Name: Class: Date: ----- ID: A

Implicit Differentiation

Multiple ChoiceIdentify the choice that best completes the statement or answers the question.

1. The slope of the line tangent to the curve y2 + (xy + 1) 3 = 0 at (2, -1) is

a.32

b.3

c.4

o3432

d.

e.

dy r.---; d2 Y2. If- = '" 1-y- then - =

dx 'dx2

a. -2yb. -y

-yC.

~1_y2

d. y1

e. 2

~ 2 dy.:>. If y = xy + x + 1 , then when x = -1, dx is

1a. -

2

b. 12

c. -1d. -2e. nonexistent

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l'1X-~ ~~f~~?L --( 1~

--------- ----- -------------------

2008 AB 6 Form BConsider the closed curve in the xy-plane given by

X2 + 2x + y4 + 4y = 5.

(a) Show that dy = -(x+l)dx 2(y3+1)'

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

(c) Find the coordinates of the two points on the curve where the line tangent to the curve isvertical.

(d) Is it possible for this curve to have a horizontal tangent at points where it intersects the x-axis?

Explain your reasoning.

2005 AB 5 Form BConsider the curve given by y2 = 2 + xy.

(a) Show that dy = -y-.dx 2y-x

(b) Find all points (x,y) on the curve where the line tangent to the curve has slope~.2(c) Show that there are no points (x, y) on the curve where the line tangent to the curve is

horizontal.(d) Let x and y be functions of time t that are related by the equation y2 = 2 + xy. At time t = 5,

the value of y is 3 and dy = 6. Find the value of dx at time t = 5.dt dt

- ------ --- -------2001AB6

The function/is differentiable for all real numbers. The point (3,±) is on the graph of y = I(x) , and

the slope at each point (x, y) on the graph is given by dy = y2(6 - 2x).dx

d2

( 1)(a) Find:' and evaluate it at the point 3, '4 .

(b) Find y = I(x) by solving the differential equation : = y2(6_ 2x) with the initial condition

1(3) =1..4

Ap® CALCULUS AB2008 SCORING GUIDELINES (Form B)

Question 6

Consider the closed curve in the xy-plane given byx2 + 2x + y4 + 4y = 5.

d -(x+l)y - )(a) Show that dx - 2(i + 1

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

(c) Find the coordinates of the two points on the curve where the line tangent to the curve is vertical.(d) Is it possible for this curve to have a horizontal tangent at points where it intersects the x-axis?

Explain your reasoning.

(a) 2x + 2+ 4i : + 4: = 0

(4i+4): =-2x-2

dy -2(x + 1) -(x + 1)dx = 4(i +1) = 2(y3+1)

(b) dyl = -(-2+1) =!dx (-2,1) 2(1+ 1) 4

Tangent line: y = 1+ ±(x + 2)

(c) Vertical tangent lines occur at points on the curve wherei + 1 = 0 (or y = -1 ) and x *- -1.

On the curve, y = -1 implies that x2 + 2x + I - 4 = 5,so x = -4 or x = 2.

Vertical tangent lines occur at the points (-4, -1) and(2, -1).

(d) Horizontal tangents occur at points on the curve wherex = -1 and y *- -1.

The curve crosses the x-axis where y = o.(_1)2 + 2 (-1) + 04 + 4 . 0 *- 5

No, the curve cannot have a horizontal tangent where itcrosses the x-axis.

2 : { 1 : implicit differentiation1 : verification

{I: slope

2: 1: tangent line equation

{

I: y =-13: I: substitutes y = -1 into the

equation of the curve1 : answer

{I: works with x = -lor y = 0

2'. I: answer with reason

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Ap® CALCULUS AB2005 SCORING GUIDELINES (Form B)

Question 5

Consider the curve given by i = 2 + ~.

(a) Show that dy = _y_.dx 2y - x

(b) Find all points (x, y) on the curve where the line tangent to the curve has slope ~.

(c) Show that there are no points (x, y) on the curve where the line tangent to the curve is horizontal.

(d) Let x and y be functions of time t that are related by the equation y2 = 2 +~. At time t = 5, the

value of y is 3 and ~ = 6. Find the value of;; at time t = 5.

(a) 2yy' = y + xy'(2y - x)y' = y

, yy=--2y-x

(b)y _ 1

2y - x -"22y =2y-xx=Oy=±J2(0, J2), (0, -J2)

(c) -y_=O2y-xy=OThe curve has no horizontal tangent since

02 * 2 + x-O for any x.

(d) When y = 3, 32 = 2 + 3x so x = 1.dy_dy dx_ y dxdt - dx . dt - 2y - x . dt

3 dx 9 dxAt t=5 6 =-_._=_.-, 6 _ 7... dt 11 dt

3

;;Ls = 232

2 : {I : implicit diff~rentiation1 : solves for y

{

l.-y =.12: . 2y - x 2

1 : answer

{l:y=O2·

. 1: explanation

{

I: solves for x3: 1: chain rule

1 : answer

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Ap® CALCULUS AB2001 SCORING GUIDELINES

Question 6

The function f is differentiable for all real numbers. The point (3, ~) is on the graph of

y = f(x) , and the slope at each point (x,y) on the graph is given by dy = y2 (6 - 2x).dx

2(a) Find d; and evaluate it at the point (3,.!.).

dx 4(b) Find y = f(x) by solving the differential equation ~~ = y2 (6 - 2x) with the initial

condition f(3) = ~.

(a) d2y dy 2-. = 2y-(6 - 2x) - 2ydx2 dx

= 2y3(6 - 2x? - 2y2

d2y I (1)2

dx2 (3 ~ \ = 0 - 2 "4'41

18

1(b) 2dy = (6 - 2x)dxy

1 2- - = 6x -x + Cy

- 4 = 18 - 9 + C = 9 + C

C = -13

1y = x2 - 6x + 13

3:

1: value at (3,~)

d2y2: -2

dx< - 2 > product rule or

chain rule error

6:

1 : separates variables

1 : antiderivative of dy term

1 rantiderivative of dx term

1 : constant of integration

1 : uses initial condition f(3) = ~1 : solves for y

Note: max 3/6 [1-1-1-0-0-0] if no

constant of integration

Note: 0/6 if no separation of variables

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