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=1/27((9x^2)+(6))^3/2 ,from x=3 to x=6

Answer 1

The length of the curve $y = \frac{1}{27}(9x^2 + 6)^\frac 32$ from x = 3 to x = 6 is 192 units

How to determine the length of the curve?

The curve is given as:

$y = \frac{1}{27}(9x^2 + 6)^\frac 32$ from x = 3 to x = 6

Start by differentiating the curve function

$y' = \frac 32 * \frac{1}{27}(9x^2 + 6)^\frac 12 * 18x$

Evaluate

$y' = x(9x^2 + 6)^\frac 12$

The length of the curve is calculated using:

$L =\int\limits^a_b {\sqrt{1 + y'^2}} \, dx$

This gives

$L =\int\limits^6_3 {\sqrt{1 + [x(9x^2 + 6)^\frac 12]^2}\ dx$

Expand

$L =\int\limits^6_3 {\sqrt{1 + x^2(9x^2 + 6)}\ dx$

This gives

$L =\int\limits^6_3 {\sqrt{9x^4 + 6x^2 + 1}\ dx$

Express as a perfect square

$L =\int\limits^6_3 {\sqrt{(3x^2 + 1)^2}\ dx$

Evaluate the exponent

$L =\int\limits^6_3 {3x^2 + 1} \ dx$

Differentiate

$L = x^3 + x|\limits^6_3$

Expand

L = (6³ + 6) - (3³ + 3)

Evaluate

L = 192

Hence, the length of the curve is 192 units

Read more about curve lengths at:

https://brainly.com/question/14015568

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