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Math 53: Quiz 2 Solutions - Circle Equation, Arc Length, and Derivatives, Lecture notes of Mathematics

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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2021/2022

Uploaded on 09/12/2022

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Name: Solutions
Math 53: Quiz 2
September 9, 2014
1. (1 point) Find the Cartesian equation of the curve and sketch it: x= cos(θ), y =
1 + sin(θ),0θ < 2π.
We have cos(θ) = xand sin(θ) = y1. Thus,
cos2(θ) + sin2(θ) = 1 x2+ (y1)2= 1.
This represents a circle of radius 1 centered at (0,1).
−3 −2 −1 0 1 2 3
−1
−0.5
0
0.5
1
1.5
2
2.5
3
3.5
4
Figure 1
2. (1 point) Find the length of the curve defined by: x=t2
2, y =(2t+1)3/2
3,0t4.
Note that dx
dt =tand dy
dt = (2t+ 1)1/2. We then have from the arc length formula
L=t=4
t=0 (t)2+ (2t+ 1) dt
=t=4
t=0
t2+ 2t+ 1 dt
=t=4
t=0 (t+ 1)2dt
=t=4
t=0
t+ 1 dt (t0 |t+ 1|=t+ 1)
=[t2
2+t]4
0
=[16
2+ 4]0 = 12.
1
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Name: Solutions

Math 53: Quiz 2

September 9, 2014

  1. (1 point) Find the Cartesian equation of the curve and sketch it: x = cos(θ) , y = 1 + sin(θ), 0 ≤ θ < 2 π. We have cos(θ) = x and sin(θ) = y − 1. Thus,

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

  1. (1 point) Find the length of the curve defined by: x = t 22 , y = (2t+1) 3 / 2 3 ,^0 ≤^ t^ ≤^4. Note that dxdt = t and dydt = (2t + 1)^1 /^2. We then have from the arc length formula

L =

∫ (^) 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

  • t

] 4

0

[

]

  1. (1 point) Find dydx and d (^2) y dx^2 :^ x^ =^ e t (^) , y = te−t. We have dy dx

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.