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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.

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    $\begingroup$ You're better off using $\operatorname{Arc Length}=\int_C\sqrt{1+(f'(x))^2}\,dx,$ since you don't have a parametrization. Also, do you mean $-3\le x\le 3?$ $\endgroup$ – Adrian Keister Dec 2 '19 at 16:43
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    $\begingroup$ Your parameterization would be $x(t)=t$ and $y(t)=f(t)$. Then this reduces to what @AdrianKeister suggests. $\endgroup$ – MPW Dec 2 '19 at 16:44
  • $\begingroup$ @AdrianKeister No, that's just how it was written. I am confused by $\int_C$ do you mean $\int_5$ since $f(3)=5$? Where is the upper bound? EDIT: Nevermind, I read the answer and I see $\int_C=\int_3^{3.1}$. $\endgroup$ – LuminousNutria Dec 2 '19 at 17:03
  • $\begingroup$ $\int_C$ denotes the integral over some curve $C$. Here we integrate over the real interval $[3,3.1]$, hence the equality. But in general $\int_C$ does not require that $C$ be a one-dimensional path. $\endgroup$ – user170231 Dec 2 '19 at 17:38
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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

$$\int_C\mathrm ds=\int_3^{3.1}\sqrt{1+f'(x)^2}\,\mathrm dx\approx\sqrt{37}\int_3^{3.1}\mathrm dx\approx0.6083$$

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