Calculus Question

Question

A curve C is described by parametric equations x= 1/3 t³ +1 and y= t²-3.

a.) Find the values of t corresponding to points P (4/3,-2) and Q (10,6) on C.

b.) Find the length of the arc of C from point P to point Q.

Answer 1

The curve is x= 1/3 t³ +1 and y= t²-3.

We are to find the values of t corresponding to points P (4/3,-2) and Q (10,6) .

When x = 4/3 we get 4/3 = 1/3 t^3 + 1 = >1/3 t^3 = 1/3 = > t = 1 ; when y = -2 then -2 = t^2 -3 = > t^2 = 1 = > t =1 ;

when x = 10 we get 10 = 1/3 t^3 +1 = > 1/3 t^3 = 9 => t^3 = 27 => t => 3;

when y = 6 we get 6 = t^2 -3 = > t ^2 = 9 => t = 3;

So for point P the value of t = 1

and for point Q the value of t = 3

Now the arc length between P and Q will be given by

Integrate (SQRT [(dx/dt)^2 + (dy/dt)^2]) from t = 1 to 3

Now dx/dt = t^2 and dy/dt = 2t hence the above form will be

Integrate(SQRT[t^4 + 4t^2]) from t = 1 to t = 3

Using a Graphing calculator TI - 84 Plus we have to use MATH function first and then using function fnInt

we get required arc length = 11.89727556

Answer 2

x-1 = (1/3) t^3

3(x-1) = t^3

cube_root( 3(x-1)) = t

y = ( 3(x-1) )^(2/3) - 3

for point P, X = 4/3 ---> t = CUBE_ROOT ( 3 ( 4/3-1)) = cube_root(1) = 1

AND Y=1^2-3 = 1-3 = -2 so yes, it checks. t=1

for point Q, cube_root( 3(10-1)) = cube_root(3(9)) = cube_root(27) = 3

AND y = 3^2 - 3 = 9-3 = 6, so yes, it checks.... t=3

y = ( 3(x-1) )^(2/3) - 3

= (3x-3)^(2/3) - 3

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the derivative is (2/3)(3x-3)^(-1/3)(3) = 2(3x-3)^(-1/3) = 3/cube_root(3x-3)

arclength function is sqrt ( 1 + 9/ (3x-3)^(2/3) )

this function must be integrated numerically from 4 to 10.

Simpson's 3/8 rule with n=0.01 gives an approximation of 9.23905