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The solutions to quiz 2 of math 53. It includes the calculation of the cartesian equation of a curve, the length of a curve, and the derivatives dy/dx and d2y/dx2. The curve is represented by x = cos(θ), y = 1 + sin(θ) and x = t^2, y = (2t+1)^(3/2).
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Name: Solutions
Math 53: Quiz 2
September 9, 2014
cos^2 (θ) + sin^2 (θ) = 1 ⇒ x^2 + (y − 1)^2 = 1.
This represents a circle of radius 1 centered at (0, 1).
−1−3 −2 −1 0 1 2 3
−0.
0
1
2
3
4
Figure 1
∫ (^) t=
t=
(t)^2 + (
2 t + 1) dt
∫ (^) t=
t=
t^2 + 2t + 1 dt
∫ (^) t=
t=
(t + 1)^2 dt
∫ (^) t=
t=
t + 1 dt (∵ t ≥ 0 ⇒ |t + 1| = t + 1)
t^2 2
dy/dt dx/dt
e−t^ − te−t et^ = e−^2 t^ − te−^2 t
and
d^2 y dx^2
d dt
( (^) dy dx
dx/dt
d dt (e − 2 t (^) − te− 2 t) et^
− 2 e−^2 t^ + 2te−^2 t^ − e−^2 t et^ = 2te−^3 t^ − 3 e−^3 t.