I am not sure how to estimate the length of $y=f(x)$ for $3 \leq x \leq 3.1$.
Let $C$ be the graph of $y=f(x)$ for $3 \leq x \leq 3.1$ and suppose $f(3)=5$ and $f'(3)=6$. Give an estimate for the length of $C$.
I know that Arc Length = $\int^b_a\sqrt{(x')^2+(y')^2}dt$.
Since the problem only gives me the value of $y'$, I am not sure how to find the arc length, and I'm not sure how to estimate it. I'm not completely sure if it's referring to Arc Length either.
Use a linear approximation to $f(x)$ at $x=3$:
$$f(x)\approx L(x)=f(3)+f'(3)(x-3)=5+6(x-3)$$
$$f(x)\approx6x-13\implies f'(x)\approx6$$
Then the arc length is approximately
$$\int_C\mathrm ds=\int_3^{3.1}\sqrt{1+f'(x)^2}\,\mathrm dx\approx\sqrt{37}\int_3^{3.1}\mathrm dx\approx0.6083$$