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perimeter
- Thread starter Raju
- Start date
D
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Raju said:Please help in solving
Find the perimeter of the cord r = a (1+cos ?)
Note:
? means "Theeta".
First define the problem fully!
Is it a circle you are talking about?
What is 'r' and 'a' and 'theta'?
Where are these located (is there a picture)?
Hello, Raju!
\(\displaystyle \text{Arc length for polar functions: }\;L \;=\;\int^{\beta}_{\alpha} \sqrt{r^2 + \left(\tfrac{dr}{d\theta}\right)^2}\,d\theta\)
\(\displaystyle \text{We have: }\;r \:=\:a(1+\cos\theta) \qquad \tfrac{dr}{d\theta} \:=\:-a\sin\theta\)
\(\displaystyle \text{Then: }\;r^2 + \left(\tfrac{dr}{d\theta}\right)^2 \;=\;\bigg[a(1+\cos\theta)\bigg]^2 + \bigg[-a\sin\theta\bigg]^2 \;=\; a^2(1 + 2\cos\theta + \cos^2\!\theta) + a^2\sin^2\theta\)
. . \(\displaystyle = \;a^2\bigg[1 + 2\cos\theta + \underbrace{\cos^2\!\theta + \sin^2\!\theta} \bigg] \;=\;a^2[2 + 2\cos\theta] \;=\;2a^2(1 + \cos\theta)\)
. . . . . . . . . . . . . . . . . \(\displaystyle ^{\text{This is 1}}\)
. . \(\displaystyle = \;4a^2\cdot\frac{1+\cos\theta}{2} \;=\; 4a^2\cos^2\!\tfrac{\theta}{2}\)
\(\displaystyle \text{Hence: }\;\sqrt{r^2 + \left(\tfrac{dr}{d\theta}\right)^2} \;=\;\sqrt{4a^2\cos^2\!\tfrac{\theta}{2}} \;=\; 2a\cos\!\tfrac{\theta}{2}\)
\(\displaystyle \text{Due to the symmetry, we can find the length from 0 to }\pi\text{ and multiply by 2.}\)
. . \(\displaystyle \text{Therefore: }\;L \;=\;2 \times 2a\int^{\pi}_0 \cos\!\tfrac{\theta}{2}\,d\theta\)
Go for it!
\(\displaystyle \text{Find the perimeter of the cardioid: }\;r \:=\:a(1+\cos\theta)\)
\(\displaystyle \text{Arc length for polar functions: }\;L \;=\;\int^{\beta}_{\alpha} \sqrt{r^2 + \left(\tfrac{dr}{d\theta}\right)^2}\,d\theta\)
\(\displaystyle \text{We have: }\;r \:=\:a(1+\cos\theta) \qquad \tfrac{dr}{d\theta} \:=\:-a\sin\theta\)
\(\displaystyle \text{Then: }\;r^2 + \left(\tfrac{dr}{d\theta}\right)^2 \;=\;\bigg[a(1+\cos\theta)\bigg]^2 + \bigg[-a\sin\theta\bigg]^2 \;=\; a^2(1 + 2\cos\theta + \cos^2\!\theta) + a^2\sin^2\theta\)
. . \(\displaystyle = \;a^2\bigg[1 + 2\cos\theta + \underbrace{\cos^2\!\theta + \sin^2\!\theta} \bigg] \;=\;a^2[2 + 2\cos\theta] \;=\;2a^2(1 + \cos\theta)\)
. . . . . . . . . . . . . . . . . \(\displaystyle ^{\text{This is 1}}\)
. . \(\displaystyle = \;4a^2\cdot\frac{1+\cos\theta}{2} \;=\; 4a^2\cos^2\!\tfrac{\theta}{2}\)
\(\displaystyle \text{Hence: }\;\sqrt{r^2 + \left(\tfrac{dr}{d\theta}\right)^2} \;=\;\sqrt{4a^2\cos^2\!\tfrac{\theta}{2}} \;=\; 2a\cos\!\tfrac{\theta}{2}\)
\(\displaystyle \text{Due to the symmetry, we can find the length from 0 to }\pi\text{ and multiply by 2.}\)
. . \(\displaystyle \text{Therefore: }\;L \;=\;2 \times 2a\int^{\pi}_0 \cos\!\tfrac{\theta}{2}\,d\theta\)
Go for it!