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Related Rates. Objective: To find the rate of change of one quantity knowing the rate of change of another quantity. Related Rates. - PowerPoint PPT Presentation
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Related Rates
Objective: To find the rate of change of one quantity knowing the rate of
change of another quantity.
Related Rates
• When looking at water draining from a cone, we notice that the radius and height of the water changes as the volume changes. If we are interested in the rate of change of the volume with respect to time, we need to take the derivative of the volume function.
hrV 2
3
Related Rates
• This is called a related rates problem because the goal is to find an unknown rate of change by relating it to other variables whose values and whose rates of change at time t are known or can be found.
hrV 2
3
]2[3
2
dt
drrh
dt
dhr
dt
dV
Example 1
• Suppose that x and y are differentiable functions of t and are related by the equation . Find
at time t = 1 if x = 2 and at time t = 1.
3xy dtdy /
4/ dtdx
dt
dxx
dt
dy
xy
2
3
3
48)4()2(3 2 dt
dy
Strategies for Solving Related Rates
1) Assign letters to all quantities that vary with time2) Identify the rates of change that are known and the
rate of change that is to be found3) Find an equation (Area, Volume, Pythagorean
Theorem, Trig Ratio) that relates variables 4) Differentiate both sides with respect to time5) AFTER differentiating both sides, substitute all
known values and solve
Example 2
• Assume that oil spilled from a ruptured tanker spreads in a circular pattern whose radius increases at a constant rate of 2ft/s. How fast is the area of the spill increasing when the radius of the spill is 60ft?
Example 2
• Assume that oil spilled from a ruptured tanker spreads in a circular pattern whose radius increases at a constant rate of 2ft/s. How fast is the area of the spill increasing when the radius of the spill is 60ft?
• We are looking for dA/dt.• We know dR/dt = 2 ft/s.• We know that r = 60 ft.
?60rdt
dA
2dt
dR
Example 2
• Assume that oil spilled from a ruptured tanker spreads in a circular pattern whose radius increases at a constant rate of 2ft/s. How fast is the area of the spill increasing when the radius of the spill is 60ft?
• We are looking for dA/dt.• We know dR/dt = 2 ft/s.• We know that r = 60 ft. dt
drr
dt
dA
rA
2
2
Example 2
• Assume that oil spilled from a ruptured tanker spreads in a circular pattern whose radius increases at a constant rate of 2ft/s. How fast is the area of the spill increasing when the radius of the spill is 60ft?
• We are looking for dA/dt.• We know dR/dt = 2 ft/s.• We know that r = 60 ft. dt
drr
dt
dA
rA
2
2
s
ft
dt
dA 2
240)2)(60(2
Example 3
A 25-foot ladder is leaning against a wall. If the top of the ladder slips down the wall at a rate of 2ft/s, how fast will the foot be moving away from the wall when the top of the ladder is 7 feet above the ground?
Example 4
• A camera is mounted at a point 3000 ft from the base of a rocket launching pad. If the rocket is rising vertically at 880 ft/s when it is 4000 ft above the launching pad, how fast must the camera elevation angle change at that instant to keep the camera aimed at the rocket?
Example 4
• A camera is mounted at a point 3000 ft from the base of a rocket launching pad. If the rocket is rising vertically at 880 ft/s when it is 4000 ft above the launching pad, how fast must the camera elevation angle change at that instant to keep the camera aimed at the rocket?
• t = time in seconds• = angel of elevation• h = height of rocket
3000tan
h
Example 4
• A camera is mounted at a point 3000 ft from the base of a rocket launching pad. If the rocket is rising vertically at 880 ft/s when it is 4000 ft above the launching pad, how fast must the camera elevation angle change at that instant to keep the camera aimed at the rocket?
• t = time in seconds• = angel of elevation• h = height of rocket
?4000 hdt
d
sftdt
dhh /8804000
3000tan
h
Example 4
• We need to take the derivative to find the answer.
3000tan
h
dt
dh
dt
d
3000
1sec2
Homework
• Pages 221-223• 1 – 9 odd• 13 – 17 odd• #21
• Let’s do 9 and 13 NOW together