Question

point on the curve 2y^3 + x^2 = 12y at which the tangent to the curve is vertical is

Answer 1

vertical lines have slope =.... ?

Answer 2

you get slope of tangent if you find dy/dx
so first find it using implicit differentiation and equate it to slope of vertical line.

Answer 3

can u plse solve it, i don;t know

Answer 4

you don't know whats slope of vertical line ?? :O

Answer 5

you know how to differentiate implicitly ?

Answer 6

i don'tknow

Answer 7

i know how to find dy/dx

Answer 8

Slope of vertical line is either positive or negative infinity if I recall.

Answer 9

\[\frac{dy}{dx}=\frac{x}{6-3y^2}\]We seek the derivative to be infinity, so we have\[6=3y^2\]\[y=\pm\sqrt{2}\]Plugging the values of y into the original equation, we have\[2(\sqrt{2})^3+x^2=12\sqrt{2}\]and\[2(-\sqrt{2})^3+x^2=-12\sqrt{2}\]Solve for x for each corresponding value of y.

Answer 10

Further calculation shows that \[12\sqrt{2}>2\sqrt{8}\] the first equation yield imaginary solution, which we don't really want. This implies only the second equation holds for real solution of x.

Answer 11

Hope this helps