Draw the function
$$ f(x)=2x^4+3x^3-2x+8 $$ into a cartesian plane.
$\endgroup$ 2How to measure the exact length of that curve from $x=2.2$ to $x=33.64$ ? Is it true that we can measure the exact length of that curve just using the differential/calculus function or some sort?
5 Answers
$\begingroup$$$ ds = \text{infinitesimal increment of arc length} = \sqrt{(dx)^2+(dy)^2}. $$ That is an application of the Pythagorean theorem.
Therefore $$ \text{arc length} = \int_{x=2.2}^{x=33.64} \sqrt{(dx)^2+(dy)^2}. $$
Then do a bit of algebra: \begin{align} & dy = f'(x)\,dx, \text{ so } (dy)^2 = f'(x)^2(dx)^2, \text{ and consequently } \\[10pt] & \sqrt{(dx)^2+(dy)^2} = \sqrt{(dx)^2+f'(x)^2(dx)^2} = \sqrt{1+f'(x)^2} \sqrt{(dx)^2} = \sqrt{1+f'(x)^2} \ dx. \end{align}
$\endgroup$ 2 $\begingroup$A length of a curve $y=f(x)$ from point $a$ to point $b$, so called Arc Lenght, is given by $$L = \int_a^b \sqrt{1+(f'(t))^2}dt$$
The formula is of an exact length of course. However, this is elliptical integral (hyperelliptical in your case) which is often have to be approximated numerically, like in your case, so the end of story the length won't really be exact.
The approximation of your specific problem obtained using Mathematica gives $$\int_{2.2}^{33.64} \sqrt{1+\left(8 x^3+9 x^2-2\right)^2}\, dx = 2.67533\times 10^6$$
$\endgroup$ 10 $\begingroup$Since Ms Lawrence is sceptical, in SIA we have Taylor's formula: $$f(x + \varepsilon) = f(x) + \varepsilon f'(x)$$ If s gives the curve length: $$s(x + \varepsilon) - s(x) = \varepsilon s'(x)$$ Where the RHS is a microstraight portion of the curve. Therefore: $$\varepsilon s'(x) = \sqrt{\varepsilon^2 + (\varepsilon f'(x))^2 }$$ From Pythagoras, which simplifies to: $$s'(x) = \sqrt{1 + f'(x)^2 }$$ $$s(x) = \int\sqrt{1 + f'(x)^2 }$$ In non-SIA a dx is added to the RHS by convention, but to numerically evaluate the length you would use the above equation, unless you use very small secants of course.
$\endgroup$ 4 $\begingroup$Ever since Newton introduced " fluxions" there is only one answer.
If the derivative $ 8 x^3 + 9 x^2 - 2 $ is labelled $ y^{'}$, then the length $$ = \int_{2.2 }^ {33.64} \sqrt{1+ y^{'2}} dx $$
To get exactness the curve must be accurately drawn, measuring contact sides raised out of the sketched plane using some hardenable plastic material of proper rigidity and flexibility, a flexible tape used ( like one used for tailoring , steel tape wont flex around corners)... but before all that, may be better to measure on available circular arcs to verify theory .
$\endgroup$ $\begingroup$The exact length of that curve must be greater than $f(33.64)-f(2.2)=2675326.03$. Because we can picture that curve roughly as a hypotenuse of a right triangle and the vertical line from $y=f(2.2)$ to $y=f(33.64)$ as the vertical side of that right triangle. So the the exact length of that curve from $x=2.2$ to $33.64$ must be greater than $2675326.03$.(A MUST)
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