Here's the problem:
Integrate $f$ over the given curve. $$ f(x,y) = \frac{x+y^2}{\sqrt{1+x^2}}\qquad C: y=\frac{x^2}{2} \text{ from } (1,1/2) \text{ to } (0,0) $$
In the solutions manual it says that $$ r(t) = (1-t)i + \frac{(1-t)^2}{2}j $$
I can deal with this after that. The only thing I don't understand is how to get the $r(t)$ vector above. I can't find it anywhere on the internet.
Thanks in advance! And sorry, I don't know Latex.
It seems that you'd like a curve $r(t)$ that traces the curve $y = \frac 12 x^2$ from $(1,1/2)$ to $(0,0)$.
In general, it's fairly easy to make a curve that follows a function $y = f(x)$ between two $x$-values. All you need to do is:
In this case, we want to graph $y = \frac 12 x^2$ from $x = 1$ to $x=0$. First, we need an $x(t)$ that starts at $1$ and ends at $0$. One easy solution is to set $x(t) = 1-t$, letting $t$ go from $0$ to $1$.
In general, we can find a linear function for which $x(0) = a$ and $x(1) = b$ by setting $x(t) = a + (b - a)t$.
Once you've done that, you need to set $y(t) = \frac 12 [x(t)]^2 = \frac 12 (1-t)^2$. Now, we can just set $$ r(t) = (1-t)i + \frac 12 (1-t)^2 j $$ and we're done.