Wednesday

April 16, 2014

April 16, 2014

Posted by **Leanna** on Wednesday, January 26, 2011 at 3:10pm.

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)

- AP Calculus -
**Damon**, Wednesday, January 26, 2011 at 3:37pma)

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

- AP Calculus -
**Leanna**, Wednesday, January 26, 2011 at 3:47pmThank you so much! It makes perfect sense and when I calculated it, it matched up.

- AP Calculus -
**Brian**, Thursday, January 12, 2012 at 10:24pmI'm confused as to how to do part C), I don't remember how to find the second derivative of an implicit differentiation problem!

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