Question 1: Find the exact length of the curve. x=y^4/8+1/4y^2, 1<=y<=2
SOLUTION:
First, find dx/dy
dx/dy = 4y^3 / 8 + (1/4) * (-2) / y^3 = y^3 / 2 – 1/(2y^3) = (1/2) * (y^3 – y^(-3))
1 + (dx/dy)^2 =>
1 + (1/4) * (y^6 – 2 + y^(-6)) =>
(1/4) * (4 + y^6 – 2 + y^(-6)) =>
(1/4) * (y^6 + 2 + y^(-6)) =>
(1/4) * (y^3 + 1/y^3)^2
sqrt((1/4) * (y^3 + (1/y^3))^2) =>
(1/2) * (y^3 + (1/y^3))
Integrate with respect to y
(1/2) * (1/4) * y^4 + (1/2) * (-1/2) * y^(-2) =>
(1/8) * y^4 – (1/4) * y^(-2)
From 1 to 2
(1/8) * (16 – 1) – (1/4) * (1/4 – 1) =>
(1/8) * (15) + (1/4) * (3/4) =>
15/8 + 3/16 =>
30/16 + 3/16 =>
33/16
Question 2: find the exact length of the curve. Use a graph to determine the parameter interval. r = cos2(θ/2)
SOLUTION:
The polar arc length formula is
ds = √[r^2 + (dr/dθ)^2] dθ
Was r meant to be this?
r = [cos(θ/2)]^2
If so, interval is from -π to π for this cardiod
Using cos(2t) = 2[cos(t)]^2 – 1
r = (1/2) [1 + cos(θ)]
r^2 = (1/4){1 + 2cos(θ) + [cos(θ)]^2} ……………(1)
dr/dθ = – sin(θ/2)cos(θ/2) = -(1/2)sin(θ)
(dr/dθ)^2 = (1/4)[sin(θ)]^2 ……………………………….(2)
r^2 + (dr/dθ)^2 = (1/4){1 + 2cos(θ) + [cos(θ)]^2 + [sin(θ)]^2 }
r^2 + (dr/dθ)^2 = (1/4)[2 + 2cos(θ)]
ds = (1/2)√[ 2 + 2cos(θ)] dθ
S = {θ: -π to π}∫√[2 + 2cos(θ)] dθ
You can do that, but you should get
S = 4
Question 3: find the exact length of the curve. y = x − x2 + sin−1 x
SOLUTION:
y =√x − x2 + sin−1√x
Okay – since they haven’t given us any limits, it’s implied that the limits are the entire domain of the curve. The domain of this function is:
(x-x^2) >= 0 AND
x >= 0 AND
-1 <= sqrt(x) <= 1
The first part comes out to:
x(1-x) >= 0, 0<=x<=1
So the total domain of this function is 0 <= x <=1.
Now – the formula for length is:
L = ∫ sqrt(1 + (dy/dx)^2) dx
(L = ∫ dL = ∫ sqrt(dx^2 + dy^2) = ∫sqrt(1 + (dy/dx)^2) dx)
First, take the derivative of y:
y’ = (1-2x) (1/2) (1/sqrt(x-x^2)) + (1/sqrt(1-x)) (1/2) (1/sqrt(x))
y’ = (1-2x)/(2 sqrt(x-x^2)) + 1/(2sqrt(x-x^2))
y’ = (2-2x)/(2 sqrt(x-x^2)) = (1-x)/sqrt(x-x^2)
y’ = sqrt(x-x^2) (1-x)/(x-x^2) = sqrt(x-x^2) / x
So the integral becomes:
L = ∫ sqrt(1 + (x-x^2)/x^2) dx [taken from 0 to 1]
L = ∫ sqrt(1 + (1-x)/x) dx [0 to 1]
L = ∫ sqrt(1/x) dx = ∫ x^(-1/2) dx [0 to 1]
L = 2 x^(1/2) | [x = 0 to 1]
L = 2
The length is 2.
Question 4: find the exact length of the curve. x = 1 3 y (y − 3), 16 ≤ y ≤ 25
Solution: (THIS ANSWER WASN’T CHECKED BY OUR EXPERT)
x = 1/3 sqrt(y)(y − 3), 16 ≤ y ≤ 25
[(d/dy ((y^(3/2))/3) – sqrt(y))^2] + 1
[(1/2)*sqrt(y) – (1/2)*(y^(-1/2))][(1/2)*sqrt(y) – (1/2)*(y^(-1/2))] + 1
[(y/4) – (1/2) + (1/4)*(1/y)] + 1
(y/4) + (1/2) + (1/4)(1/y)
((y^2) + (2y) + 1)/(4y)
[(y + 1)^2/(4y)]
now sqrt the entire function
int_16^25 (y + 1)/(2sqrt(y)) dy
int_16^25 (sqrt(y) + (y^(-1/2)))/2 dy = 64/3
Question 5: find the exact length of the curve. x = 7 cos(t) − cos(7t), y = 7 sin(t) − sin(7t), 0 ≤ t ≤ π
SOLUTION:
Length of curve:
L = ∫[0,π] √[ (x'(t))² + (y'(t))² ] dt
x(t) = 7cos(t) – cos(7t)
x'(t) = -7sin(t) + 7sin(7t)
x'(t)² = 49sin²(t) – 98 sin(t)sin(7t) + 49sin²(7t)
y(t) = 7sin(t) – sin(7t)
y'(t) = 7cos(t) – 7cos(7t)
y'(t)² = 49cos²(t) – 98cos(t)cos(7t) + 49cos²(7t)
x'(t)² + y'(t)²
= 49sin²(t) – 98 sin(t)sin(7t) + 49sin²(7t) + 49cos²(t) – 98cos(t)cos(7t) + 49cos²(7t)
= 49(sin²(t)+cos²(t)) – 98(sin(t)sin(7t) + cos(t)cos(7t)) + 49(sin²(7t)+cos²(7t))
= 49(1) – 98(sin(t)sin(7t) + cos(t)cos(7t)) + 49(1)
= 98 – 98(sin(t)sin(7t) + cos(t)cos(7t))
Use angle sum formulas for sin and cos:
sin(7t) = sin(t+6t) = sin(t)cos(6t) + cos(t)sin(6t)
cos(7t) = cos(t+6t) = cos(t)cos(6t) – sin(t)sin(6t)
sin(t)sin(7t) – cos(t)cos(7t)
= sin(t)[sin(t)cos(6t)+cos(t)sin(6t)] + cos(t)[cos(t)cos(6t)-sin(t)sin(6t)]
= sin²(t)cos(6t) + sin(t)cos(t)sin(6t) + cos²(t)cos(6t) – sin(t)cos(t)sin(6t)
= sin²(t)cos(6t) + cos²(t)cos(6t)
= cos(6t) (sin²(t)+cos²(t)
= cos(6t)
Replacing this in formula for x'(t)²+y'(t)², we get:
x'(t)² + y'(t)²
= 98 – 98(sin(t)sin(7t) + cos(t)cos(7t))
= 98 – 98cos(6t)
= 98 – 98(1 – 2sin²(3t))
= 98 – 98 + 196sin²(3t)
= 196 sin²(3t)
So now we can calculate length of curve:
L = ∫[0,π] √[ (x'(t))² + (y'(t))² ] dt
L = ∫[0,π] √196 sin²(3t) dt
L = ∫[0,π] 14 | sin(3t) | dt
Now we must use |sin(3t)| to calculate length.
If we didn’t use absolute values, then we would just be calculating length of straight line from first point to last point on curve.
now sin(3t) >= 0 on intervals [0,π/3] and [2π/3,π]
and sin(3t) <= 0 on intervals [π/3,2π/3]
Therefore
L = ∫[0,π] 14 | sin(3t) | dt
L = ∫[0,π/3] 14sin(3t) dt – ∫[π/3,2π/3] 14sin(3t) dt + ∫[2π/3,π] 14sin(3t) dt
L = -14/3 cos(3t) |[0,π/3] + 14/3 cos(3t) |[π/3,2π/3] – 14/3 cos(3t) |[2π/3,π]
L = -14/3(-1-1) + 14/3(1-(-1)) – 14/3(-1-1)
L = 28/3 + 28/3 + 28/3
L = 28
Question 6: find the exact length of the curve. y = x3 3 + 1 4x , 1 ≤ x ≤ 2
SOLUTION:
y = x^3/3 + 1/(4x), what is the exact length of the curve from 1 to 2?
integral of (sqrt(1 + (f'(x)^2))?
f'(x) = x^2 – 1/4(x^-2)
Squaring that gives you a gnarly thing
X^4 – 1/2 + 1/16(x^-4) + 1
So x^4 + 1/2 + 1/16(x^-4)
That can be factored into (x^2 + 1/4(x^-2))^2
sqrt that term to get
x^2 + 1/4(x^-2)
Integrate that
x^3/3 – 1/4(x^-1)
Evaluate at 1 and at 2
(8/3 – 1/8) – (1/3 – 1/4) — 2.4833 so some fraction.
