+0  
 
+1
404
3
avatar

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
 #1
avatar+89972 
+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"

 

 

 

cool cool cool

CPhill  Jun 8, 2017
 #2
avatar+89972 
+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

 

 

 

cool cool cool

CPhill  Jun 8, 2017
 #3
avatar+89972 
+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

 

 

cool cool cool

CPhill  Jun 8, 2017

20 Online Users

avatar
avatar
avatar
avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.