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ANSWER

[tex] =102 \sqrt{ 17}[/tex]

EXPLANATION

The equation of the curve is:

[tex]x = 9 + 9 {t}^{2} [/tex]

[tex]y = 6 + 6 {t}^{3} [/tex]

We differentiate to obtain:

[tex] \frac{dx}{dt} = 18t[/tex]

[tex] \frac{dy}{dt} = 18t[/tex]

The length of the arc

[tex]l = \int _0 ^{4} \sqrt{ {( \frac{dx}{dt} )}^{2} + {( \frac{dy}{dt} )}^{2} } dt[/tex]

This implies that:

[tex] = \int _0 ^{4} \sqrt{ {( 18t )}^{2} + {( {18 {t}^{2} } )}^{2} } dt[/tex]

[tex] = \int _0 ^{4} 18t \sqrt{ 1+ { {18 {t}^{2} } }} dt[/tex]

[tex] =102 \sqrt{ 17}[/tex]

azikennamdi

The exact length of the curve is given in standard form as 102√17. See the solution below.

What is the explanation to the above solution?

Recall that the equation for the curve is given as:

x = 9 + 9t²

y = 6 + 6t³

The next step is to differentiate them. Thus we have:

dx/dt = 18t

dy/dt = 18t

To determine the length of the curve, we state:

t = [tex]\int_0^4 \sqrt{(\frac{dx}{dt})^{2} + (\frac{dy}{dt})^{2} dt }[/tex]

This must result in

= [tex]\int_0^4 \sqrt{(18t)^{2} + (18t})^{2} dt }[/tex]

= [tex]\int_0^4 18t\sqrt{1 + 18t^{2} dt }[/tex]

= 102 √17

Learn more about curves at;
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