+0

# PLEASE HELP!!! CALCULUS BC

+1
79
3

Help would be appreciated, thank you! Bolded are my answers..

1. what is the length of the curve with parateric equtaions x=1-2cos(t), y=2-sin(t) from t=0 to tot=pi?

a. 8, b. 4, c. 8pi, d. 2pi

2. What is the length of the path along the graph of y=sqrt(9-x^2) between x=0 and x=2?

a. 3arcsin(1/3), b. 3arcsin(2/3), c. 2arcsin(2/3), d. 2arcsin(1/3)

3. what is the length of the curve x=3y^2/3(-1) from y=0 to y=8?

a. 16(sqrt 2), b. 16 (sqrt 2)-8, c. 16 (sqrt 2)+8, d. none of these

4. find the arc length of the curve from t=0 to t=2 whose derivatives in paramertic form are dx/dt cos(t) and dy/dt=ln(t+1). Write answer to two decimal places. Not sure about this one!!

5. Find the arc length on the interval for t between 0 and 1 inclusive for the curve described with the parametric equations x=1+3t^2, y=2t^3+4.

a. 2(sqrt 2)-1, b. 4 (sqrt 2)-2, c. 2(sqrt 2)-2, d. (sqrt 2)-2

Guest Jun 8, 2017
Sort:

### 3+0 Answers

#1
+75376
+1

What is the length of the path along the graph of y=sqrt(9-x^2) between x=0 and x=2?

dy/dx  =   (1/2) (9 - x^2)^(-1/2) ( -2x)  =  -x / ( 9 - x^2)^(1/2)

[dy/dx]^2  =  x^2 / [ 9 - x^2]

The arc length  is given by

2

∫      √ ( 1 + [dy/dx]^2  )  dx   =

0

2

∫   √ ( 1 +    x^2 / [ 9 - x^2]  ) dx   =

0

2

∫   √ ( 9  / [ 9 - x^2]  ) dx  =

0

2

3  ∫   √ ( 1 / [ 9 - x^2]  ) dx  =

0

let  x  =  3sin (θ)   →    x^2    =  9sin^2( θ)

dx  =  3cos (θ) dθ

Change the limits of integration

when x  = 2               when x  =  0

2/3  = sin (θ)              0  = sin (θ)

θ = arcsin(2/3)           θ =  0

So  we have

arcsin(2/3)

3   ∫  √ ( 1 / [ 9  -  9sin^2( θ) ] ) 3 cos ( θ) dθ   =

0

arcsin(2/3)

3   ∫ (1/3)  √ ( 1 / [ 1  -  sin^2( θ) ] )  3 cos ( θ) dθ   =

0

arcsin(2/3)

3   ∫  √ ( 1 / [ 1  -  sin^2( θ) ] ) cos ( θ) dθ   =

0

arcsin(2/3)

3   ∫   √ ( 1 / [cos^2 ( θ) ]  ) cos ( θ) dθ   =

0

arcsin(2/3)

3   ∫    ( 1 / [cos ( θ) ] ) cos ( θ) dθ   =

0

arcsin(2/3)

3   ∫    dθ   =

0

arcsin(2/3)

3 [ θ ]                       =     3 [  arcsin(2/3)  -  0 ]   =    3  arcsin(2/3)

0

Answer "b"

CPhill  Jun 8, 2017
#2
+75376
+1

1. what is the length of the curve with parateric equtaions x=1-2cos(t), y=2-sin(t) from t=0 to tot=pi?

dx/dt = 2sin(t)    ....  [ dx/dt]^2  =  4sin^2(t)

dy/dt  = -cos(t).....    [dy/dt]^2  =  cos^2(t)

The arc length  is  found as

pi

∫   √ (  [ dx/dt]^2  + [ dy/dt]^2 ) dt   =

0

pi

∫   √ (  4sin^2(t)  + cos^2 (t) ) dt

0

pi

∫   √ (  4sin^2(t)  + 1 - sin^2(t) ) dt

0

pi

∫   √ (  3sin^2(t)  + 1 ) dt   ≈         4.84422

0

CPhill  Jun 8, 2017
#3
+75376
+1

4. find the arc length of the curve from t=0 to t=2 whose derivatives in paramertic form are dx/dt = cos(t) and dy/dt=ln(t+1). Write answer to two decimal places. Not sure about this one!!

2

∫   √    ( dx/dt)^2  +  (dy/dt)^2  )   dt

0

2

∫   √  [ cos^2(t)  +  (ln (t + 1])^2  ] dt      ≈  1.91

0

CPhill  Jun 8, 2017

### 22 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details