Find the Exact Length of the Curve Answers

by Quizaza team | Last Updated: January 4, 2021

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.