1.) Find the arc length y=2/3x ^3/2 + 2 over (14,16)
2.) Find the arc length y=8x^3/2+7 over (0,6)
The two questions are very similar in setup, I will do the first one, you do the second
length = integral √( 1 + (y')^2) dx from a to b
yours:
y = (2/3)x^(3/2) + 2
y' = x^(1/2)
(y')^2= x
length = integral √( 1 + x) dx from 14 to 16
= [ (2/3)(1+x)^(3/2) ] from 14 to 16
= (2/3)(17^(3/2) ) - (1/2)(15^(3/2) )
= appr 7.999
To find the arc length of a curve, we will use the formula:
L = ∫[a,b] sqrt(1 + (dy/dx)^2) dx
where dy/dx represents the derivative of y with respect to x.
1.) Find the arc length of y = (2/3)x^(3/2) + 2 over the interval (14, 16):
First, let's find the derivative of y:
dy/dx = (2/3) * (3/2)x^(3/2 - 1) = (2/3) * (3/2)x^(1/2)
dy/dx = √(2/3) * √x
Now, substitute the values from the given interval into the formula:
∫[14, 16] sqrt(1 + (√(2/3) * √x)^2) dx
∫[14, 16] sqrt(1 + (2/3) * x) dx
To integrate this expression, we can use the substitution method. Let u = 1 + (2/3) * x, then du = (2/3)dx, and dx = (3/2)du. Substituting these values back into the integral:
∫[14, 16] sqrt(u) * (3/2) du
(3/2) ∫[14, 16] √u du
(3/2) * (2/3) *[u^(3/2)] | from 14 to 16
[u^(3/2)] | from 14 to 16
Evaluating the integral:
[(16 + (2/3)*16^(3/2)] - [(14 + (2/3)*14^(3/2)]
Finally, simplify the expression:
[(16 + (2/3)*64)^(1/2)] - [(14 + (2/3)*14^(3/2))^(1/2)]
2.) Find the arc length of y = 8x^(3/2) + 7 over the interval (0, 6):
First, find the derivative of y:
dy/dx = 8 * (3/2)x^(3/2 - 1) = 12x^(1/2)
dy/dx = 12√x
Now, substitute the values from the given interval into the formula:
∫[0, 6] sqrt(1 + (12√x)^2) dx
∫[0, 6] sqrt(1 + 144x) dx
Using the same substitution method as in the previous example, let u = 1 + 144x, then du = 144dx, and dx = (1/144)du:
∫[0, 6] √u * (1/144) du
(1/144) ∫[0, 6] √u du
(1/144) * (2/3) * [u^(3/2)] | from 0 to 6
(1/216) * [u^(3/2)] | from 0 to 6
Evaluating the integral:
(1/216) * [(6^(3/2)) - (0^(3/2))]
Finally, simplify the expression:
(1/216) * (6)^(3/2)
To find the arc length of the given curves, we will use the formula:
L = ∫[a,b] sqrt(1 + (dy/dx)^2) dx
where L represents the arc length, [a,b] is the interval over which we want to find the arc length, and dy/dx is the derivative of y with respect to x.
Let's proceed with each question step-by-step:
1.) Find the arc length of y = (2/3)x^(3/2) + 2 over the interval (14,16):
Step 1: Find the derivative dy/dx:
dy/dx = (2/3)(3/2)x^(3/2 - 1) = (2/3)(3/2)x^(1/2) = x^(1/2)
Step 2: Calculate the integrand sqrt(1 + (dy/dx)^2):
sqrt(1 + (dy/dx)^2) = sqrt(1 + x)
Step 3: Integrate sqrt(1 + (dy/dx)^2) with respect to x over the interval (14,16):
L = ∫[14,16] sqrt(1 + x) dx
To evaluate the integral, we need to use a technique called substitution. Let u = 1 + x, so du = dx. As a result, the integral becomes:
L = ∫[15,17] sqrt(u) du
We integrate (√u) using the power rule, which yields (2/3)u^(3/2):
L = (2/3) * [u^(3/2)] evaluated from 15 to 17
L = (2/3) * [(17)^(3/2) - (15)^(3/2)]
Now, substitute the values and evaluate to find the arc length.
2.) Find the arc length of y = 8x^(3/2) + 7 over the interval (0,6):
Step 1: Find the derivative dy/dx:
dy/dx = 8(3/2)x^(3/2 - 1) = 8(3/2)x^(1/2) = 12x^(1/2)
Step 2: Calculate the integrand sqrt(1 + (dy/dx)^2):
sqrt(1 + (dy/dx)^2) = sqrt(1 + 144x)
Step 3: Integrate sqrt(1 + (dy/dx)^2) with respect to x over the interval (0,6):
L = ∫[0,6] sqrt(1 + 144x) dx
To evaluate the integral, we can use another substitution. Let u = 1 + 144x, so du = 144 dx. The integral becomes:
L = (1/144) ∫[1,865] sqrt(u) du
Integrating (√u) using the power rule, we get (2/3)u^(3/2):
L = (2/3) * [u^(3/2)] evaluated from 1 to 865
L = (2/3) * [(865)^(3/2) - (1)^(3/2)]
Substitute the values and evaluate to find the arc length.