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Find the length of a curve

renyikouniao

Member
Jun 1, 2013
41
find the length of a curve given by:
f(x)=integral(upper bound: x lower bound: 0) (cos^2(x)+4cos(x)+1)^0.5 dx

Here's my solution:
I use the equation L=integral ( upper bound: a lower bound: b ) sqrt[1+(f'(x))^2]

f'(x)=(cos^2(x)+4cos(x)+1)^0.5

1+[f'(x)]^2=2+cos^(x)+4cos(x)=[cos(x)+2]^2-2

L=integral( upper bound: x lower bound: 0 ) [2+cos^2(x)+4cos(x)]^0.5

This is where I am stuck...Am I right so far?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
It is probably best to let the dummy variable of integration be something other than $x$...most times I see $t$ being used in such cases:

\(\displaystyle f(x)=\int_0^x \sqrt{\cos^2(t)+4\cos(t)+1}\,dt\)

However, I am unclear what the limits of integration are to be when computing the arc-length along this curve. If you are to use parameters, I suggest something other than $x$, such as $a$ and $b$, and then we could give the arc-length $s$ as:

\(\displaystyle s=\int_a^b\sqrt{1+\left(f'(x) \right)^2}\,dx\)

You have correctly applied the derivative form of the FTOC to obtain:

\(\displaystyle f'(x)=\sqrt{\cos^2(x)+4\cos(x)+1}\)

and thus:

\(\displaystyle 1+\left(f'(x) \right)^2=1+\cos^2(x)+4\cos(x)+1=\cos^2(x)+4\cos(x)+2\)

and so we have:

\(\displaystyle s=\int_a^b\sqrt{\cos^2(x)+4\cos(x)+2}\,dx\)

Completing the square as you did, gives:

\(\displaystyle s=\int_a^b\sqrt{(\cos(x)+2)^2-2}\,dx\)

At this point, I might consider the substitution:

\(\displaystyle \cos(x)+2=\sqrt{2}\tan(\theta)\)

but, this leads to an integrand that seem far more complicated.

Are you certain you have given all the relevant information?
 

renyikouniao

Member
Jun 1, 2013
41
It is probably best to let the dummy variable of integration be something other than $x$...most times I see $t$ being used in such cases:

\(\displaystyle f(x)=\int_0^x \sqrt{\cos^2(t)+4\cos(t)+1}\,dt\)

However, I am unclear what the limits of integration are to be when computing the arc-length along this curve. If you are to use parameters, I suggest something other than $x$, such as $a$ and $b$, and then we could give the arc-length $s$ as:

\(\displaystyle s=\int_a^b\sqrt{1+\left(f'(x) \right)^2}\,dx\)

You have correctly applied the derivative form of the FTOC to obtain:

\(\displaystyle f'(x)=\sqrt{\cos^2(x)+4\cos(x)+1}\)

and thus:

\(\displaystyle 1+\left(f'(x) \right)^2=1+\cos^2(x)+4\cos(x)+1=\cos^2(x)+4\cos(x)+2\)

and so we have:

\(\displaystyle s=\int_a^b\sqrt{\cos^2(x)+4\cos(x)+2}\,dx\)

Completing the square as you did, gives:

\(\displaystyle s=\int_a^b\sqrt{(\cos(x)+2)^2-2}\,dx\)

At this point, I might consider the substitution:

\(\displaystyle \cos(x)+2=\sqrt{2}\tan(\theta)\)

but, this leads to an integrand that seem far more complicated.

Are you certain you have given all the relevant information?
Yes,and this is where I am stuck..