Calculus Tangent line problems?

Implicit differentiation w.r.t/ x gives cos (x - y)(1 - y') = y + xy'

At (0, pi) cos (0 - pi)(1 - y') = pi + 0 i.e. -1 + y' = pi

Tangent line y - pi = (pi + 1)x

D: x^2 y' / [2(y-2)^0.5 ] + 2x(y-2)^0.5 = 2y y' - 3

At (1, 2) 1/(y-2)^0.5 will be undefined so multiply it out first:

x^2 y' / [2 ] + 2x(y-2) = 2y(y-2)^0.5 y' - 3(y-2)^0.5

At (1, 2) y' /2 + 0 = 0 - 0 so y' = 0 Normal is vertical,

equation is x = 1

The third curve does not pass through (1, -9)

The deriv is 3x^2 - 9 which is = -6 at (1, -8)

equation is y + 8 = -6(x -1)

y = 4x + 3 is tangent to the curve y = x^2 + c. to establish that this to be authentic, the line could go the curve at precisely one element. this implies we could consistently get carry of purely one answer, could we equate those purposes to a minimum of one yet another. enable's do this. 4x + 3 = x^2 + c circulate each thing to the perfect hand facet. 0 = x^2 - 4x + c - 3 bear in mind that we might understand the kind of ideas via fixing a quadratic for its discriminant, "b^2 - 4ac". If this value is 0, then we've one answer. As a reminder: For a quadratic equation ax^2 + bx + c = 0 we can determine the character of roots via fixing b^2 - 4ac. a million) If b^2 - 4ac > 0, then we've 2 (unique) genuine ideas. 2) If b^2 - 4ac = 0, then we've one genuine answer (or 2 ideas repeated two times). 3) If b^2 - 4ac < 0, then we've 2 complicated(imaginary) ideas. Our purpose is to make it such that #2 is authentic, simply by fact a line tangent to a curve crosses the curve precisely as quickly as. fixing for "b^2 - 4ac" provides us (-4)^2 - 4(a million)(c - 3) And we might desire to equate this to 0. (-4)^2 - 4(a million)(c - 3) = 0 sixteen - 4(c - 3) = 0 All we could do now could be resolve for c. -4(c - 3) = -sixteen Divide the two aspects via (-4), c - 3 = 4 upload 3 to the two aspects, c = 7.