NAME:
Calculus 3
Quiz # 3
September 6, 2012
Solutions
Instructions.
• SHOW WORK! and write clearly.
• Try to write in the form of full sentences. I may penalize you for the failure to use full sentences. A sentence
must contain a verb. For example “5” or “x2 + y 2 are NOT sentences. “x2 + y 2 = 5” is a sentence; the symbol
“=” would be read “is equal,” and “is” is a verb.
• Do not use an equal symbol to connect quantities that are not equal, except when setting up equations.
• Use extra paper as necessary. There is no prize, more likely a penalty, for cramming the maximum amount
of material into the minimum amount of space.
The Test.
1. (45 points) Find the unit tangent vector T(t) at the point where the parameter t equals 1 (that is, t = 1)
for the curve of vector equation
r(t) = ht3 + 3t, t2 + 1, 3t + 4i.
Solution.
The unit tangent vector at a point of parameter t is T(t) = r0 (t)/|r0 (t)|. In our case,
p
√
r0 (t) = h3t2 + 3, 2t, 3i, thus r0 (1) = h6, 2, 3i, |r0 (1)| = 62 + 22 + 32 = 49 = 7.
The answer is T(t) =
1
h6, 2, 3i.
7
2. (45 points) Find parametric equations for the tangent line to the curve described parametrically by
x = e−t cos t,
y = e−t sin t,
z = e−t
at the point (1, 0, 1).
Solution. For the equation of a line we need a point, and a direction vector. We are given a point. Since
the line is supposed to be tangent to the given curve at the point (1, 0, 1), we need to find a tangent vector to
the curve at that point. The only parameter value for which the curve passes through the point∗ (1, 0, 1) is
t = 0. In vector notation the curve is r(t) = e−t cos ti + e−t sin tj + e−t k, thus r0 (t) = (−e−t cos t − e−t sin t)i +
(−e−t sin t + e−t cos t)j − e−t k; thus r0 (0) = −i + j − k is a vector tangent at the point in question From this
we get the parametric equation AT ONCE:
x = 1 − t,
y = t,
z = 1 − t.
3. (10 points) Sketch the plane curve with the vector equation
r(t) = e2t i + et j,
and sketch the position vector r(t) and the tangent vector r0 (t) for t = 0.
Solution. The curve is given parametrically by x = e2t , y = et . Eliminating t from the equations, we
get x = y 2 , which is the equation of a parabola with the x-axis as it axis. However, since y = et is always
non-negative, it can only be the upper half of such a parabola. All positive values of y being assumed, we get
the whole upper half. Now r(0) = i + j; r0 (t) = 2e2t i + et j so that r0 (0) = 2i + j. A sketch follows.
∗ If there were two or more parameter values, which happens if the curve goes through a point more than once, the problem, as
stated, could have more than one solution.
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