from a handpicked tutor in LIVE 1-to-1 classes
Find the exact length of the curve: y = 4 + 4x3/2, 0 ≤ x ≤ 1?
We will be using the formula of the exact length of the curve to solve this.
Answer: The exact length of the curve y = 4 + 4x3/2, 0 ≤ x ≤ 1, is 4.149 units.
Let's solve this question step by step.
Explanation:
Given function is y = 4 + 4x3/2
Now, differentiate the given function (4 + 4x3/2) with respect to “x”
dy/dx = d(4 + 4x3/2)/dx
dy/dx = 0 + (3/2) × 4 × x1/2
dy/dx = 6x(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 + (6x1/2)2]dx
L= \(\int_{a}^{b}\)√[1 + 36x] dx
Subsitute, a = 0 and b = 1
L=\(\int_{0}^{1}\)√[1 + 36x] dx.
Now, Integrate using substitution method
1+ 36x = u
Differentiate both side w.r.t.x
d(1)/dx + d(36x)/dx = d(u)/dx
dx = du/36
L=\(\int_{0}^{1}\)√[u]/36 du.
Since we have substituted the function in terms of u, therefore we have to change the limits.
If x = 0 then, u= 1+36(0) = 1
if x = 1 then, u= 1+36(1) = 37
Substitute the value the limits.
L=\(\int_{1}^{37}\)√[u]/36 du.
⇒ L= (1/36)\(\int_{1}^{37}\)√[u] du
⇒ L= (1/36)\(\int_{1}^{37}\) u(1/2) du
⇒ L= (1/36)[ u(3/2) / (3/2) \(]_{1}^{37}\) [Since, \(\int\) xn = (xn+1)/(n +1) ]
⇒ L= (1/36)×(2/3)[ u(3/2) \(]_{1}^{37}\)
⇒ L= (2/108)[ u(3/2) \(]_{1}^{37}\)
Put the upper and lower limits.
⇒ L= (2/108)[ 37(3/2) - 1(3/2)]
L= 4.149 units.
Hence, the exact length of the curve y = 4 + 4x3/2, 0 ≤ x ≤ 1, is 4.149 units.
visual curriculum