Question

Find the length of the following curve. If you have a grapher, you may want to graph the curve to see what it looks like. y = (2/3) (x^2 + 1) ^ (3/2) from x = 0 to x = 6 The length of the curve is ___. (Type an exact answer, using radicals as needed.)

Answer 1

Answer:

Length of curve=150 units

Step-by-step explanation:

We are given that a curve

$y=\frac{2}{3}(x^2+1)^{\frac{3}{2}}$

x=0 to x=6

We have to find the length of curve.

Differentiate w.r.t x then we get

$\frac{dy}{dx}=\frac{2}{3}\times \frac{3}{2}\times (x^2+1)^{\frac{1}{2}}\times 2x=2x(x^2+1)^{\frac{1}{2}}$

Using formula:$\frac{dx^n}{dx}=nx^{n-1}$

Length of curve=$\int_{a}^{b}\sqrt{1+(\frac{dy}{dx})^2}dx$

Substitute the value

Then, we get

Length of curve=$\int_{0}^{6}\sqrt{1+(2x(x^2+1)^{\frac{1}{2}})^2}dx$

Length of curve=$\int_{0}^{6}\sqrt{1+4x^2(x^2+1)}dx$

Length of curve=$\int_{0}^{6}\sqrt{1+4x^4+4x^2}dx=\int_{0}^{6}\sqrt{(2x^2+1)^2dx$

Length of curve=$\int_{0}^{6}(2x^2+1)dx=[\frac{2x^3}{3}+x]^{6}_{0}$

Using formula:$\int_{a}^{b}f(x)dx=f(b)-f(a)$

Length of curve=$\frac{2}{3}(6)^3+6-0-0=150$ units