How do I determine the equation of the tangents in the points in this exercise?

Answer

This is a so-called parameter function in two dimensions where x and y are both functions of t. For each value of t you get a point in the (x,y) plane. If t changes, that point will ‘walk around’ in the (x,y) plane. Example: if x = cos

But first, your chart doesn’t show what’s happening near (0,0) because the scale you’re using is too big for that. If you limit the graph to near (0,0) to a smaller area you can see what is really happening (see accompanying figure).

So there are two t-values ​​for which y = 0, namely the values ​​for which y

Eg with t = -1 : by filling in x and y see that it is about the point (0, 0)

with t = 1 : that is about the point (2,0)

How do you find the tangent line in such a point?

Well, for that you need the derivative of the followed curve, and because that curve lies in the (x,y)-plane that is a derivative dy/dx

However… you cannot derive y from x because y is a function of t, not of x (at least not directly). But the chain rule does this:

dy/dx = ( dy/dt) . (dt/dx) = (dy/dt) / ( dx/dt) = y’_t /x’_t = (in this case) = -2t / 5t⁴

For example, for t = 1 that becomes -2/5

the tangent at the point reached by t = 1 is thus: y – 0 = -2/5 ( x – 2)

Note that dy/dt can be calculated as y’_t /x’_t but….

the second derivative d²y/dx² is NOT y”_t /x”_t but yes (y’_t /x’_t ) ‘/x’_t