>>7618646This a terribly written solution (how do you solve a math problem without words ?), who gave you this ?
There's a whole bunch of tangent vectors at each point of the circle (a whle line, actually) so the questions that ask you to "find the tangent vector" aren't very well posed..
Now, how do you find the tangent line at a point of a curve, given a parameterization ? Well, having a parameterization is like being given the position e(t) of a particle traveling on the plane. The direction of the tangent line to its path at any point is given by its speed e'(t) (if it's nonzero).
That's how you find the answer to question c): (1,0) is e(0), therefore a tangent vector is e'(0) = (0,1) (but you could have given any other vector collinear to it)
To solve question d), note that if (x,y) is on T(Gamma), there is a (u,v) on Gamma such that (x,y)= T(u,v). Now solve for u,v in terms of x and y and write the equation and replace u and v by the expressions in terms of x and y. That should tell you that x and y are on a certain circle. Conversely, you should check that if x and y satisfy that equation, then u and v are on the circle centered at O of radius 1 (to make sure that you have an equivalence).
For question e), note that given a curve C and an invertible linear transformation T, the tangent line to T(C) at a point T(p) is the image by T of the tangent line at C at point p.
Here, a tangent vector to Gamma at (0,1) = e(pi/2) is e'(pi/2) = (-1,0), therefore, a tangent vector to T(Gamma) at T(0,1) is T(-1,0) = (-2,3)
