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Drill: find th e derivative of the following. 2xy + y 2 = x + y 2xy’ +2y + 2yy’ = 1 + y’ 2xy’ + 2yy’ – y’ = 1 – 2y y’(2x + 2y – 1) = 1 – 2y y’ = (1-2y)/(2x + 2y -1). xsin (y)= 1 – xy xcos (y)y’ + sin (y) = -( xy ’ +y) xcos (y)y’ + sin (y) = - xy ’ –y xcos (y) y ’+xy ’= -y – sin (y) - PowerPoint PPT Presentation
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Drill: find the derivative of the following
• 2xy + y2 = x + y• 2xy’ +2y + 2yy’ = 1 + y’• 2xy’ + 2yy’ – y’ = 1 – 2y• y’(2x + 2y – 1) = 1 – 2y• y’ = (1-2y)/(2x + 2y -1)
• xsin(y)= 1 – xy• xcos(y)y’ + sin (y) = -(xy’ +y)• xcos(y)y’ + sin (y) = -xy’ –y• xcos(y)y’+xy’= -y – sin (y)• y’ (xcos (y) + x) = -y – sin (y)• y’ = - (y + sin(y))/(xcos(y) + x)
Related Rates
Section 4.6
Objectives
• Students will be able to– solve related rate problems.
Strategy for Solving Related Rate Problems
• Understand your problem• Develop a mathematical model of the problem • Write an equation relating the variable whose rate of
change you seek with the variables whose rate of change you know
• Differentiate both sides of the equation implicitly with respect to time t.
• Substitute values for any quantities that depend on time
• Interpret the solution
Finding related rate equations
• Assume the radius r of a sphere is a differentiable function of t and let V be the volume of the sphere. Find an equation that related dV/dt and dr/dt
3
34 rV
dtdrr
dtdV 24
Finding related rate equations
• Assume that the radius and height h of a cone are differentiable functions of t and let V = volume of the cone. Find an equation that relates dV/dt, dr/dt and dh/dt.
hrV 2
31
)(31 2hrV
)2(31 2
dtdrrh
dtdhr
dtdV
Example 1 Finding Related Rate EquationsAssume that the width a and the length b of a rectangle are functions of t and let A be the area of the rectangle. Find an equation that relates . and ,,
dtdb
dtda
dtdA
Example 1 Finding Related Rate EquationsAssume that the width a and the length b of a rectangle are functions of t and let A be the area of the rectangle. Find an equation that relates . and ,,
dtdb
dtda
dtdA
abA
abdtdA
dtd
adtdbb
dtda
dtdA
Example 2 A Rising BalloonA hot-air balloon rising straight up from a level field is tracked by a range finder 700 feet from the lift-off point. At the moment the range finder’s elevation angle is , the angle is increasing at the rate of 0.16 radians per minute. How fast is the balloon rising at that moment?
6
Example 2 A Rising BalloonA hot-air balloon rising straight up from a level field is tracked by a range finder 700 feet from the lift-off point. At the moment the range finder’s elevation angle is , the angle is increasing at the rate of 0.16 radians per minute. How fast is the balloon rising at that moment?
6
feet 700
h
dtdh :find want toWe
rad/min 16.0 :know We dtd
700tan h
Example 2 A Rising Balloon
feet 700
h
dtdh :find want toWe rad/min 16.0 :know We
dtd
700tan h
tan700h
tan700dtd
dtdh
dtd
dtdh 2sec700
16.06
sec700 2
dtdh
33.149 ft/min
Example 3 A Highway ChaseA police cruiser approaching a right-angled intersection from the north is chasing a speeding car that has turned the corner and is now moving straight east. When the cruiser is 0.70 mile north of the intersection and the car is 0.80 mile to the east, the police determine with radar that the distance between them and the car is increasing at 15 mph. If the cruiser is moving at 45 mph at the instant of measurement, what is the speed of the car?
Example 3 A Highway ChaseA police cruiser approaching a right-angled intersection from the north is chasing a speeding car that has turned the corner and is now moving straight east. When the cruiser is 0.70 mile north of the intersection and the car is 0.80 mile to the east, the police determine with radar that the distance between them and the car is increasing at 15 mph. If the cruiser is moving at 45 mph at the instant of measurement, what is the speed of the car?
Example 3 A Highway ChaseA police cruiser approaching a right-angled intersection from the north is chasing a speeding car that has turned the corner and is now moving straight east. When the cruiser is 0.70 mile north of the intersection and the car is 0.80 mile to the east, the police determine with radar that the distance between them and the car is increasing at 15 mph. If the cruiser is moving at 45 mph at the instant of measurement, what is the speed of the car?
Let x = distance of the speeding car from the intersectionLet y = distance of the police cruiser from the intersectionLet z = distance between the car and the cruiser
Example 3 A Highway ChaseWe want to know:
We know:
x
y
z
dtdx
mph 15dtdz
mph 45dtdy
222 zyx
222 zdtdyx
dtd
dtdzz
dtdyy
dtdxx 222
dtdzz
dtdyy
dtdxx
Example 3 A Highway Chasex = .8 and y = .6
x
y
z
157.08.0457.08.0 22 dtdx
9.155.318.0 dtdx 4.478.0
dtdx
25.59dtdx
At that moment, the car’s speed was about 59.3 mph.
mph 15dtdz mph 45
dtdy
We can now substitute, remembering that z2 = x2 + y2 ; so z =(x2 + y2)1/2
dtdzz
dtdyy
dtdxx
Example 4 Filling a Conical TankWater runs into a conical tank at the rate of 12 ft3 /min. The tank stands point down and has a height of 12 ft and a base radius of 9 ft. How fast is the water level rising when the water is 8 ft deep?
Example 4 Filling a Conical TankWater runs into a conical tank at the rate of 12 ft3/min. The tank stands point down and has a height of 12 ft and a base radius of 9 ft. How fast is the water level rising when the water is 8 ft deep?
rdtdh :find want toWe
h
min/ft 12 :know We 3dtdV
hrV 2
31
129
hr
hr43
hhV2
43
31
Example 4 Filling a Conical Tank
r
dtdh :find want toWe
h
min/ft 12 :know We 3dtdV
129
hr
hr43
hhV2
43
31
3
163 hV
3
163 h
dtdV
dtd
dtdhh
dtdV 2
169
Example 4 Filling a Conical Tank
r
dtdh :find want toWe
h
min/ft 12 :know We 3dtdV
129
hr
hr43
dtdhh
dtdV 2
169
dtdh28
16912
dtdh3612
dtdh
31 106.0 ft/min
Homework
• Page 251: 9-21 odd
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