Question

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


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

(b) Write an equation for the line tangent to the curve at the point (−1,−2).

(c) Evaluate d2y/dx2 at the point (−1,−2).

(d) The point (16,0) is on the curve. Find the value of (y−1)′(0).

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

Answer 2

Answer:

(A) The value of  $dy/dx=\frac{4}{(2y+1)^2}$.

(B) The equation of the tangent is : $y=(4/9)x-(14/9)$

(C) The value of  $\frac{d^2y}{dx^2}=-64/243$

(D) The point (16,0) is not on curve so it can not be determined by the given equation.

Step-by-step explanation:

Given information:

The equation $(2y+1)^3-24x=-3$

(A) For the first derivative of the given equation:

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

$3(2y+1)^2(dy/dx)-24=0$

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

Hence , from the above equation it is shown that the value of

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

(B) The equation of the tangent to the curve is given by:

$y-y_o=m(x-x_0)$

On putting the given values in the above equation

We get:

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

$m=4/9$

Hence, the equation of the tangent can be written as :

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

So, the equation of the tangent is :

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

(C) Now ,

To find $d^2y/dx^2$ for the equation

We have to find double derivative of the equation;

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

On putting the values from the given information in the above equation;

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

$\frac{d^2y}{dx^2}=-64/243$

(D) For the equation $(2y+1)^3-24x=-3$

First check for the given points (16,0) if it satisfies the given equation or not.

Now on checking for the same the point is not satisfying the given equation hence, we can not find the value of $(y-1)'(0)$.

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