# Find the arc length of the cuve r(t)=(cos^3 t) i +  (sin^3 t) j +  2k     ,     0=< t =<phi/2solve this early

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to determine the length of the vector function `bar r(t) =`  `ltcos^3 t, sin^3 t, 2gt, ` over interval `[0,pi/2], ` hence you need to find the magnitude of the tangent vector such that:

`|bar r'(t)| = sqrt((3cos^2 t*(-sin t))^2 + (3sin^2 t*cos t)^2)`

`|bar r'(t)| = sqrt(9cos^4 t*sin^2 t + 9sin^4 t*cos^2 t) `

`|bar r'(t)| = sqrt(9cos^2 t*sin^2 t(cos^2 t+ sin^2 t) `

You need to remember that `cos^2 t + sin^2 t = 1`  such that:

`|bar r'(t)| = sqrt(9cos^2 t*sin^2 t)`

`|bar r'(t)| = 3 sin t*cos t`

You need to remember the formula of arc length of a curve such that:

`L = int_0^(pi/2) |bar r'(t)| dt`

`L = int_0^(pi/2) 3 sin t*cos t dt`

You need to multiply and divide the integrand `sin t*cos t`  by 2 to obtain the formula of sine of double angle such that:

`L = (3/2) int_0^(pi/2)2 sin t*cos t dt`

`L = (3/2) int_0^(pi/2) sin(2t) dt `

`L = -(3/2) (cos(2t))/2 |_0^(pi/2)`

`L = -(3/4)(cos pi - cos 0)`

`L = -(3/4)(-1- 1)`

`L = 3/2`

Hence, evaluating the arc length of the curve given in vectorial form yields `L = 3/2` .

neela | High School Teacher | (Level 3) Valedictorian

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r(t) =(cos^3t)i+sin^3t*j+2*k.To find the arc length from t=0 to t=pi/2.

We know that  if r(t) = x(t)*i+y(t)*j +z(t)*k, then

S =  Integral {sqrt[(x'(t))^2+(y'(t))^2+(z'(t))^2]} dt, t = a to b is the arc length from t =a to t = b.

Therefore here,

S = Int { sqrt[((3cos^3t)' )^2 + ((3sin^3t)')^2+ (2')^2]} dt , t= 0o to pi/2

S = Int { sqrt[(3cos^2t*(-sint))^2+(3sin^2t*cost)^2+(0)]}dt, t = 0 to pi/2

=Int {3sintcost sqrt[cos^2+sin^t)} dt, t= 0 to pi/2

=Int (3sintcost) dt , t = 0 to 1.

=3/2 Int (2sintcost) dt , t = 0, to pi/2

= (3/2 ) Int sin2t dt, t = 0 to pi/2

= 3/2{ (-cos2t)/2 at t =pi/2 ] - [(-cos2t)/2 art t=0] }

=3/4 { -cos pi + cos 0}

=3/4 ( -(-1)+(1)}

=(3/4)(2)

=3/2