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
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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