Vector Calc: Can you verify my answer?

**rsq_a** · Jun7-09, 04:31 PM

Just looking for a yes/no. I worked out the following question:

(A) Calculate the arc length of the curve $LaTeX Code: r(t) = (\\log t, 2t, t^2)$ where $LaTeX Code: 1 \\leq t \\leq e$

(B) Let C be the ellipsed form by intersecting the cylinder and the plane and let $LaTeX Code: \\textbf{f}(x,y,z) = (y,z,x)$ . What is $LaTeX Code: \\int_C \\textbf{f} d\\textbf{r}$ .

(C) Let C be the hyperbola formed by intersecting the cone and the plane and let $LaTeX Code: \\textbf{f}(x,y,z) = \\textbf{k}/z^2$ . What is $LaTeX Code: \\int_C \\textbf{f} \\times d\\textbf{r}$

I got,

(A)

(B) $LaTeX Code: \\pi$
(C) $LaTeX Code: -(1+3\\pi/2, 1+3\\pi/2, 0)$

Did I hit the mark?

**HallsofIvy** · Jun7-09, 05:13 PM

I haven't done (C) yet but I don't get anything like you have for either (A) or (B). Show how you got those answers.

**rsq_a** · Jun7-09, 05:25 PM

Originally Posted by HallsofIvy

I haven't done (C) yet but I don't get anything like you have for either (A) or (B). Show how you got those answers.

Okay. So for (A):

$LaTeX Code: \\begin{align*}L &= \\int_1^e |rsingle-quote(t)| dt \\\\ &= \\int_1^e |(1/t, 2, 2t)| dt \\\\ <BR>&= \\int_1^e \\sqrt{\\frac{(2t^2+1)^2}{t^2}} dt = \\int_1^e 2t + \\frac{1}{t} dt \\\\<BR>&= e^2 - 1 + 1 = e^2\\end{align*}$

For (B), the contour is,

$LaTeX Code: \\gamma(t) = (\\cos t, \\sin t, 2\\cos t + 1)$

So the relevant integral is,

$LaTeX Code: \\int_0^{2\\pi} (\\sin t, 2 \\sin t + 1, \\cos t) \\cdot (-\\sin t, \\cos t, -2 \\cos t) \\ dt = \\pi$

Does that look wrong?

**daudaudaudau** · Jun8-09, 05:36 AM

In (B), if you say that y=sin(t), then you should find z=2*y+1=2*sin(t)+1, right?

**rsq_a** · Jun8-09, 05:41 AM

Originally Posted by daudaudaudau

In (B), if you say that y=sin(t), then you should find z=2*y+1=2*sin(t)+1, right?

Doh! But I think the answer is the same. I've corrected it in the post above.