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Use implicit differentiation
W-up Use implicit differentiation
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14.6 Related Rates SWBAT solve related rate problems
Problems involving rates of related variables are related rate problems. Example: The rate at which the volume of a balloon is changing at a specific radius
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Ex1: If xy + 6x + y3 = -2 find ππ¦ ππ‘ π€βππ π₯=2, π¦=β3 πππ ππ₯ ππ‘ =3
Taking the derivative with respect to βtβ β there are no βtβsβ in the equation When you take the derivative multiply by π (ππππππππ) ππ‘ (just like implicit differentiation) Product rule for xy Substitute in given values Simplify and solve for ππ¦ ππ‘ Factored out a ππ¦ ππ‘
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Steps for solving related rate problem
Draw a picture (if possible) Identify / assign the variables Identify what you want_____ when____ List what is known, rates Write formula that relates variables in problem Differentiate Substitute numerical values for the variables and rate Solve.
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A child throws a stone into a still pond causing a circular ripple to spread. If the radius of the circle increases at the constant rate of 0.5 feet/ second, how fast is the area of the ripple increasing when the radius is 30 feet? 1) r = radius, A = area , t = seconds 2) Want rate area is increasing or π
π¨ π
π when r = 30 feet 3) rate of change of radius π
π π
π = .5 ft/sec 4) A = pr2 5) Differentiate π
π¨ π
π =ππ
π π
π π
π 6) π
π¨ π
π =ππ
ππ (.π) 7) Solve π
π¨ π
π =πππ
βππ.π πππ/ππππππ
Remember: When you take the derivative multiply by π (ππππππππ) ππ‘ Must label answer
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A balloon in the form of a sphere is being inflated at the rate of 10 cubic meters per minute. Find the rate at which the surface area of the sphere is increasing at the instant when the radius of the sphere is 3 meters. 1) r = radius, A = area , V = volume, t = minutes 2) Want rate area is increasing or π
π¨ π
π when r = 3 meters 3) rate change of volume or π
π½ π
π = 10m3/ min 4) A = 4pr2 and V = π π pr3 5) Differentiate π
π¨ π
π =ππ
π π
π π
π π
π½ π
π =ππ
ππ π
π π
π 6) π
π¨ π
π =ππ
π π
π π
π oh no we donβt know π
π π
π , what can we use to find it?
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ππ ππ‘ = 10 36π = 5 18π now plug, into formula in step 6
ππ=ππ
ππ π
π π
π ππ=ππ
(π)π π
π π
π I want the rate when radius is 3, solve for π
π π
π ππ ππ‘ = π = 5 18π now plug, into formula in step 6 π
π¨ π
π =ππ
π 5 18π 7) Solve π
π¨ π
π = ππ π βπ.ππππ/ππππππ
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Homework: # 1-8 all, 9,11,13
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