Unit #5 - Implicit Di erentiation, Related Rates Implicit ... - Queen's U

Unit #5 - Implicit Differentiation, Related Rates

Some problems and solutions selected or adapted from Hughes-Hallett Calculus.

Implicit Differentiation

(c) Setting the y value in both equations equal to

each other and solving to find the intersection

point, the only solution is x = 0 and y = 0, so

the two normal lines will intersect at the origin,

(x, y) = (0, 0).

Solving process:

1. Consider the graph implied by the equation

xy 2 = 1.

What is the equation of the line through ( 14 , 2)

which is also tangent to the graph?

Differentiating both sides with respect to x,





dy

2

=0

y + x 2y

dx

3

3

(x ? 4) + 3 = ? (x ? 4) ? 3

4

4

3

3

x?3+3=? x+3?3

4

4

3

3

x=? x

4

4

6

x=0

4

x=0

so

?y 2

dy

=

dx

2xy

dy

= ?4, so we can use the

dx

point/slope formula to obtain the tangent line

At the point ( 41 , 2),

y = ?4(x ? 1/4) + 2

2

Subbing back into either equation to find y, e.g.

y = 34 (0 ? 4) + 3 = ?3 + 3 = 0.

2

2. Consider the circle defined by x + y = 25

3. Calculate the derivative of y with respect to x,

given that

(a) Find the equations of the tangent lines to

the circle where x = 4.

x4 y + 4xy 4 = x + y

(b) Find the equations of the normal lines to

this circle at the same points. (The normal

line is perpendicular to the tangent line.)

Let x4 y + 4xy 4 = x +

 y. Then



dy

dy

3

4 dy

4x y + x

+ (4x) 4y 3

+ 4y 4 = 1 +

dx

dx

dx

hence

dy

4yx3 + 4y 4 ? 1

=

.

dx

1 ? x4 ? 16xy 3

(c) At what point do the two normal lines intersect?

Differentiating both sides,




d

d

x2 + y 2 =

(25)

dx

dx

dy

2x + 2y

=0

dx

dy

?2x

?x

=

=

dx

2y

y

4. Calculate the derivative of y with respect to x,

given that

xey = 4xy + 5y 4

dy

To solve for dx

, we must think of y as a function of x

and differentiate both sides of the equation, using the

chain rule where appropriate:

(a) The points at x = 4, satisfying x2 +y 2 = 25, would

be y = 3 and y = ?3.

Using the point/slope formula for a line, and our

dy

calculated

, through (4, 3): y = ?1.33333 ?

dx

(x ? 4) + 3, and

through (4, -3): y = 1.33333 ? (x ? 4) + (?3)

ey + xey

dy

dy

dy

= 4y + 4x

+ 20y 3

dx

dx

dx

dy

Now, we simplify and move the terms with a dx

to the

dy

right, and keep the terms without a dx to the left:

dy

ey ? 4y = (4x + 20y 3 ? xey ) dx

dy

Finally, we can solve for dx :

dy

ey ? 4y

=

dx

4x + 20y 3 ? xey

(b) If you have a line with slope m, the slope of a

perpendicular line will be ?1/m.

Through (4, 3): y = 34 (x ? 4) + 3, and

through (4, -3): y = 34 (x ? 4) + (?3)

1

To simplify a little we multiply both sides through by

¡Ì

2 x + y:

5. Use implicit differentiation to find the equation

of the tangent line to the curve xy 3 + xy = 14

at the point (7, 1).




¡Ì

¡Ì

dy

1 ? 4x2 y x + y = 4xy 2 x + y ? 1

dx

¡Ì

dy

(4xy 2 x + y) ? 1

¡Ì

=

dx

1 ? 4x2 y x + y

d

To get the slope, we take the derivative

of both

dx

sides.




d

d

xy 3 + xy =

(14)

dx

dx





 

 

dy

dy

(1)y 3 + x 3y 2

+ (1)y + x

=0

dx

dx

Gathering terms with

7. Find all the x-coordinates of the points on the

curve x2 y 2 + xy = 2 where the slope of the tangent line is ?1.

dy

We need to find the derivative

by implicit differendx

tiation. Differentiating with respect to x on both sides

of the equation,





dy

dy

2

2

+y+x

=0

2xy + x 2y

dx

dx

dy

, and those without it,

dx







dy

3xy 2 + x + y 3 + y = 0

dx

dy

y3 + y

=?

dx

3xy 2 + x

dy

, but we actually know the

dx

dy

slope we want this time:

= ?1, so let¡¯s just sub

dx

that in now:

Here, we could solve for

At the point (x, y) = (7, 1),

13 + 1

dy

=?

dx

3(7)(1)2 + 7

?2

?1

=

=

28

14

2xy 2 + x2 (2y(?1)) + y + x(?1) = 0

or 2xy 2 ? 2x2 y + y ? x = 0

factoring first terms:

Factoring common y ? x:

To build a line that goes through the point (7, 1), and

?1

, we can use the point-slope line formula.

with slope

14

We obtain the formula for the tangent line:

y=

Subbing each possibility into the equation for the curve,

x2 y 2 + xy = 2

?1

(x ? 7) + 1

14

If y = x, x2 (x2 ) + x(x) = 2

x4 + x2 = 2

x4 + x2 ? 2 = 0

(x2 + 2)(x2 ? 1) = 0

So x = ?1, 1 are two solutions, with the corresponding

y values, using y = x being y = ?1 and 1 as well.

d

of both sides,

dx

The other possible case, 2xy + 1 = 0, leads to



2





?1

?1

If y = ?1/2x, x2

+x

=2

2x

2x

1 ?1

or +

=2

4

2







d ¡Ì

d

9 + x2 y 2

x+y =

dx

 dx

 

1

1

dy

dy

2

2

¡Ì

1+

= 2xy + x 2y

2 x+y

dx

dx

Expanding left side:









 

1

1

dy 1

1

dy

2

2

¡Ì

¡Ì

+

= 2xy + x 2y

2 x+y

dx 2 x + y

dx

Gathering terms with and without

dy

dx



1

¡Ì

? 2x2 y

2 x+y

(y ? x)(2xy + 1) = 0

Meaning either y ? x = 0 (so y = x), or (2xy + 1) = 0.

6. Find dy/dx by implicit differentiation.

¡Ì

x + y = 9 + x2 y 2

Taking

2xy(y ? x) + (y ? x) = 0



which is impossible, so the assumption that y = ?1/2x

must be impossible to use with this curve.

dy

,

dx

Therefore the only two points on the curve x2 y 2 +xy =

dy

2 which have slope

= ?1 are (x, y) = (1, 1) and

dx

(?1, ?1).

1

= 2xy 2 ? ¡Ì

2 x+y

2

8. Where does the normal line to the ellipse

x2 ? xy + y 2 = 3

at the point (?1, 1) intersect the ellipse for the

second time?

To obtain a normal (perpendicular) line, we find a line

perpendicular to the tangent line on the ellipse at the

point (-1, 1). (Linear algebra students may have other

ways to do this.)

To get the slope of the tangent line, we use implicit

differentiation, with respect to x:

9. The curve with equation 2y 3 + y 2 ? y 5 = x4 ?

2x3 + x2 has been likened to a bouncing wagon

(graph it to see why). Find the x-coordinates

of the points on this curve that have horizontal

tangents.




d

d

x2 ? xy + y 2 =

(3)

dx

dx





dy

dy

+ 2y

=0

2x ? y + x

dx

dx

dy

(?x + 2y) = ?2x + y

dx

dy

?2x + y

=

dx

(?x + 2y)

Taking an implicit







d

d

2y 3 + y 2 ? y 5 =

x4 ? 2x3 + x2

dx

dx

dy

4 dy

2 dy

+ 2y

? 5y

= 4x3 ? 6x2 + 2x

6y

dx

dx

dx

At the point on the ellipse (?1, 1),

dy

?2(?1) + 1

=

dx

(?(?1) + 2(1))

= 3/3 = 1

dy

Since we know we want points where dx

= 0 (horizondy

tal tangents), we¡¯ll set dx = 0 immediately:

So the slope of the tangent line to the ellipse at (?1, 1)

is 1.

6y 2 (0) + 2y(0) ? 5y 4 (0) = 4x3 ? 6x2 + 2x

The slope of the normal line will be perpendicular to

that, or ?1/(1) = ?1.

factoring:

0 = 2x(2x2 ? 3x + 1)

0 = 2x(2x ? 1)(x ? 1)

The normal line, which also passes through (-1, 1), will

therefore be

So x = 0, 12 and 1 are the points where the tangent to

the graph will be horizontal.

y = ?1(x ? (?1)) + 1

or y = ?1(x + 1) + 1

Note that the question only asked for the x coordinate

of these points. As you can see on the graph below, all

of these x coordinates correspond to multiple y values,

but all of those points have horizontal tangents.

or y = ?x

To find the intersections of this normal line with the ellipse to find a second crossing, we sub in this expression

for y into the ellipse formula:

x2 ? xy + y 2 = 3

x2 ? x(?x) + (?x)2 = 3

x2 + x2 + x2 = 3

x2 = 1x

d

of both sides,

dx

= ¡À1

Since x = ?1 is the point we started at, x = +1 must

be the other intersection. At that point, y = ?x, so

the point is (x, y) = (1, ?1). Below is a graph of the

scenario.

3

Substituting x = 1 back into the circle equation to

find the matching y coordinates gives

10. Use implicit differentiation to find an equation

of the tangent line to the curve

y 2 (y 2 ? 4) = x2 (x2 ? 5)

(1)2 + y 2 ? 2(1) ? 4y = ?1

y 2 ? 4y = 0

at the point (x, y) = (0, ?2).

y(y ? 4) = 0

y = 0, 4

The points with horizontal tangents are (1,0) and

(1,4).

(b) The points with vertical tangents are those where

dy

is zero (making the slope

the denominator of dx

undefined). From part (a), we have

dy

?2x + 2

=

dx

2y ? 4

(The devil¡¯s curve)

Setting the denominator equal to zero gives

This point happens to be at the bottom of the loop on

the y axis, so the slope there is zero. Therefore, the

tangent line equation is simply:

2y ? 4 = 0

y = ?2

Substituting y = 2 back into the circle equation to

find the matching x coordinates gives

y=2

11. Use implicit differentiation to find the (x, y)

points where the circle defined by

x2 + (2)2 ? 2x ? 4(2) = ?1

x2 ? 2x ? 3 = 0

x2 + y 2 ? 2x ? 4y = ?1

(x ? 3)(x + 1) = 0

has horizontal and vertical tangent lines.

x = 3, ? 1

(a) Find the points where the curve has a horizontal tangent line.

This gives points on the circle with vertical tangents at (3,2), and (-1,2).

(b) Find the points where the curve has a vertical tangent line.

(a) Use implicit differentiation, then set

12. The relation

x2 ? 2xy + y 2 + 6x ? 10y + 29 = 0

dy

= 0.

dx

defines a parabola.




d

d

x2 + y 2 ? 2x ? 4y =

(?1)

dx

dx

dy

dy

2x + 2y

?2?4

=0

dx

dx

dy

(2y ? 4) = ?2x + 2

dx

dy

?2x + 2

=

dx

2y ? 4

Setting

(a) Find the points where the curve has a horizontal tangent line.

(b) Find the points where the curve has a vertical tangent line.

dy

= 0.

dx

The only point with horizontal tangent is (2, 5).

(a) Use implicit differentiation, then set

dy

now equal to zero gives

dx

(b) From the implicit derivative, get the equation in

dy

...

the form

=

.

dx

...

The point with vertical tangent is the point where

dy

the denominator of

is zero, or (1, 6).

dx

?2x + 2

2y ? 4

?2x + 2 = 0

0=

x=1

4

Solving for y 0 gives

13. The graph of the equation

x2 + xy + y 2 = 9

y0 = ?

is a slanted ellipse illustrated in this figure:

2x + y

.

x + 2y

Setting

y0 = 0

and solving for x gives

2x + y = 0

y

x=? .

2

=?

Substituting in the original equation gives

y2

y2

?

+ y 2 = 9.

4

2

Thus

3y 2

=9

4

or

y 2 = 12.

Think of y as a function of x.

Hence the horizontal tangent lines have the equations

¡Ì

y = ¡À 12.

(a) Differentiating implicitly, find a formula for

the slopes of this shape. (Your answer will

depend on x and y.)

The plus sign gives the upper tangent line, the minus

sign the lower. By symmetry the vertical tangent lines

have the equations

¡Ì

x = ¡À 12.

(b) The ellipse has two horizontal tangents.

The upper one has the equation characterized by y 0 = 0. To find the vertical tangent

use symmetry, or think of x as a function

of y, differentiate implicitly, solve for x0 and

then set x0 = 0.

Taking

The rightmost vertical tangent line with the equation

¡Ì

x = 12

d

of both sides of the relation

dx

x2 + xy + y 2 = 9

touches the ellipse in the point where

¡Ì

¡Ì

x

12

y=? =?

= ? 3.

2

2

gives

2x + y + xy 0 + 2yy 0 = 0.

Related Rates

14. Gravel is being dumped from a conveyor belt at

a rate of 30 cubic feet per minute. It forms a

pile in the shape of a right circular cone whose

base diameter and height are always equal. How

fast is the height of the pile increasing when the

pile is 17 feet high?

Recall that the volume of a right circular cone

with height h and radius of the base r is given

by V = 13 ¦Ðr2 h.

r = h/2. We have

dV

dh

, and want

.

dt

dt

Always true:

but since r = h/2 for this cone,

Taking

d

dt

of both sides:

so

Let V (t) = volume of the conical pile, and h and r be

the height and bottom radius of the cone respectively.

The cone has a base diameter and h that are equal, so

Subbing in

5

dV

dt

1 2

¦Ðr h

3

¦Ð h2

V =

h

3 4

dV

¦Ð 2 dh

=

3h

dt

12

dt

dh

4 dV

=

dt

¦Ðh2 dt

V =

= 30 ft3 /min, and h = 17,

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