Math201: Calculus III
Quiz 1
13th September 2022
Semester 221
Dr. Slim Belhaiza
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Problem 1
Consider the curve C : x = 3t + 1; y = t2 − t.
Find the equation of the tangent line at P (−2, 2).
Solution 1
On one hand x = 3t + 1 = −2, therefore t = −1. On the other hand, y = t2 − t = 2, gives
t = −1 or t = 2. Therefore, t = −1.
∂x
We have: ∂y
∂t = 2t − 1 and ∂t = 3.
∂y
2t−1
3
Hence, ∂x = 3 = − 3 = −1, at t = −1.
There is a unique tangent (T) to C at P : (y − 2) = −(x + 2); or y = −x.
Problem 2
What is the surface area obtained by rotating the curve C given by: x = cos2 (t); y = sin2 (t);
0 ≤ t ≤ π4 , about the y-axis?
Solution 2
We have: ∂x
= −sin(2t) and ∂y
= sin(2t).
∂t = −2sin(t)cos(t)
∂t = 2sin(t)cos(t)
p
√ Rπ
R π4
2
4
2
Hence, the surface area S = 0 2πcos (t) 2sin (2t)dt = 2 2π 0 cos2 (t)sin(2t)dt.
Note that sin(2t) is positive since 0 ≤ t ≤ π4 .
√
π
√ Rπ
√
√
Therefore, S = 2 2π 04 2sin(t)cos3 (t)dt = 2 2π[− 12 cos4 (t)]04 = 2 2π(− 81 + 21 ) = 3 4 2 π.
Problem 3
Find the points where the tangent to the polar curve r = 3cos(θ) is horizontal.
Solution 3
We know that x = rcos(θ) and y = rsin(θ).
We have
dr
dθ
= −3sin(θ).
3−6sin2 (θ)
−6sin(θ)cos(θ) .
√
5π 7π
The tangent is horizontal when sin2 (θ) = 21 . Hence sin(θ) = ± 22 . Hence, θ = π4 ; 3π
4 ; 4 ; 4 .
3 π
3 3π
3 5π
3 7π
√
√
√
√
There are many possible answers including the points ( 2 , 4 ); (− 2 , 4 ); (− 2 , 4 ); ( 2 , 4 ).
Notice that for all these points dx
dθ 6= 0.
Therefore,
1 Department
dy
dx
=
−3sin(θ)sin(θ)+3cos(θ)cos(θ)
−3sin(θ)cos(θ)−3cos(θ)sin(θ)
=
−3sin2 (θ)+3cos2 (θ)
−6sin(θ)cos(θ)
of Mathematics, (c) KFUPM: slimb@kfupm.edu.sa
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