Quick Homework Help

# im stumped on this one:The slope of the curve y^3 - xy^2 = 4 at the point where y = 2 is what??? ⚑ Flag

by eunju001 at November 24, 2009

Well, if we differentiate the slope implicitly:3(y^2)y' - [x*(2yy') + y^2] = 0-->       y'(3y^2-2xy) + y^2 = 0-->      y' = y^2/(3y^2 - 2xy)and if y = 2, first of all, going from the original equation we find x:2^3 - x*2^2 = 4          8 - 4x = 4               4x = 4                 x = 1So we plug x = 1, y = 2 into our slope equation:y' = 2^2/(3*2^2 -2*1*2)    =4/(12-4)    =4/8    =1/2Yay!  We found our slope

Alison037 November 25, 2009

2^3-X x 2^2= 42 x 2 x 2-X x 2 x2= 4 8-X x 4 = 48-7=1 x 4= 4 so x = 7CHECK:2^3-7 x 2^28-7 x 4=?1x 4 = 4

Amber_Malaine November 25, 2009

* first, use implicit derivative to slove the problem. 3y^2-y^2-2xyy' =0 -2xyy' = -2y^2 -2xy(y')/-2xy = -2y^2/-2xy y' =y/x when y=2, solve for x by plotting 2 in the equation given (2)^3-x(2)^2 =4 8-4x=4 x= -1 at the point (-1,2) plot those points in y' = -2

The_Engineer November 26, 2009

@ The_Engineer, you forgot to cancel the negatives when solving for x on this step:8 - 4x = 4If you subtract 8 across you get -4 on the right side, so you have:-4x = -4    or   x = 1

Alison037 November 26, 2009

@ Alison036.....Thanks for the correction, I even made a mistake in the derivative...I will solve the problem again for you. Given f(x) = y^3-xy^2 = 4: at the point y = 2 3y^2 y'-y^2-2xyy' = 0  3y^2 y' - 2xyy' = y^2   ...adding (y^2) to both sides y' (3y^2 - 2xy) / (3y^2 - 2xy) = y^2 / (3y^2-2xy) y' = y^2 / y (3y - 2x) y' = y / (3y-2x)  ...they gave you y=2, plot it in here! y' = 2 / (6-2x)  This is not enough to answer the question. Plot in y=2 into the equation  y^3-xy^2  = 4 (2)^3-x(2)^2 = 4 8 - 4x = 4 -4x = -4 x = 1 going back to y' = 2 / (6 - 2x) y' = 2/4 y' = 1/2

The_Engineer November 26, 2009