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+1
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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
+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

CPhill  Jun 8, 2017
#2
+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

CPhill  Jun 8, 2017
#3
+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

CPhill  Jun 8, 2017