section 2.6 – related rates
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Section 2.6 – Related Rates. Introduction to Related Rates. We have seen a lot of relations (whether implicit or explicit) that involve two variables (frequently x and y ). It is possible these two variables are themselves functions of another variable, such as t . For instance:. - PowerPoint PPT PresentationTRANSCRIPT
Section 2.6 – Related Rates
Introduction to Related RatesWe have seen a lot of relations (whether implicit or
explicit) that involve two variables (frequently x and y). It is possible these two variables are themselves functions of another variable, such as t. For instance:
x 2 y 2 1if
x cos ty sin t
Introduction to Related RatesLet’s investigate what occurs when t changes:
x 2 y 2 1if
x cos ty sin t
t x y Equation
0
1
0
12 02
1
4
22
22
22 2
22 2
1
0
1
02 12
1
2
23
12
32
12 2
32 2
1
76
32 2
12 2
1
12
32
Notice how when t changes, both the x and y change in relation to the value of t.
As x and y change, their rates of change are related to each other. But how are
they related?
Introduction to Related RatesIn order to take the derivative of the relation using x and y , it must be done with the respect to t. For instance:
x 2 y 2 1if
x cos ty sin t
Chain Rule Twice
u1
f1 u1
f1' u1
x
u12
dxdt
2u1
ddt x
2 y 2 ddt 1
ddt x
2 ddt y
2 ddt 1
u2
f2 u2
y
u22
u2 '
f2 ' u2
dydt
2u2
2u1dxdt 2u2
dydt 0
Differentiate both sides
2xdxdt 2ydydt 0
In our exercises, we will not need to know the exact relations.
x 2 y 2 1if
x f t y g t Now we know how the rate of
change for x and y are related to each other.
u1'
Example 1Suppose x and y are both differentiable functions of t and are related by
. Find when x = 10, if when x = 10.
Chain Rule
dxdt 15
ddt 5x 2 y d
dt 100
ddt 5x 2 d
dt y ddt 100
5 ddt x
2 ddt y d
dt 100
52x dxdt dydt 0
52 10 15 dydt 0
Find the derivative by differentiating
both sides.
5x 2 y 100
1500 dydt 0
dydt 1500
dydt
Substitute the known information
Solve for the unknown
Example 2Suppose x and y are both differentiable functions of t and
are related by . Find when x = 9, y>0, and .
Chain Rule
dydt 5 2 2 2d d
dt dty x
2 2 2d d ddt dt dty x
2 1 22 2d d ddt dt dty x
1 2122 2 0dy dx
dt dty x 1 21
22 5 2 9 dxdty
Find the derivative by differentiating
both sides.
y 2 2 x 2
1320 dx
dt 60dx
dt
dxdt
Substitute the known
information
Solve for the unknown
y 2 2 9 2
y 2 42y
Find other important values:
x
1 2122 2 5 2 9 dx
dt
Example 3A spherical balloon is being filled with a gas in such a way that when the
radius is 2ft, the radius is increasing at the rate 1/6 ft/min. How fast is the volume ( ) changing at this time?
Chain Rule
ddtV d
dt43 r
3
ddtV 4
3 ddt r
3
dVdt 4
3 3r2 drdt
dVdt 4 2 2 1
6
Find the derivative by differentiating
both sides.
V 43 r
3
83
dVdt
Substitute the known information
Solve for the unknown ft3 per minute
dVdt 4r2 dr
dt
Related Rates Guidelines1. Draw a figure, if appropriate, and assign variables to the
quantities that vary. (Be careful not to label a quantity with a number unless it never changes in the problem)
2. Find a formula or equation that relates the variables. (Eliminate unnecessary variables)
3. Differentiate the equations. (typically implicitly)
4. Substitute specific numerical values and solve algebraically for any required rate. (The only unknown value should be the one that needs to be solved for.)
ft/s
Example 1A person 6 ft tall is walking towards a streetlight 20 ft high
at the rate of 7 ft/s. At what rate is the length of the person’s shadow changing?
20 ft
6 ft
xyChain Rule
xy20 x
6
ddtxy20 d
dtx6
1 120 6
d ddt dtx y x
1 120 6
d d ddt dt dtx y x
1 120 6
dydx dxdt dt dt
Find the rates by
differentiating both sides.
71 120 20 6dx dxdt dt
3dxdt
Substitute the known information
Solve for the
unknown
1 120 67dx dx
dt dt
7 720 60
dxdt
Using similar triangles, the equation is:
The y value is getting smaller. So
the rate needs to be negative when substituting.
Example 2A bag is tied to the top of a 5 m ladder resting against a vertical wall. Suppose
the ladder begins sliding down the wall in such a way that the foot of the ladder is moving away from the wall. How fast is the bag descending at the instant the foot of the ladder is 4 m from the wall and the foot is moving away at the rate of 2 m/s?
5 mLadder
x
y Chain Rule
x 2 y 2 52
ddt x
2 y 2 ddt 25
ddt x
2 ddt y
2 ddt 25
2xdxdt 2ydydt 0
2 4 2 2ydydt 0
Find the rates by
differentiating both sides.
16 6dydt 0
dydt 8
3
Substitute the known information
Solve for the
unknown
m/s
2 4 2 2 3 dydt 0
6dydt 16
Using The Pythagorean Theorem, the equation
is:
4 2 y 2 25
y 2 9
y 3
Find other important values:
x
Example 3A trough 10 ft long has a cross section that is an isosceles
triangle 3 ft deep and 8 ft across. If water flows in at the rate 2 ft3/min, how fast is the surface rising when the water is 2 ft deep?
Nothing is known
about b…
Vwater Abhprism
ddt V d
dt12 bh10
ddt V d
dt 583 hh
ddt V 40
3ddt h
2
2 803 2 dhdt
Find the rates by
differentiating both sides.
dhdt 3
80
Substitute the known information
Solve for the unknown
ft/min
2 1603dhdt
Using the volume of a prism, the equation is:
83 b
h
b 83 h
Using similar triangles:
10 ft
b8 ft
h3 ft
Vwater 12 bh10
Chain Rule
dVdt 40
3 2h dhdt
Example 4A rocket launches with a velocity of 550 miles per hour. 25 miles away there
is a photographer filming the launch. At what rate is the angle of elevation of the camera changing when the rocket achieves an altitude of 25 miles?
25 mi
Θ
x Chain Rule
25tan x
25tand d xdt dt
125tand d
dt dt x 2 1
25sec d dxdt dt
2 125sec 550d
dt
Find the rates by
differentiating both sides.
22 22d
dt
11ddt
Substitute the known information
Solve for the
unknown
rad/h
24sec 22d
dt
2 22ddt
Using The Trigonometry, the equation is:
2525tan
1tan 1 4
Use “x” to find other important values:
This is “x” and there is no “x” in the derivative…