 6s+ 13 =s+ 8 or 5s =−5 or s =−1. Then t+ 1 =−3 + 7 = 4 so t= 3 and the point is (4,10,7).
4.(6pts) Compute the curvature of the curve r(t) = ht−sin(t),1−cos(t), ti att= π 3. (a) κ=
√ 10
8 (b) κ=
1 2
s
11−4√3
3−√3 (c) κ=
√ 5 4 (d) κ=
√ 5
2 (e) κ=
1 √ 2
Solution.
Solution. r0 =h1−cos(t),sin(t),1i and r00=hsin(t),cos(t),0iso κ= |r
0×r00| |r0|3
= | h−cos(t),sin(t),cos(t)−1i | | h1−cos(t),sin(t),1i |3
=
s
1 + (cos(t)−1)2
3−2 cos(t)3 Att=π/3, cos(t) = 1/2 and so
κ(π/3) =
r
5/4 23 =
r
10 26 =
√ 10 8
5.(6pts) Given three vectors u, v and w in R3, which of the following statements is not necessarily true?
(a) Ifu×v=0, then we must have either u=0 or v=0. (b) u×v=−v×u.
(c) (u+v)×w=u×w+v×w. (d) u×u=0.
(e) u×v is perpendicular to u and v. Solution.
Solution. All the listed properties are true, except for u×v implying u = 0 or v = 0. We could have, e.g.,u=v=h1,0,0i.
6.(6pts) Suppose that two vectorsu and v∈ R3 are such that|u|= 7, |v|= 3 and the angle
between them is θ = 90◦. Compute |u×v|.
(a) 21. (b) 3
7. (c)
7
3. (d) 21π.
(e) There is not enough information. Solution.
Solution. Using the identity|u×v|=|u| |v|sin(θ), we get |u×v|= 21.
7.(6pts) If f(x, y) =x2+exy and x=s+t, y= sin(t), compute ∂ f
∂ t at (s, t) = (1,0).
Solution.
Solution. The chain rule says that ∂ f ∂ t =
∂ f ∂ x
∂ x ∂ t +
∂ f ∂ y
∂ y
∂ t. In this case, ∂ f
∂ t = (2x+ye
xy)·1 +xexy·cos(t)
At (s, t) = (1,0), we havex= 1, y= 0, so that ∂ f
∂ t(1,0) = 2 + 1 = 3 OR
∇f = h2x+yexy, xexyi and if r(t) = hs+t,sin(t)i and ∂r
∂ t = h1,cos(t)i. When s = 1, t = 0, x= 1, y= 0 so
∇f(1,0)•∂r
∂ t(1,0) =h2,1i • h1,1i= 3
8.(6pts) The planeS contains the points (0,1,3), (2,2,2), and (3,2,1). Which of the following is an equation for S?
(a) x−y+z = 2 (b) x+y+z = 6 (c) x+ 3y+z = 6
(d) y+z = 4 (e) 2x−4y+ 3z = 5
Solution.
Solution. There are two ways to solve this problem. You can calculate the normal vector as in the worksheet and find a plane with that normal and one of those points and get x−y+z = 2. Alternately, you can test each equation to see if all three given points satisfy it, and the only one which works for all three is x−y+z = 2.
9.(6pts) Let f(x, y, z) =exy +z2y. Find the sum of partial derivatives fx+fy +fz.
(a) (x+y)exy + 2yz+z2 (b) ey +ex+z2+ 2zy (c) yexy (d) 2exy +z2 + 2zy (e) ex+ey + 2z+ 1
Solution.
Solution. fx+fy +fz =yexy +xexy +z2 + 2zy.
10.(6pts) Which one of the following functions has level curves as concentric circles? (a) f(x, y) =e−(x2+y2) (b) f(x, y) = 8−x−y (c) f(x, y) =x2−y2 (d) f(x, y) =e4x2−y (e) f(x, y) = sin(x+y)
Solution.
Solution. The contour at level cis given by f(x, y) = c. If f(x, y) =e−(x2+y2) the contour at level cis x2+y2 =−lnc, which is a circle centered at (0,0). (Note c∈(0,1])
Forf(x, y) = 8−x−ythe level curves arex+y = 8−c, which are lines. Forf(x, y) =x2−y2
the level curves are x2−y2 = c, which are hyperbolas and two lines. For f(x, y) = e4x2−y the level curves are 4x2−y = lnc, which are parabolas. For f(x, y) = sin(x+y) the level curves are x+y= arcsincwhich are lines.
Solution.
Solution. We know that the area of the parallelogram determined by the vectors−P Q−* and −*
QR equals
−−*
P Q×−QR*
, so the area of the triangle is just half this quantity. From here we
just compute
−−*
P Q=h−3,1,−7i −QR*=h3,−6,2i −−*
P Q×−QR*=
i j k
−3 1 −7
3 −6 2
=h−40,−15,15i= 5h−8,−3,3i and then
Area of ∆ = | −−*
P Q×−QR|*
2 =
5√82 2
12.(10pts) Let z =z(x, y) be the function of x, y given implicitly by the equation x2+y3+z4+ 2xyz = 1
Find ∂z ∂x and
∂z ∂y. Solution.
Solution. Let F(x, y) = x2 +y3 +z4 + 2xyz. The function G(x, y) given by G(x, y) = F(x, y, z(x, y)) is constant equal to 1, hence both ∂G
∂x and ∂G
∂y are zero. Using the Chain Rule for ∂G
∂x, we may solve for ∂z ∂x:
∂z ∂x =
−∂F ∂x ∂F
∂z
= −(2x+ 2yz) 4z3+ 2xy
Using the Chain Rule for ∂G
∂y, we may solve for ∂z ∂y: ∂z
∂y = −∂F
∂y ∂F
∂z
= −(3y
2+ 2xz)
4z3+ 2xy
13.(10pts) The line L passes through the point (2,7,8) and is perpendicular to the plane R whose equation is 3x−2y+z = 14. Find the point where L intersects R.
Solution.
Solution. Since L is perpendicular to R, its direction vector is R’s normal, h3,−2,1i. Since we have a direction and a point, we can write an equation for L:
r(t) = h3,−2,1it+h2,7,8i=h3t+ 2,−2t+ 7, t+ 8i
To find the intersection ofL and R, we substitute these coordinates into the equation for R.
3(3t+ 2)−2(−2t+ 7) + (t+ 8) = 14 9t+ 6 + 4t−14 +t+ 8 = 14 14t= 14 t= 1
Plugging this back intoL, we get that the intersection is at (5,5,9). A quick check confirms that this point is indeed on R.
14.(10pts) A space curve is described by the equation r(t) =
*√
3 2 t
2, tcos(t)−sin(t), tsin(t) + cos(t) +
for (0,∞).
(a) Write an equation for the unit tangent vector to the curve at time t. (b) Write an equation for the binormal to the curve at timet.
(c) For what values of t ∈ (0,2π) is the osculating plane at t of the curve perpendicular to the line with equations
x−1
1 =
z−3 √
3 y= 2 ?
Solution.
Solution. r0 =D√3t,−tsin(t), tcos(t)E,|r0|= 2t,
T = D√3/2,−sin(t)/2,cos(t)/2E, T0 = h0,−cos(t)/2,−sin(t)/2i, |T0| = 1/2, N = h0,−cos(t),−sin(t)i,B =T×N=D1/2,√3 sin(t)/2,−√3 cos(t)/2E.
OR
r0(t) =D√3t,−tsin(t), tcos(t)E
r00(t) = D√3,−sin(t)−tcos(t),cos(t)−tsin(t)E r0 ×r00=
i j k
√
3t −tsin(t) tcos(t)
√
3 −sin(t)−tcos(t) cos(t)−tsin(t)
=
D
t2,√3t2sin(t),−√3t2cos(t)
E
. and a parallel vector is D1,√3 sin(t),−√3 cos(t)E. Hence a formula for the binormal is B =D1/2,√3 sin(t)/2,−√3 cos(t)/2E.
OR
A vector parallel to r0(t) = D√3t,−tsin(t), tcos(t)E, is p(t) =D√3,−sin(t),cos(t)E. p0(t) =h0,−cos(t),−sin(t)iand
p×p0 =
i j k
√
3 −sin(t) cos(t) 0 −cos(t) −sin(t)
= D1,√3 sin(t),−√3 cos(t)E. Hence a formula for the binormal is B=D1/2,√3 sin(t)/2,−√3 cos(t)/2E.
The binormal has to be parallel tou=D1,0,√3E, the vector of the line. When sin(t) = 0 and cos(t) = −1, B = 1
2
D
1,0,√3
E
= 1
2u. But in (0,2π) the only value(s) of t for which sin(t) = 0, cos(t) =−1 is t =π.