# [SOLVED] FINDING RADIUS OF CURVATURE OF CARDIODE r=a(1-cost)

• Jan 25th 2010, 03:12 AM
avengerevenge
[SOLVED] FINDING RADIUS OF CURVATURE OF CARDIODE r=a(1-cost)
question:- r=a(1-cost)
answer :- 4/3a(sin(t/2)) (which i have to prove)

Please help me to find the radius of curvature, i know the formula everything but i have to prove the answer as :- 4/3a(sin(t/2)), which i am unable to prove it,
i cannot prove the answer as above, pls help

here is the formula for finding the radius of curvature :-

radius of curvature =((r^2+(r1)^2 )^3/2)/r^2+2(r1)^2-r(r2) were
r1 is first derivative of r
and

r2 is second derivative of r

Please any body help me this problem is driving me crazy,

• Jan 25th 2010, 04:42 AM
HallsofIvy
Quote:

Originally Posted by avengerevenge
question:- r=a(1-cost)
answer :- 4/3a(sin(t/2)) (which i have to prove)

Please help me to find the radius of curvature, i know the formula everything but i have to prove the answer as :- 4/3a(sin(t/2)), which i am unable to prove it,
i cannot prove the answer as above, pls help

here is the formula for finding the radius of curvature :-

radius of curvature =((r^2+(r1)^2 )^3/2)/r^2+2(r1)^2-r(r2) were
r1 is first derivative of r
and

r2 is second derivative of r

Please any body help me this problem is driving me crazy,

What, exactly, is your difficulty? The derivatives should be very easy. From r= a(1- cos t), r'= a sin t and r"= a cos t.

The numerator is $\displaystyle (a^2(1- cos t)^2+ a^2 sin^2 t)^{3/2}$$\displaystyle = (a^2- 2a^2 cos t+ a^2 cos^2 t+ a^2 sin^2 t)^{3/2}$$\displaystyle = (2a^2- 2a^2 cos t)^{3/2}= 2^{3/2}a^3 (1- cos t)^{3/2}$.

The denominator is $\displaystyle a^2(1- cos t)^2+ 2a^2 sin^2 t- a^2(1- cos t)(cos t)$$\displaystyle = a^2 - 2a^2 cos t+ a^2 cos^2 t+ 2a^2 sin^2 t- a^2 cos t+ a^2 cos^2 t$$\displaystyle = 3a^2- 3a^2 cos t= 3a^2(1- cos t)$

So the fraction is $\displaystyle \frac{2^{3/2}a^3 (1- cos t)^{3/2}}{3 a^2 (1- cos 6)}= \frac{2^{3/2}}{3}a (1- cos t)^{1/2}= \frac{2^{3/2}}{3}a \sqrt{1- cos t}$

Now use a "half-angle" formula: $\displaystyle sin(t/2)= \sqrt{\frac{1}{2}(1- cos t)}$
• Jan 25th 2010, 05:17 AM
$\displaystyle r = a - a cos t$ Differentiating we get: $\displaystyle r_1 = a sin t$ and again we get: $\displaystyle r_2 = a cos t$

$\displaystyle Curvature = \frac{(r^2+r_1^2)^{\frac{3}{2}}}{r^2+2r_1^2-rr_2}$

Let's look at top and bottom of fraction seperately:
$\displaystyle Top = (r^2+r_1^2)^{\frac{3}{2}}=(a^2(1-cost)^2+a^2sin^2t)^{\frac{3}{2}}=(a^2)^\frac{3}{2} (1-2cost+cos^2t+sin^2t)^\frac{3}{2}$

Using: $\displaystyle cos^2t+sin^2t=1$

$\displaystyle Top=a^3(2-2cost)^\frac{3}{2}=a^3.2\sqrt2(1-cost)^\frac{3}{2}$

$\displaystyle Bottom = r^2+2r_1^2-rr_2$

$\displaystyle = a^2(1-cost)^2+2a^2sin^2t-a^2cost(1-cost)= a^2(1-2cost+cos^2t+2sin^2t-cost+cos^2t$

Again using: $\displaystyle cos^2t+sin^2t=1$

$\displaystyle Bottom= a^2(3-3cost)=3a^2(1-cost)$

So bringing them together:

$\displaystyle Curvature=\frac{Top}{Bottom}=\frac{a^3.2\sqrt2(1-cost)^\frac{3}{2}}{3a^2(1-cost)}=\frac{2\sqrt2}{3}a\sqrt{1-cost}$

Now we need to use formula:
$\displaystyle cos(A+B)=cosAcosB-sinAsinB$

$\displaystyle cos(2A)=cos^2A-sin^2A=1-2sin^2A$

$\displaystyle sin^2A= \frac{1}{2}(1-cos2A)$

$\displaystyle sinA= \frac{1}{\sqrt2}\sqrt{1-cos2A}$

$\displaystyle Let \ A=\frac{t}{2}$

$\displaystyle sin(\frac{t}{2})=\frac{1}{\sqrt2}\sqrt{1-cost}$

$\displaystyle Or: \sqrt{1-cost}=\sqrt2sin(\frac{t}{2})$

Substituring back in our formula for curvature:
$\displaystyle Curvature=\frac{2\sqrt2}{3}a\sqrt{1-cost}=\frac{4}{3}asin(\frac{t}{2})$

Which is the result you wanted!
Voila!
• Jan 25th 2010, 05:19 AM
Well done HallsofIvy you beat me to it!
• Jan 25th 2010, 07:42 AM
avengerevenge
Quote:

Originally Posted by HallsofIvy
What, exactly, is your difficulty? The derivatives should be very easy. From r= a(1- cos t), r'= a sin t and r"= a cos t.

The numerator is $\displaystyle (a^2(1- cos t)^2+ a^2 sin^2 t)^{3/2}$$\displaystyle = (a^2- 2a^2 cos t+ a^2 cos^2 t+ a^2 sin^2 t)^{3/2}$$\displaystyle = (2a^2- 2a^2 cos t)^{3/2}= 2^{3/2}a^3 (1- cos t)^{3/2}$.

The denominator is $\displaystyle a^2(1- cos t)^2+ 2a^2 sin^2 t- a^2(1- cos t)(cos t)$$\displaystyle = a^2 - 2a^2 cos t+ a^2 cos^2 t+ 2a^2 sin^2 t- a^2 cos t+ a^2 cos^2 t$$\displaystyle = 3a^2- 3a^2 cos t= 3a^2(1- cos t)$

So the fraction is $\displaystyle \frac{2^{3/2}a^3 (1- cos t)^{3/2}}{3 a^2 (1- cos 6)}= \frac{2^{3/2}}{3}a (1- cos t)^{1/2}= \frac{2^{3/2}}{3}a \sqrt{1- cos t}$

Now use a "half-angle" formula: $\displaystyle sin(t/2)= \sqrt{\frac{1}{2}(1- cos t)}$

First of all thank you very much for your post
actually i know how to find derivatives, but i didnt get the idea to use half angle formula, i was stuck in that step,
but be more clear, like you posted this

Now use a "half-angle" formula: sin(t/2)= \sqrt{\frac{1}{2}(1- cos t)}

actually it made no sense like how sin(t/2)= \sqrt{\frac{1}{2}(1- cos t)}.

thanks to suhada for more clear explanation.
• Jan 25th 2010, 07:43 AM
avengerevenge
Quote:

Originally Posted by HallsofIvy
What, exactly, is your difficulty? The derivatives should be very easy. From r= a(1- cos t), r'= a sin t and r"= a cos t.

The numerator is $\displaystyle (a^2(1- cos t)^2+ a^2 sin^2 t)^{3/2}$$\displaystyle = (a^2- 2a^2 cos t+ a^2 cos^2 t+ a^2 sin^2 t)^{3/2}$$\displaystyle = (2a^2- 2a^2 cos t)^{3/2}= 2^{3/2}a^3 (1- cos t)^{3/2}$.

The denominator is $\displaystyle a^2(1- cos t)^2+ 2a^2 sin^2 t- a^2(1- cos t)(cos t)$$\displaystyle = a^2 - 2a^2 cos t+ a^2 cos^2 t+ 2a^2 sin^2 t- a^2 cos t+ a^2 cos^2 t$$\displaystyle = 3a^2- 3a^2 cos t= 3a^2(1- cos t)$

So the fraction is $\displaystyle \frac{2^{3/2}a^3 (1- cos t)^{3/2}}{3 a^2 (1- cos 6)}= \frac{2^{3/2}}{3}a (1- cos t)^{1/2}= \frac{2^{3/2}}{3}a \sqrt{1- cos t}$

Now use a "half-angle" formula: $\displaystyle sin(t/2)= \sqrt{\frac{1}{2}(1- cos t)}$

First of all thank you very much for your post
actually i know how to find derivatives, but i didnt get the idea to use half angle formula, i was stuck in that step,
but be more clear, like you posted this

Now use a "half-angle" formula: sin(t/2)= \sqrt{\frac{1}{2}(1- cos t)}

actually it made no sense like how sin(t/2)= \sqrt{\frac{1}{2}(1- cos t)}.

thanks to suhada for more clear explanation.