Question

Consider the curve given by the equation (2y+1)^3 − 24x = −3.

Answer 1

Answer:

(a) dy/dx = 4/(2y+1)^2.

(b) y = 4/9 x - 14/9

(c) d2y/dx2 = -64/243

Step-by-step explanation:

You have the following equation

$(2y+1)^3-24x=-3$   (1)

(a) You first derivative implicitly the equation (1) respect to x:

$\frac{d}{dx}[(2y+1)^3-24x]=\frac{d}{dx}[-3]\\\\3(2y+1)^2(2\frac{dy}{dx})-24=0$

next, you solve the last result for dy/dx:

$6(2y+1)^2\frac{dy}{dx}=24\\\\\frac{dy}{dx}=\frac{4}{(2y+1)^2}$(2)

(b) The equation for the tangent line is given by:

$y-y_o=m(x-x_o)$    (3)

with yo = -2 and xo = -1

To find the slope m you use the result of the equation (2), because dy/dx evaluated in (-1,-2) is the slope at such point:

m = $\frac{dy}{dx}=\frac{4}{(2(-2)+1)^2}=\frac{4}{9}$

Hence, by replacing in the equation (3) you obtain:

$y-(-2)=\frac{4}{9}(x-(-1))\\\\y+2=\frac{4}{9}x+\frac{4}{9}\\\\y=\frac{4}{9}x-\frac{14}{9}$

hence, the equation for the tangent line is y = 4/9 x - 14/9

(c) To find d2y/dx2 you derivative the result obtain in the equation (2):

$\frac{d^2y}{dx^2}=\frac{d}{dx}[4(2y+1)^{-2}]\\\\\frac{d^2y}{dx^2}=-8(2y+1)^{-3}(2\frac{dy}{dx})\\\\\frac{d^2y}{dx^2}=-16(2y+1)^{-3}\frac{dy}{dx}$     (4)

the second derivative for the point (-1,-2) is obtained by replacing y=-2 and dy/dx=m=4/9 in the equation (4):

$\frac{d^2y}{dx^2}=-16(2(-2)+1)^{-3}(\frac{4}{9})=-\frac{64}{243}$

hence, d2y/dx2 evaluated in (-1,-2) is -64/243