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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