Answer to Question #134205 in Calculus for Promise Omiponle

Question

Let C be the curve of intersection of the parabolic cylinder y=x^2/2 and z=xy/3. What is the length of this curve between the origin and the point (6,18,36)? You'll want to parameterize by t as simply as possible (e.g. x(t) = t ).

Answer 1

$$Solution$$

$y= \frac{1}{2}x^2 .............eq 1$

$z= \frac{1}{3}xy.............eq2$

Change equation 1 into a parametric form replacing x with t i.e.

$when, x=t, y= \frac{1}{2}t^2$

Solving equation 2 for z in terms of t, we have;

$z= \frac{1}{3}(x.y) = \frac{1}{3} (t. \frac{1}{2}t^2 ) = \frac{1}{6}t^3$

plugging in the x, y, and z coordinates into a vector equation for the curve r(t), we have;

$r(t) = < t, \frac{1}{2}t^2 , \frac{1}{6}t^3 >$

Now calculating the 1^st derivative of r(t)

$r'(t) = < 1, t, \frac{1}{2}t^2 >$

Calculating and simplifying the magnitude of r'(t), we have;

$\begin{vmatrix} r'(t) \\ \end{vmatrix} = \sqrt{\frac{1}{4}t^4 + t^2 +1}$

$=$ $\frac{1}{2} \sqrt{ t^4 + 4 t^2 +4}$

$=$ $\frac{1}{2} \sqrt{( t^2 +2)^2}$

$= \frac{1}{2}t^2 +1$

Now finding the range of t along the curve between the origin and the point (6, 18, 36) i.e.

At the origin: $(0,0,0) → t=0$

At the point: $(6,18, 36) t → 6$

Therefore, the range of t is; $0 ≤ t ≤ 6$

Now length of the curve, C is obtained by taking the integral for the arclength from 0 to 6, i.e.

$C=\int _{0}^6 (\frac{1}{2}t^2 + 1 ) dt$

$C = (\frac{1}{6}t^3) | _{0}^6 = 42$

Therefore the length of the curve is 42 Units