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
![{\displaystyle {\bigl |}P_{i}P_{i+1}{\bigr |}={\sqrt {(y_{i+1}-y_{i})^{2}+(x_{i+1}-x_{i})^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1e935417cba5e210c0469260a045dcc707b0fa13)
So an estimate of the length of the curve
is the sum
![{\displaystyle \sum _{i=0}^{n-1}{\bigl |}P_{i}P_{i+1}{\bigr |}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1559f8f4d0e74eeb7133458293a757a211f74adf)
As we divide the interval
into more pieces this gives a better estimate for the length of
. In fact we make that a definition.
Suppose that
is continuous on
. Then the length of the curve given by
between
and
is given by
![{\displaystyle L=\int \limits _{a}^{b}{\sqrt {1+f'(x)^{2}}}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a934cf4f030a85c6db9c46ad52e12968e851ef98)
And in Leibniz notation
![{\displaystyle L=\int \limits _{a}^{b}{\sqrt {1+\left({\tfrac {dy}{dx}}\right)^{2}}}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0a3ff348626defe37621aa7dad8a06c60fb64e3c)
Proof: Consider
. By the Mean Value Theorem there is a point
in
such that
![{\displaystyle y_{i+1}-y_{i}=f(x_{i+1})-f(x_{i})=f'(z_{i})(x_{i+1}-x_{i})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/39103a400def789b2bdd969c091df1fca0305c61)
So
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Putting this into the definition of the length of
gives
![{\displaystyle L=\lim _{n\to \infty }\sum _{i=0}^{n-1}{\sqrt {1+f'(z_{i})^{2}}}\Delta x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a73267a77ce5b5006d7c3c49117b3c8abb392370)
Now this is the definition of the integral of the function
between
and
(notice that
is continuous because we are assuming that
is continuous). Hence
![{\displaystyle L=\int \limits _{a}^{b}{\sqrt {1+f'(x)^{2}}}dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a934cf4f030a85c6db9c46ad52e12968e851ef98)
as claimed.
Example: Length of the curve ![{\displaystyle y=2x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/582d8299e827a8ee042bff79fd37ead41199f7ef) from ![{\displaystyle x=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/953917eaf52f2e1baad54c8c9e3d6f9bb3710cdc) to ![{\displaystyle x=1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ee42176e76ae6b56d68c42ced807e08b962a2b54)
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1. Find the length of the curve
![{\displaystyle y=x{\sqrt {x}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b8e06ce8082d40bb729d7ea236fc35b78cbac1dc)
from
![{\displaystyle x=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/953917eaf52f2e1baad54c8c9e3d6f9bb3710cdc)
to
![{\displaystyle x=1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ee42176e76ae6b56d68c42ced807e08b962a2b54)
.
:![{\displaystyle {\frac {13{\sqrt {13}}-8}{27}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2dfc47eda1e1ff7e5212b43f81b265a30c5a6088)
:![{\displaystyle {\frac {13{\sqrt {13}}-8}{27}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2dfc47eda1e1ff7e5212b43f81b265a30c5a6088)
2. Find the length of the curve
![{\displaystyle y={\frac {e^{x}+e^{-x}}{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3a7c864ed63fa34292d8eb0685bdbdaedfd3dc14)
from
![{\displaystyle x=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/953917eaf52f2e1baad54c8c9e3d6f9bb3710cdc)
to
![{\displaystyle x=1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ee42176e76ae6b56d68c42ced807e08b962a2b54)
.
:![{\displaystyle {\frac {e-{\frac {1}{e}}}{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f8f3c20e93e2209268c46b2a0f62f12dc66a020e)
:![{\displaystyle {\frac {e-{\frac {1}{e}}}{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f8f3c20e93e2209268c46b2a0f62f12dc66a020e)
Solutions
For a parametric curve, that is, a curve defined by
and
, the formula is slightly different:
![{\displaystyle L=\int \limits _{a}^{b}{\sqrt {f'(t)^{2}+g'(t)^{2}}}\,dt}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8cbce625c0a0481643382a0b1d2234531b24888b)
Proof: The proof is analogous to the previous one:
Consider
and
.
By the Mean Value Theorem there are points
and
in
such that
![{\displaystyle y_{i+1}-y_{i}=g(t_{i+1})-g(t_{i})=g'(c_{i})(t_{i+1}-t_{i})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f47ca8e602e60dcc0c473f6474428bd16ac1beab)
and
![{\displaystyle x_{i+1}-x_{i}=f(t_{i+1})-f(t_{i})=f'(d_{i})(t_{i+1}-t_{i})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/94014653b0f0741b5bd0c69fa06e69b04fb869db)
So
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Putting this into the definition of the length of the curve gives
![{\displaystyle L=\lim _{n\to \infty }\sum _{i=0}^{n-1}{\sqrt {f'(d_{i})^{2}+g'(c_{i})^{2}}}\Delta t}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ad585b2fe59904c1cd2bdc38b2f7fa8c7e83baf3)
This is equivalent to:
![{\displaystyle L=\int \limits _{a}^{b}{\sqrt {f'(t)^{2}+g'(t)^{2}}}\,dt}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8cbce625c0a0481643382a0b1d2234531b24888b)
3. Find the circumference of the circle given by the parametric equations
![{\displaystyle x(t)=R\cos(t)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1e372f5b8369461b974dc8e2c330b14764051364)
,
![{\displaystyle y(t)=R\sin(t)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/96213b01f40c2fb5926d0cb097a2dcce2741ac59)
, with
![{\displaystyle t}](https://wikimedia.org/api/rest_v1/media/math/render/svg/65658b7b223af9e1acc877d848888ecdb4466560)
running from
![{\displaystyle 0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2aae8864a3c1fec9585261791a809ddec1489950)
to
![{\displaystyle 2\pi }](https://wikimedia.org/api/rest_v1/media/math/render/svg/73efd1f6493490b058097060a572606d2c550a06)
.
:![{\displaystyle 2\pi R}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a56efcb39c0744370094e561e7630cd3069a5371)
:![{\displaystyle 2\pi R}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a56efcb39c0744370094e561e7630cd3069a5371)
4. Find the length of one arch of the
cycloid given by the parametric equations
![{\displaystyle x(t)=R{\bigl (}t-\sin(t){\bigr )}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8f5e88f179336b5040bcc1bcb98004d4f008292a)
,
![{\displaystyle y(t)=R{\bigl (}1-\cos(t){\bigr )}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4ce527acae1186ebfa8ee6a45a66b60c64411dda)
, with
![{\displaystyle t}](https://wikimedia.org/api/rest_v1/media/math/render/svg/65658b7b223af9e1acc877d848888ecdb4466560)
running from
![{\displaystyle 0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2aae8864a3c1fec9585261791a809ddec1489950)
to
![{\displaystyle 2\pi }](https://wikimedia.org/api/rest_v1/media/math/render/svg/73efd1f6493490b058097060a572606d2c550a06)
.
:![{\displaystyle 8R}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c5939ef2b6c3579a6f1ff23aabc9eaeeeb723822)
:![{\displaystyle 8R}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c5939ef2b6c3579a6f1ff23aabc9eaeeeb723822)
Solutions