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

1. ## [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,

2. 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 $(a^2(1- cos t)^2+ a^2 sin^2 t)^{3/2}$ $= (a^2- 2a^2 cos t+ a^2 cos^2 t+ a^2 sin^2 t)^{3/2}$ $= (2a^2- 2a^2 cos t)^{3/2}= 2^{3/2}a^3 (1- cos t)^{3/2}$.

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

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

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

3. $r = a - a cos t$ Differentiating we get: $r_1 = a sin t$ and again we get: $r_2 = a cos t$

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

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

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

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

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

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

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

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

So bringing them together:

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

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

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

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

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

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

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

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

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

Which is the result you wanted!
Voila!

4. Well done HallsofIvy you beat me to it!

5. 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 $(a^2(1- cos t)^2+ a^2 sin^2 t)^{3/2}$ $= (a^2- 2a^2 cos t+ a^2 cos^2 t+ a^2 sin^2 t)^{3/2}$ $= (2a^2- 2a^2 cos t)^{3/2}= 2^{3/2}a^3 (1- cos t)^{3/2}$.

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

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

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

6. 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 $(a^2(1- cos t)^2+ a^2 sin^2 t)^{3/2}$ $= (a^2- 2a^2 cos t+ a^2 cos^2 t+ a^2 sin^2 t)^{3/2}$ $= (2a^2- 2a^2 cos t)^{3/2}= 2^{3/2}a^3 (1- cos t)^{3/2}$.

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

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

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