MATH 2001 FALL 2012: MIDTERM
Q.1. Find the exact length of the curve,
x = et + e−t , y = 5 − 2t, 0 ≤ t ≤ 3.
A.1. To determine the exact length, we use the arc length formula:
s
Z 3p
Z b 2 2
dx
dy
(et − e−t )2 + (−2)2 dt
+
dt =
dt
dt
0
a
Z 3p
=
e2t − 2 + e−2t + 4dt
0
Z
3
p
=
(et + e−t )2 dt
0
Z
3
=
=
et + e−t dt
0
t
[e − e−t ]30 = e3 − e−3
Q.2. Find the area of the region that lies inside both curves
r = 3 + 2cosθ, r = 3 + 2sinθ.
A.2. To determine the area bounded by both polar curves, we must know where
the two curves intersect; equating the two we find an equation for θ: tan(θ) = 1.
5π
7π
This occurs when θ = − π4 , π4 , 3π
4 , 4 and 4
For each region we must determine which curve lies further from the origin than
the other to determine the area. By plotting sine and cosine on the interval [− π4 , 7π
4 ]
7π
,
),
sine
is
larger
than
we see that: cosine is larger than sine on (− π4 , π4 ) and ( 5π
4
4
or equal to cosine on ( π4 , 5π
)
.
4
Thus we may determine the area by considering the sum of integrals
Z π4
Z 7π
4
2
2
[(3 + 2cos(θ)) − (3 + 2sin(θ)) ]dθ +
[(3 + 2cos(θ))2 − (3 + 2sin(θ))2 ]dθ
A =
−π
4
Z
5π
4
5π
4
+
[(3 + 2sin(θ))2 − (3 + 2cos(θ))2 ]dθ
π
4
At this point full marks would be given for identifying these integrals in the
midterm. Evaluating the integrals is a straightforward exercise, this is left as a
practice problem.
Q.3. Find a vector that has the same direction as < −2, 4, 2 > but has length 6.
1
2
MATH 2001 FALL 2012: MIDTERM
√
A.3.
√ Consider a new vector v =< −2c, 4c, 2c > with magnitude
√ |v| = c 4 + 16 + 4 =
c 32 if we choose that this vector has magnitude six then c 32 = 6 and we may
3
then v is a vector who has the same direction as the
solve for c to find c = 2√
2
original vector.
Q.4.
(1) Find a nonzero vector orthogonal to the plane through the points P, Q, and
R:
P (0, 0, −3), Q(4, 2, 0), R(3, 3, 1).
(2) Find the area of the triangle P QR.
A.4.
(1) Producing vectors from the directed line segments between P,Q and R we
have u =< 4, 2, 3 > and v =< 3, 3, 4 > the vector orthogonal to this plane
will be w = u × v =< −1, −7, 6 >
(2) We have seen from class that the area of the parallelogram formed by the
vectors
u and v√is the magnitude of the cross product w which is |w| =
√
1 + 49 + 36 = 86. By cutting this parallelogram in half we produce the
required triangle, and√hence its area is half that of the parallelogram, the
area of P QR is then 286 .
Q.5. Find parametric equations for the tangent line to the curve with the given
parametric equations at the point (3, 0, 2).
√
x = 1 + 2 t, y = t3 − t, z = t3 + t,
√
A.5. In this case the position vector for the curve is r(t) =< 1+2 t, t3 −t, t3 +t >,
1
with tangent vector r0 (t) =< t− 2 , 3t2 − 1, 3t2 + 1 >. The vector equation for the
tangent line at the point r(t0 ) is rT (s) = r(t0 ) + sr0 (t0 ). In this case t0 = 1 (solve
for y(t)=0) and so the vector equation for the tangent line is
rT (s) =< 3, 0, 2 > +s < 1, 2, 4 >=< 3 + s, 2s, 2 + 4s >