Calculus- Early Transcendentals, 2021a

406 Polar Coordinates, Parametric Equations
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Figure 11.6: Points of vertical and horizontal tangency for r = 1 + cosθ.
We know that the second derivative f ′′ (x) is useful in describing functions, namely, in describing
concavity. We can compute f ′′ (x) in terms of polar coordinates as well. We already know how to write
dy/dx = y ′ in terms of θ, then
d dy
dx dx = dy′
dx = dy′ dθ
dθ dx = dy′ /dθ
dx/dθ .
Example 11.8: Second Derivative of Cardioid
Find the second derivative for the cardioid r = 1 + cosθ.
Solution.
d cosθ + cos 2 θ − sin 2 θ
dθ −sinθ − 2sinθ cosθ ·
1
dx/dθ = ··· = 3(1 + cosθ)
(sinθ + 2sinθ cosθ) 2 ·
=
−3(1 + cosθ)
(sinθ + 2sinθ cosθ) 3 .
1
−(sinθ + 2sinθ cosθ)
The ellipsis here represents rather a substantial amount of algebra. We know from above that the cardioid
has horizontal tangents at ±π/3; substituting these values into the second derivative we get y ′′ (π/3) =
− √ 3/2 andy ′′ (−π/3)= √ 3/2, indicating concave down and concave up respectively. This agrees with
the graph of the function.
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Exercises for 11.2
Exercise 11.2.1 Compute y ′ = dy/dx and y ′′ = d 2 y/dx 2 .
(a) r = θ
(b) r = 1 + sinθ
(c) r = cosθ
(d) r = sinθ
(e) r = secθ
(f) r = sin(2θ)
Exercise 11.2.2 Sketch the curves over the interval [0,2π] unless otherwise stated.