Find the exact length of the curve: y = 5 + 2x3/2, 0 ≤ x ≤ 1?
We will be using the formula of the exact length of the curve and also, use the substitution method to solve this.
Answer: The Exact Length of the Curve y = 5 + 2x3/2, 0 ≤ x ≤ 1 is 2.268 units.
Let's solve this question step by step.
Explanation:
Given function, y = 5 + 2x3/2
Now, differentiate the given function y = (5 + 2x3/2) with respect to “x”
dy/dx = d(5 + 2x3/2)/dx
dy/dx = 0 + (3/2) × 2 × x1/2
dy/dx = 3x(1/2) ---- (1)
To find arc length, we use the following formula for the length of the arc(L),
L = \(\int_{a}^{b}\) √[1 + (dy/dx)2] dx.
Putting the value of dy/dx in length of curve formula from (1)
L= \(\int_{a}^{b}\) √[1 + (3x1/2)2] dx.
L= \(\int_{a}^{b}\)√[1 + 9x] dx.
Subsitute, a = 0 and b = 1
L = \(\int_{0}^{1}\)√[1 + 9x] dx.
Now, Integrate using substitution method
1+ 9x = u2
Differentiate both sides of 1+ 9x = u2 w.r.t.x
d(1)/dx + d(9x)/dx = d(u2)/dx
9dx = 2u du
⇒ dx = (2/9)u du
Since we have substituted the function in terms of u, therefore we have to change the limits.
If x = 0 then, u = √[1+9(0)] = 1
if x = 1 then, u = √[1+9(1)] = √10
Substitute the value the limits.
L = (2/9) \(\int_{1}^{\sqrt{10}}\)u.u du.
⇒ L = (2/9)\(\int_{1}^{\sqrt{10}}\) u2 du
⇒ L = (2/9)[ u3/3 \(]_{1}^{\sqrt{10}}\) [Since, \(\int\) xn = (xn+1)/(n +1) ]
Put the upper and lower limits.
⇒ L = (2/27) [10√10 - 1]
⇒ L= (2/27) × 30.62277
⇒ L = 2.268 units.