The curvature of a plane parametric curve x = f(t), y = g(t) is $ \kappa = \dfrac{|\dot{x} \ddot{y} - \dot{y} \ddot{x}|}{[\dot{x}^2 + \dot{y}^2]^{3/2}}$ where the dots indicate derivatives with respect to t. Use the above formula to find the curvature. x = 6et cos(t), y = 6et sin(t)

The curvature is modelled by .

Step-by-step explanation:

The equation of the curvature is:

The parametric componentes of the curve are:

and

The first and second derivative associated to each component are determined by differentiation rules:

First derivative

and

and

Second derivative

Now, each term is replaced in the the curvature equation:

And the resulting expression is simplified by algebraic and trigonometric means:

The curvature is modelled by .

given that triangle abc is transformed two times to get triangle

a"b"c"

first abc is transformed in to a'b'c' as follows:

the coordinates of a are (1,-1) and transformed into a'(-1,1)

similarly b (4,-2) became b'(-4,2) and

c(7,-2) became c'(-7,2)

i.e. (x,y) becomes (-x,-y)

this is nothing but reflection about a point here origin.

thus first transformation is reflection on the origin.

next is exactly shifting 3 units down

source:

answer: i'm not sure

step-by-step explanation: