Difference between revisions of "Arc Length and Surface Area"

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Suppose that we are given a function <math>f</math> that is continuous on an interval <math>[a,b]</math> and we want to calculate the length of the curve drawn out by the graph of <math>f(x)</math> from <math>x=a</math> to <math>x=b</math> . If the graph were a straight line this would be easy — the formula for the length of the line is given by Pythagoras' theorem. And if the graph were a piecewise linear function we can calculate the length by adding up the length of each piece.
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 +
The problem is that most graphs are not linear. Nevertheless we can estimate the length of the curve by approximating it with straight lines. Suppose the curve <math>C</math> is given by the formula <math>y=f(x)</math> for <math>a\le x\le b</math> . We divide the interval <math>[a,b]</math> into <math>n</math> subintervals with equal width <math>\Delta x</math> and endpoints <math>x_0,x_1,\ldots,x_n</math> . Now let <math>y_i=f(x_i)</math> so <math>P_i=(x_i,y_i)</math> is the point on the curve above <math>x_i</math> . The length of the straight line between <math>P_i</math> and <math>P_{i+1}</math> is
 +
:<math>\bigl|P_iP_{i+1}\bigr|=\sqrt{(y_{i+1}-y_i)^2+(x_{i+1}-x_i)^2}</math>
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 +
So an estimate of the length of the curve <math>C</math> is the sum
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:<math>\sum_{i=0}^{n-1}\bigl|P_iP_{i+1}\bigr|</math>
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 +
As we divide the interval <math>[a,b]</math> into more pieces this gives a better estimate for the length of <math>C</math> . In fact we make that a definition.
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 +
{{Calculus/Def | title=Length of a Curve | text=The length of the curve <math>y=f(x)</math> for <math>a\le x\le b</math> is defined to be
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:<math>L=\lim_{n\to\infty}\sum_{i=0}^{n-1}\bigl|P_{i+1}P_i\bigr|</math>
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}}
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==The Arclength Formula==
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Suppose that <math>f'</math> is continuous on <math>[a,b]</math> . Then the length of the curve given by <math>y=f(x)</math> between <math>a</math> and <math>b</math> is given by
 +
:<math>L=\int\limits_a^b \sqrt{1+f'(x)^2}dx</math>
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And in Leibniz notation
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:<math>L=\int\limits_a^b \sqrt{1+\left(\tfrac{dy}{dx}\right)^2}dx</math>
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'''Proof:''' Consider <math>y_{i+1}-y_i=f(x_{i+1})-f(x_i)</math> . By the [[Calculus/Mean Value Theorem for Functions|Mean Value Theorem]] there is a point <math>z_i</math> in <math>(x_{i+1},x_i)</math> such that
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:<math>y_{i+1}-y_i=f(x_{i+1})-f(x_i)=f'(z_i)(x_{i+1}-x_i)</math>
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 +
So
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:{|
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|<math>\bigl|P_iP_{i+1}\bigr|</math>
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|<math>=\sqrt{(x_{i+1}-x_i)^2+(y_{i+1}-y_i)^2}</math>
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|-
 +
|
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|<math>=\sqrt{(x_{i+1}-x_i)^2+f'(z_i)^2(x_{i+1}-x_i)^2}</math>
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|-
 +
|
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|<math>=\sqrt{\bigl(1+f'(z_i)^2\bigr)(x_{i+1}-x_i)^2}</math>
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|-
 +
|
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|<math>=\sqrt{1+f'(z_i)^2}\Delta x</math>
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|}
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 +
Putting this into the definition of the length of <math>C</math> gives
 +
 +
:<math>L=\lim_{n\to\infty}\sum_{i=0}^{n-1}\sqrt{1+f'(z_i)^2}\Delta x</math>
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 +
Now this is the definition of the integral of the function <math>g(x)=\sqrt{1+f'(x)^2}</math> between <math>a</math> and <math>b</math> (notice that <math>g</math> is continuous because we are assuming that <math>f'</math> is continuous). Hence
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:<math>L=\int\limits_a^b \sqrt{1+f'(x)^2}dx</math>
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as claimed.
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{{ExampleRobox|title=Example: Length of the curve <math>y=2x</math> from <math>x=0</math> to <math>x=1</math>}}
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As a sanity check of our formula, let's calculate the length of the "curve" <math>y=2x</math> from <math>x=0</math> to <math>x=1</math> . First let's find the answer using the Pythagorean Theorem.
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:<math>P_0=(0,0)</math>
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and
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:<math>P_1=(1,2)</math>
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so the length of the curve, <math>s</math> , is
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:<math>s=\sqrt{2^2+1^2}=\sqrt5</math>
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Now let's use the formula
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:<math>s=\int\limits_0^1 \sqrt{1+\left(\tfrac{d(2x)}{dx}\right)^2}\,dx=\int\limits_0^1 \sqrt{1+2^2}\,dx=\sqrt5x\bigg|_0^1=\sqrt5</math>
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{{Robox/Close}}
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===Exercises===
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{{question-answer|question=1. Find the length of the curve <math>y=x\sqrt{x}</math> from <math>x=0</math> to <math>x=1</math> .|answer={{noprint|
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:<math>\frac{13\sqrt{13}-8}{27}</math>}}}}
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{{question-answer|question=2. Find the length of the curve <math>y=\frac{e^x+e^{-x}}{2}</math> from <math>x=0</math> to <math>x=1</math> .|answer={{noprint|
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:<math>\frac{e-\frac{1}{e}}{2}</math>}}}}
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{{noprint|[[Calculus/Arc_length/Solutions|Solutions]]}}
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==Arclength of a parametric curve==
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For a parametric curve, that is, a curve defined by <math>x=f(t)</math> and <math>y=g(t)</math> , the formula is slightly different:
 +
:<math>L=\int\limits_a^b \sqrt{f'(t)^2+g'(t)^2}\,dt</math>
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 +
'''Proof:''' The proof is analogous to the previous one:
 +
Consider <math>y_{i+1}-y_i=g(t_{i+1})-g(t_i)</math> and <math>x_{i+1}-x_i=f(t_{i+1})-f(t_i)</math> .
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By the Mean Value Theorem there are points <math>c_i</math> and <math>d_i</math> in <math>(t_{i+1},t_i)</math> such that
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:<math>y_{i+1}-y_i=g(t_{i+1})-g(t_i)=g'(c_i)(t_{i+1}-t_i)</math>
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and
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:<math>x_{i+1}-x_i=f(t_{i+1})-f(t_i)=f'(d_i)(t_{i+1}-t_i)</math>
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So
 +
:{|
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|<math>\bigl|P_iP_{i+1}\bigr|</math>
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|<math>=\sqrt{(x_{i+1}-x_i)^2+(y_{i+1}-y_i)^2}</math>
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|-
 +
|
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|<math>=\sqrt{f'(d_i)^2(t_{i+1}-t_i)^2+g'(c_i)^2(t_{i+1}-t_i)^2}</math>
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|-
 +
|
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|<math>=\sqrt{\bigl(f'(d_i)^2+g'(c_i)^2\bigr)(t_{i+1}-t_i)^2}</math>
 +
|-
 +
|
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|<math>=\sqrt{f'(d_i)^2+g'(c_i)^2}\Delta t</math>
 +
|}
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 +
Putting this into the definition of the length of the curve gives
 +
:<math>L=\lim_{n\to\infty}\sum_{i=0}^{n-1}\sqrt{f'(d_i)^2+g'(c_i)^2}\Delta t</math>
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This is equivalent to:
 +
:<math>L=\int\limits_a^b \sqrt{f'(t)^2+g'(t)^2}\,dt</math>
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===Exercises===
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{{question-answer|question=3. Find the circumference of the circle given by the parametric equations <math>x(t)=R\cos(t)</math> , <math>y(t)=R\sin(t)</math> , with <math>t</math> running from <math>0</math> to <math>2\pi</math> .|answer={{noprint|
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:<math>2\pi R</math>}}}}
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{{question-answer|question=4. Find the length of one arch of the [[w:Cycloid|<u>cycloid</u>]] given by the parametric equations <math>x(t)=R\bigl(t-\sin(t)\bigr)</math> , <math>y(t)=R\bigl(1-\cos(t)\bigr)</math> , with <math>t</math> running from <math>0</math> to <math>2\pi</math> .|answer={{noprint|
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:<math>8R</math>}}}}
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==Resources==
 
<strong>Arc Length</strong>
 
<strong>Arc Length</strong>
 
* [https://youtu.be/seoFxrNL85c Arc Length - Part 1 of 2] by James Sousa, Math is Power 4U
 
* [https://youtu.be/seoFxrNL85c Arc Length - Part 1 of 2] by James Sousa, Math is Power 4U

Revision as of 13:13, 6 October 2021

Suppose that we are given a function that is continuous on an interval and we want to calculate the length of the curve drawn out by the graph of from to . If the graph were a straight line this would be easy — the formula for the length of the line is given by Pythagoras' theorem. And if the graph were a piecewise linear function we can calculate the length by adding up the length of each piece.

The problem is that most graphs are not linear. Nevertheless we can estimate the length of the curve by approximating it with straight lines. Suppose the curve is given by the formula for . We divide the interval into subintervals with equal width and endpoints . Now let so is the point on the curve above . The length of the straight line between and is

So an estimate of the length of the curve is the sum

As we divide the interval into more pieces this gives a better estimate for the length of . In fact we make that a definition.

Template:Calculus/Def

The Arclength Formula

Suppose that is continuous on . Then the length of the curve given by between and is given by

And in Leibniz notation

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L=\int\limits_a^b \sqrt{1+\left(\tfrac{dy}{dx}\right)^2}dx}

Proof: Consider Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y_{i+1}-y_i=f(x_{i+1})-f(x_i)} . By the Mean Value Theorem there is a point Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle z_i} in Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x_{i+1},x_i)} such that

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y_{i+1}-y_i=f(x_{i+1})-f(x_i)=f'(z_i)(x_{i+1}-x_i)}

So

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bigl|P_iP_{i+1}\bigr|} Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\sqrt{(x_{i+1}-x_i)^2+(y_{i+1}-y_i)^2}}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\sqrt{(x_{i+1}-x_i)^2+f'(z_i)^2(x_{i+1}-x_i)^2}}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\sqrt{\bigl(1+f'(z_i)^2\bigr)(x_{i+1}-x_i)^2}}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\sqrt{1+f'(z_i)^2}\Delta x}

Putting this into the definition of the length of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C} gives

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L=\lim_{n\to\infty}\sum_{i=0}^{n-1}\sqrt{1+f'(z_i)^2}\Delta x}

Now this is the definition of the integral of the function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g(x)=\sqrt{1+f'(x)^2}} between Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle b} (notice that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g} is continuous because we are assuming that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'} is continuous). Hence

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L=\int\limits_a^b \sqrt{1+f'(x)^2}dx}

as claimed.

Template:ExampleRobox As a sanity check of our formula, let's calculate the length of the "curve" Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y=2x} from Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=0} to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=1} . First let's find the answer using the Pythagorean Theorem.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P_0=(0,0)}

and

so the length of the curve, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s} , is

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s=\sqrt{2^2+1^2}=\sqrt5}

Now let's use the formula

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s=\int\limits_0^1 \sqrt{1+\left(\tfrac{d(2x)}{dx}\right)^2}\,dx=\int\limits_0^1 \sqrt{1+2^2}\,dx=\sqrt5x\bigg|_0^1=\sqrt5}

Template:Robox/Close

Exercises

Template:Question-answer

Template:Question-answer Template:Noprint

Arclength of a parametric curve

For a parametric curve, that is, a curve defined by Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=f(t)} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y=g(t)} , the formula is slightly different:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L=\int\limits_a^b \sqrt{f'(t)^2+g'(t)^2}\,dt}

Proof: The proof is analogous to the previous one: Consider Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y_{i+1}-y_i=g(t_{i+1})-g(t_i)} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_{i+1}-x_i=f(t_{i+1})-f(t_i)} .

By the Mean Value Theorem there are points Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c_i} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle d_i} in Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (t_{i+1},t_i)} such that

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y_{i+1}-y_i=g(t_{i+1})-g(t_i)=g'(c_i)(t_{i+1}-t_i)}

and

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_{i+1}-x_i=f(t_{i+1})-f(t_i)=f'(d_i)(t_{i+1}-t_i)}

So

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bigl|P_iP_{i+1}\bigr|} Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\sqrt{(x_{i+1}-x_i)^2+(y_{i+1}-y_i)^2}}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\sqrt{f'(d_i)^2(t_{i+1}-t_i)^2+g'(c_i)^2(t_{i+1}-t_i)^2}}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\sqrt{\bigl(f'(d_i)^2+g'(c_i)^2\bigr)(t_{i+1}-t_i)^2}}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\sqrt{f'(d_i)^2+g'(c_i)^2}\Delta t}

Putting this into the definition of the length of the curve gives

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L=\lim_{n\to\infty}\sum_{i=0}^{n-1}\sqrt{f'(d_i)^2+g'(c_i)^2}\Delta t}

This is equivalent to:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L=\int\limits_a^b \sqrt{f'(t)^2+g'(t)^2}\,dt}

Exercises

Template:Question-answer

Template:Question-answer


Resources

Arc Length


Surface Area

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