Calculus Exam 2
Instructions: Instructions: This answer key is designed as a study and practice re-
source. If you have any questions about it, feel free to reach out and let me know!
Question 1 (10 points). Compute the exact length of the following curve
4 3
f (x) = x 2 , 0 ≤ x ≤ 2
3
1 + f ′2 (x). f ′ = 2x1/2 by the power
p
For the arclength formula, we need to compute √
′2
rule, leading to 1 + f = 1 + 4x. This leads to 1 + 4x. This will lead to the substitution
u = 1 + 4x and du = 4dx, so that 1/4du = dx. The rest of the computation is below:
√
Z 2 Z 2
1√ 1 2 3/2 1 3/2 9 1 3/2 1
1 + 4xdx = udu = u = u |1 = 9 −
0 0 4 43 6 6 6
Note that here we converted the u bounds to 1 and 9, but it’s fine to plug 1 + 4x back in
and then use 0 and 2. Additionally, this final answer simplifies to 13/3.
Question 2 (10 points). Consider the below parametric curve on the provided interval and
answer the following questions.
π
x(t) = sin(t), y(t) = 4 − sin2 (t), ≤t≤π
2
6 points Eliminate the parameter for this parametric curve.
4 points Compute both the initial and terminal points for this parametric curve
While it’s POSSIBLE to do some stuff with t = arcsin(x), it’s easier to notice that
sin2 (t) = (sin(t))2 , leading to y = 4 − x2 through a direct substitution. Additionally, the
initial and terminal points would come from plugging in t = π/2 and t = π and using unit
circle values, for an initial point of (1, 3) and terminal point of (0, 4).
1
, 2
Question 3 (10 points). Find an equation for the tangent line to the parametric curve
x(t) = t2 − t + 2 and y(t) = t3 − 2t2 + 1 at the point (4, 1).
The first step would be to find the value of t by plugging in x = 4 and y = 1. This gives
4 = t2 − t + 2 for t2 − t − 2 = 0 or (t − 2)(t + 1) = 0 for t = −1 or t = 2. Plugging those
into the second equation gives that t = 2 is the ONLY correct t value. Then, the slope is
m = y ′ /x′ = (3t2 − 4t)/(2t − 1) = 4/3. Since the slope is 4/3 and the point is (4, 1), we get
the line y − 1 = 4/3(x − 4) or y = 4/3x − 13/3.
Question 4 (8 points). Graph the point (2, 5π
6 ) in polar coordinates AND find a point with
r < 0 that has the same graph.
Roughly speaking, the idea is that the point should be in quadrant 2 because 5pi/6 is
in quadrant 2, and also should indicate a radius of 2. Additional points would come from
either (r, θ) = (−r, θ + π) or (r, θ) = (r, θ + 2π), so that would in this case mostly be
(−2, −π/6) or (−2, 11π/6).
Instructions: Instructions: This answer key is designed as a study and practice re-
source. If you have any questions about it, feel free to reach out and let me know!
Question 1 (10 points). Compute the exact length of the following curve
4 3
f (x) = x 2 , 0 ≤ x ≤ 2
3
1 + f ′2 (x). f ′ = 2x1/2 by the power
p
For the arclength formula, we need to compute √
′2
rule, leading to 1 + f = 1 + 4x. This leads to 1 + 4x. This will lead to the substitution
u = 1 + 4x and du = 4dx, so that 1/4du = dx. The rest of the computation is below:
√
Z 2 Z 2
1√ 1 2 3/2 1 3/2 9 1 3/2 1
1 + 4xdx = udu = u = u |1 = 9 −
0 0 4 43 6 6 6
Note that here we converted the u bounds to 1 and 9, but it’s fine to plug 1 + 4x back in
and then use 0 and 2. Additionally, this final answer simplifies to 13/3.
Question 2 (10 points). Consider the below parametric curve on the provided interval and
answer the following questions.
π
x(t) = sin(t), y(t) = 4 − sin2 (t), ≤t≤π
2
6 points Eliminate the parameter for this parametric curve.
4 points Compute both the initial and terminal points for this parametric curve
While it’s POSSIBLE to do some stuff with t = arcsin(x), it’s easier to notice that
sin2 (t) = (sin(t))2 , leading to y = 4 − x2 through a direct substitution. Additionally, the
initial and terminal points would come from plugging in t = π/2 and t = π and using unit
circle values, for an initial point of (1, 3) and terminal point of (0, 4).
1
, 2
Question 3 (10 points). Find an equation for the tangent line to the parametric curve
x(t) = t2 − t + 2 and y(t) = t3 − 2t2 + 1 at the point (4, 1).
The first step would be to find the value of t by plugging in x = 4 and y = 1. This gives
4 = t2 − t + 2 for t2 − t − 2 = 0 or (t − 2)(t + 1) = 0 for t = −1 or t = 2. Plugging those
into the second equation gives that t = 2 is the ONLY correct t value. Then, the slope is
m = y ′ /x′ = (3t2 − 4t)/(2t − 1) = 4/3. Since the slope is 4/3 and the point is (4, 1), we get
the line y − 1 = 4/3(x − 4) or y = 4/3x − 13/3.
Question 4 (8 points). Graph the point (2, 5π
6 ) in polar coordinates AND find a point with
r < 0 that has the same graph.
Roughly speaking, the idea is that the point should be in quadrant 2 because 5pi/6 is
in quadrant 2, and also should indicate a radius of 2. Additional points would come from
either (r, θ) = (−r, θ + π) or (r, θ) = (r, θ + 2π), so that would in this case mostly be
(−2, −π/6) or (−2, 11π/6).
