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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


Sagot :

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

How to determine the length of the curve?

The curve is given as:

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

Start by differentiating the curve function

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

Evaluate

[tex]y' = x(9x^2 + 6)^\frac 12[/tex]

The length of the curve is calculated using:

[tex]L =\int\limits^a_b {\sqrt{1 + y'^2}} \, dx[/tex]

This gives

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

Expand

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

This gives

[tex]L =\int\limits^6_3 {\sqrt{9x^4 + 6x^2 + 1}\ dx[/tex]

Express as a perfect square

[tex]L =\int\limits^6_3 {\sqrt{(3x^2 + 1)^2}\ dx[/tex]

Evaluate the exponent

[tex]L =\int\limits^6_3 {3x^2 + 1} \ dx[/tex]

Differentiate

[tex]L = x^3 + x|\limits^6_3[/tex]

Expand

L = (6³ + 6) - (3³ + 3)

Evaluate

L = 192

Hence, the length of the curve is 192 units

Read more about curve lengths at:

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