Relating circumference and area (video) | Khan Academy
=A(r)=4(%CF%80)r^{2}&line2=\mathrm{Solution:\:}\frac{\sqrt{%CF%80}\sqrt{r}}{2%CF%80},-\frac{\sqrt{%CF%80}\sqrt{r}}{2%CF%80} "inverse of f(x)=A(r)=4(pi)r^2")
inverse of f(x)=A(r)=4(pi)r^2
a) Plot the point with polar coordinates (4, 2 pi / 3). Then find its Cartesian coordinates. (b) The Cartesian coordinates of a point are (-3, 3). Find two sets of polar
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![Let f1: RrarrR ,\ \ f2:(-pi/2,pi/2)->R\ \ f3:(-1,\ e^(pi/2)-2)rarrR and f4: RrarrR be functions defined by (i)f1(x)=sin(sqrt(1-e^(-x)^2)), (ii) f2(x)={(|sinx|)/(tan^(-1)x)\ \ \ \ \ if\ x!=0 1\ \ \ \ \ \ \ \ \ \ \ \](https://d10lpgp6xz60nq.cloudfront.net/physics_images/LN_SELF_PHY_XI_ENG_JEE_C01_K1115.PNG "Let f1: RrarrR ,\ \ f2:(-pi/2,pi/2)->R\ \ f3:(-1,\ e^(pi/2)-2)rarrR and f4: RrarrR be functions defined by (i)f1(x)=sin(sqrt(1-e^(-x)^2)), (ii) f2(x)={(|sinx|)/(tan^(-1)x)\ \ \ \ \ if\ x!=0 1\ \ \ \ \ \ \ \ \ \ \ \")
Let f1: RrarrR ,\ \ f2:(-pi/2,pi/2)->R\ \ f3:(-1,\ e^(pi/2)-2)rarrR and f4: RrarrR be functions defined by (i)f1(x)=sin(sqrt(1-e^(-x)^2)), (ii) f2(x)={(|sinx|)/(tan^(-1)x)\ \ \ \ \ if\ x!=0 1\ \ \ \ \ \ \ \ \ \ \ \
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