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I got an answer of $\frac{4}{3}$, but the textbook answer says $2$. How would one solve it, and what did I do wrong?

Let $\vec r=(1+t^2)\hat i+\arctan(t)\hat j$ where $0≤t≤1$

$dx=2tdt, dy=\frac{1}{1+t^2}dt, \sec^2y=1-\tan^2y$ $\int_c\tan ydx+x\sec ^2ydy=\int_0^1\tan(\arctan(t))\cdot 2tdt+(1+t^2)(1-\tan ^2(\arctan(t))\cdot \frac{1}{1+t^2}dt$

$=\int_0^1(2t^2+(1-t^2))dt$

$=\int_0^1(t^2+1)dt=\frac{4}{3}$

(Stewart Multivariable Calculus 4e, 13.3, Q19)

David
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1 Answers1

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You have a small sign/formula mistake:

$\sec^2y=1\color{red}{-}\tan^2y$

This should be: $\sec^2y = 1\color{blue}{+}\tan^2y$; so:

$\int_0^1(2t^2+(1\color{red}{-}t^2)) \,\mbox{d}t$

becomes:

$$\int_0^12t^2+\left(1\color{blue}{+}t^2\right)\,\mbox{d}t = 2$$

Alternatively, notice that: $$\mbox{d}\left(x\tan y \right) = x\,\mbox{d}\tan y + \tan y \,\mbox{d} x = x\sec^2y\,\mbox{d}y + \tan y \,\mbox{d} x $$ so: $$\int_C x\sec^2y\,\mbox{d}y + \tan y \,\mbox{d} x = \int_C \mbox{d}\left(x\tan y \right) = \left[ x\tan y \right]_{\left( 1,0 \right)}^{( 2,\pi/4)} = \ldots$$

StackTD
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  • Thanks, this is a much more simple approach to the problem. What was the mistake I made in mine? – David May 10 '17 at 11:25
  • @David I was just adding that; see above. – StackTD May 10 '17 at 11:27
  • Thanks, this is the first time I've seen this total differential method, but it does help me better understand what the $P$ and $Q$ terms are composed of. It certainly is worth exploring. – David May 10 '17 at 11:35
  • @David Your question ("any path") seems to imply the result is path-independent, and that should ring an alarm bell to look for the gradient (or total differential) of a function. – StackTD May 10 '17 at 11:42
  • I just realized you were actually using the fundamental theorem for line integrals, my bad. Yes, the first part of the question was to prove path-independence. – David May 10 '17 at 12:15
  • @David Alright; in that case they were probably trying to 'lead you' towards this approach, using the fundamental theorem for line integrals. – StackTD May 10 '17 at 13:36