a)
2 x dx + 8 y dy = 3x dy + 3y dx
dy(3x-8y)=dx(2x-3y)
b)
Where is the slope = 0?
where 2x = 3y
if x = 3
9 + 4 y^2 = 7 + 9 y
4 y^2 - 9 y + 2 = 0
(4y-1)(y-2) = 0
y = 1/4 or y = 2
check both of those to find which works
(3,2)
3*2 - 2*3 = 0
yes, (3,2) works
(3,1/4)
3/4 -6 = 0 no way so (3,2) is it
c)
dy/dx=(3y-2x)/(8y-3x)
d^2y/dx^2 = [(8y-3x)(-2)-(3y-2x)(-3)] / (8y-3x)^2
= [(16-9)(-2)]/(16-9)^2 = -2/7
that is negative, so a maximum of the function
Consider the curve given by x^2+4y^2=7+3xy
a) Show that dy/dx=(3y-2x)/(8y-3x)
b) Show that there is a point P with x-coordinate 3 at which the line tangent to the curve at P is horizontal. Find the y-coordinate of P.
c) Find the value of d^2y/dx^2 (second derivative) at the point P found in part b). Does the curve have a local maximum, a local minimum, or neither at the point P? Justify your answer..
The only part that I did was a). I don't really know how to find Point P for b) and c)
3 answers
Thank you so much! It makes perfect sense and when I calculated it, it matched up.
I'm confused as to how to do part C), I don't remember how to find the second derivative of an implicit differentiation problem!
5 answers