Using parametric form of circle x = r.cos(t) : y = r.sin(t)
Using curvature in parametric form k=\(\left |\frac{y”x’-y’x”} {((x’)^2+(y’)^2)^{\frac{3}{2}}}\right |\) we have
k=\(\left |\frac{(-rsin(t))(-rsin(t))-(-rcos(t))(rcos(t))} {((-rsin(t))^2+(rcos(t))^2)^{\frac{3}{2}}}\right |\)
k=\(\left |\frac{r^2(sin^2(t)+cos^2(t))} {r^3(sin^2(t)+cos^2(t))^{\frac{3}{2}}}\right |\)
k=\(\left |\frac{1}{r}\right |\)

The curve itself is involute of evolute of curve with a known starting point. The or name for involute is evolvent.

w.k.t \(\frac{ds}{dt} = \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2}?(1)\)
\(\frac{dx}{dt} = ae^t (cos t + sin t), \frac{dy}{dt} = ae^t (-sin? t + cos?t)\)
substituting in (1) we get
\(\frac{ds}{dt} = ae^t \sqrt{(cos t + sin t)^2+(-sin t+cos?t)^2}\)
\(\frac{ds}{dt} = ae^t \sqrt{1+2 sin t cos + 1 – 2 sin? t cos?t} = \sqrt{2} ae^t?(cos^{2} t + sin{2} t = 1).\)

Though function f(x) is discontinuous at x = 1 we can still find f(1) as function is continuous and differentiable in interval. [0, 1).

\(=lt_{x\rightarrow 0}(\frac{x^n}{x^n})\times\frac{(a_n+a_{n-1}\frac{1}{x}+…+a_1\frac{1}{x^{n-1}}+a_0\frac{1}{x^{n}})}{(b_n+b_{n-1}\frac{1}{x}+…+b_1\frac{1}{x^{n-1}}+b_0\frac{1}{x^{n}})}\)
\(=\frac{a_n}{b_n}\)

Taking Logarithm on both side of polar curve
we get log?r = log?a + 2 log?sec?(\(\frac{?}{2}\))
differentiating w.r.t ? we get
\(\frac{1}{r} \frac{dr}{d?} = \frac{2 sec?(\frac{?}{2}).tan?(\frac{?}{2}) }{2 sec?(\frac{?}{2})} = tan?(\frac{?}{2})\)
\(cot?? = cot?(\frac{?}{2} – \frac{?}{2}) \rightarrow ? = \frac{?}{2} -\frac{?}{2}\)
w.k.t length of perpendicular is given by p=r sin ?
thus substituting ? value we get \(p = r \,sin(\frac{?}{2} – \frac{?}{2}) = r \,cos(\frac{?}{2})\)
at \( ?=\frac{?}{4}, p = r \,cos \frac{?}{4} = \frac{r}{\sqrt{2}}?.(1)\)
but \( r=a \,sec^2 (\frac{?}{2}) \,at\, ? = \frac{?}{4}, r=a \,sec^2 (\frac{?}{4}) = 4a?(2)\)
from (1) & (2) \(p=\frac{4a}{\sqrt{2}} = 2a\sqrt{2}\).

Given f(x) = ¹⁄_x
Let, x – 1 = h
Hence, x = 1 + h
Hence, f(x) = f(1 + h) = f(1) + ^h⁄_1! f’ (1) + ^h2⁄_2! f^” (1) +^h3⁄_3! f^”’ (1)+…
Now, f(1) = 1, f'(1) = -1, f”(1) = 2 ,f”'(1) = -6,…?.
Hence, f(1 + h) = 1 – h + h² – h³+….
hence, 1 – (x-1) + (x-1)² – (x-1)³ +….

\(\lim_{x\rightarrow 1}\frac{f(x)}{g(x)}=\lim_{x\rightarrow 1}\frac{x^2-3x+2}{x^3-x^2+x-1}\)=(1-3+2)/(1-1+1-1)=0/0
Hence this gives indeterminent forms hence by L’Hospital Rule
\(\lim_{x\rightarrow 1}\frac{f(x)}{g(x)}=\lim_{x\rightarrow 1}\frac{f'(x)}{g'(x)}=\lim_{x\rightarrow 1}\frac{2x-3}{3x^2-2x+1}=\frac{2-3}{3-2+1}=-\frac{1}{2}\)
Hence
\(\lim_{x\rightarrow 1}\frac{f(x)}{g(x)}=-\frac{1}{2}\)

There are 2 types of Iterative methods, (i) Interpolation methods (or Bracketing methods) and (ii) Extrapolation methods (or Open-end methods).

Given, f(x)=Cos(xSin(t))=real part of (e^ixSin(t))
=real part of(e^xCos(t) e^ixSin(t))
=real part of(e^{x[Cos(t)+iSin(t)]})
=Real part of \(\sum_{n=0}^? \frac{x^n [Cos(nt)+iSin(nt)]}{n!}\)
=\(\sum_{n=0}^? \frac{x^n [Cos(nt)]}{n!}\)