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Calc 2.6

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(1) The radius of a balloon increases faster when first starting to pump air in compared to just before bursting. When first pumping, the radius is small so a given increase in volume causes a larger increase in radius. Near bursting, the radius is large so the same volume increase causes a smaller radius increase. (2) To solve related rates problems, write an equation relating the variables, take the derivative of the equation with respect to time, and substitute given rate information. Be careful not to substitute values before differentiating, as this prevents variables from changing over time. (3) Geometry formulas are useful for setting up related rates equations involving shapes like spheres or cones. Implicit differentiation allows finding rates of

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2.6 Related Rates Solving Real-Life Problems

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You pump air at a steady rate into a deflated balloon until the balloon bursts. Does the radius of the balloon change faster when you first start pumping the air, or just before the balloon bursts? Why?

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Ex 1 p. 149 Two rates that are related Suppose x and y are two differentiable functions of t and are related by equation y = x3 – 2 Find dy/dt , given that dx/dt = 2 when x = 1 Implicitly derive x and y with respect to t y = x3 − 2 dy 2 dx = 3x Now substitute in what you know dt dt dy = 3(1) 2 (2) =6 dt In this problem you were given the equation that relates y and x. Most times you have to create the equation that relates variables. Luckily we know geometry and trig!

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Ex 2 p. 150 Ripples in a pond A pebble is dropped in a calm pond, causing ripples in the form of concentric circles. The radius r of the outer ripple is increasing at a constant rate of 1 foot per second. When the radius is 5 feet, at what rate is the total area A of the disturbed water changing? Equation: A = π r 2 r dr Given rate: = 1 ft/sec dt A = π r2 Find dA/dt when r = 5 ft dA dr Substitute what you know! = π (2r ) dA dt dt = 2π (5)(1) = 10π dt dA dr = 2π r When radius is 5 ft, the Area is changing dt dt at rate of 10π ft2/sec

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Ex 3 p. 151 An Inflating Balloon Air is being pumped into a spherical balloon at a rate of 5 cubic feet per minute. Find the rate of change of the radius when the radius is 1.5 feet. Volume is changing with time, and so is the radius Equation: 4 3 V = πr 3 Given dV/dt = 5 ft3/min, Find dr/dt when r = 1.5 ft dV  4  2 dr dV 2 dr =  π ÷3r = 4π r dt  3  dt dt dt Now plug in the given info after a cleanup dr dr 5 5 = 4π (1.5) 2 = ft / min ≈ 0.177 ft / min dt dt 9π

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Ex 5 p.152 Changing Angle of Elevation Find the rate of change in the angle of elevation of the camera 10 seconds after lift-off 2 2 hyp = s + 2000 2 Solution: Let θ be angle of elevation. Equations: tan θ = s 2000 s = 50t 2 Given t = 10 so s(10) = 5000 ft ds = 100t = velocity dt dθ 1 ds dθ (100t ) sec 2 θ = = cos 2 θ dt 2000 dt dt 2000 2000 cos θ = 2 s 2 + 20002 dθ  2000  (100t ) 2000 ( 100 ) ( 10 ) 2 = ÷ = = rad/sec dt  s 2 + 20002  2000 50002 + 20002 29

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Ex 6 Conical tanks and water going into or out of a tank When water goes into a conical tank, the volume, the radius, and the height of the water are all a function of time. We need to find an equation that relates volume, radius and height, namely π r 2h V= 3 (You might recall that a cylinder with same height and radius is 3 times as much volume as a cone.)

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A conical tank is 8 inches high and 8inches across the top. If water is flowing into the tank at rate of 5 in3/min, find the rate of change of the depth when the water is 6 inches deep. dV = 5in3 / min Given: dt Implicitly derive with respect r 4 to time: π r 2h What you can observe: = V= h 8 3 or r = 1 h dr 1 dh = 2 dt 2 dt dV π  2 dh dr  What you are solving for: dh = r + h(2r ) ÷ dt dt 3  dt dt  when h = 6 in. Replace all r’s with expressions dV π  2 dh dr  = r + 2rh ÷ in h, and all dr/dt’s with dt 3  dt dt  expressions in dh/dt Yuck!!!!

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Let’s start again. dV 1 We can figure out: = 5in3 / min r= h dt 2 π r 2h V= Substitute in for r 3 π ( 1 2 h) 2 h π h3 V= = 3 12 dV π 2 dh π h 2 dh = (3h ) = dt 12 dt 4 dt π 62 dh dh 5= = 9π 4 dt dt Solution: 5 dh = 9π dt or the height is changing 5 9π inches /minute

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•Related rate problems often involve a situation in which you are asked to calculate the rate at which one quantity changes with respect to time from the rate at which a second quantity changes with respect to time. •Related rate problems can be recognized because the rate of change of one or more quantities with respect to time is given and the rate of change with respect to time of another quantity is required. http://www.mathdemos.org/mathdemos/relatedrates/relatedrates.html

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Calc 2.6

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DO NOT MAKE ANY SUBSTITUTIONS before you differentiate! That creates a problem where things can’t change with time. If a rate is decreasing or getting smaller over time, it is a negative rate! Geometry formulas are in the back of your book. 2.6 p. 154 #1-8, 13-27 odd, 35, 36, 45, 46

More Related Content

Calc 2.6

  • 1. 2.6 Related Rates Solving Real-Life Problems
  • 2. You pump air at a steady rate into a deflated balloon until the balloon bursts. Does the radius of the balloon change faster when you first start pumping the air, or just before the balloon bursts? Why?
  • 3. Ex 1 p. 149 Two rates that are related Suppose x and y are two differentiable functions of t and are related by equation y = x3 – 2 Find dy/dt , given that dx/dt = 2 when x = 1 Implicitly derive x and y with respect to t y = x3 − 2 dy 2 dx = 3x Now substitute in what you know dt dt dy = 3(1) 2 (2) =6 dt In this problem you were given the equation that relates y and x. Most times you have to create the equation that relates variables. Luckily we know geometry and trig!
  • 4. Ex 2 p. 150 Ripples in a pond A pebble is dropped in a calm pond, causing ripples in the form of concentric circles. The radius r of the outer ripple is increasing at a constant rate of 1 foot per second. When the radius is 5 feet, at what rate is the total area A of the disturbed water changing? Equation: A = π r 2 r dr Given rate: = 1 ft/sec dt A = π r2 Find dA/dt when r = 5 ft dA dr Substitute what you know! = π (2r ) dA dt dt = 2π (5)(1) = 10π dt dA dr = 2π r When radius is 5 ft, the Area is changing dt dt at rate of 10π ft2/sec
  • 5. Ex 3 p. 151 An Inflating Balloon Air is being pumped into a spherical balloon at a rate of 5 cubic feet per minute. Find the rate of change of the radius when the radius is 1.5 feet. Volume is changing with time, and so is the radius Equation: 4 3 V = πr 3 Given dV/dt = 5 ft3/min, Find dr/dt when r = 1.5 ft dV  4  2 dr dV 2 dr =  π ÷3r = 4π r dt  3  dt dt dt Now plug in the given info after a cleanup dr dr 5 5 = 4π (1.5) 2 = ft / min ≈ 0.177 ft / min dt dt 9π
  • 6. Ex 5 p.152 Changing Angle of Elevation Find the rate of change in the angle of elevation of the camera 10 seconds after lift-off 2 2 hyp = s + 2000 2 Solution: Let θ be angle of elevation. Equations: tan θ = s 2000 s = 50t 2 Given t = 10 so s(10) = 5000 ft ds = 100t = velocity dt dθ 1 ds dθ (100t ) sec 2 θ = = cos 2 θ dt 2000 dt dt 2000 2000 cos θ = 2 s 2 + 20002 dθ  2000  (100t ) 2000 ( 100 ) ( 10 ) 2 = ÷ = = rad/sec dt  s 2 + 20002  2000 50002 + 20002 29
  • 7. Ex 6 Conical tanks and water going into or out of a tank When water goes into a conical tank, the volume, the radius, and the height of the water are all a function of time. We need to find an equation that relates volume, radius and height, namely π r 2h V= 3 (You might recall that a cylinder with same height and radius is 3 times as much volume as a cone.)
  • 8. A conical tank is 8 inches high and 8inches across the top. If water is flowing into the tank at rate of 5 in3/min, find the rate of change of the depth when the water is 6 inches deep. dV = 5in3 / min Given: dt Implicitly derive with respect r 4 to time: π r 2h What you can observe: = V= h 8 3 or r = 1 h dr 1 dh = 2 dt 2 dt dV π  2 dh dr  What you are solving for: dh = r + h(2r ) ÷ dt dt 3  dt dt  when h = 6 in. Replace all r’s with expressions dV π  2 dh dr  = r + 2rh ÷ in h, and all dr/dt’s with dt 3  dt dt  expressions in dh/dt Yuck!!!!
  • 9. Let’s start again. dV 1 We can figure out: = 5in3 / min r= h dt 2 π r 2h V= Substitute in for r 3 π ( 1 2 h) 2 h π h3 V= = 3 12 dV π 2 dh π h 2 dh = (3h ) = dt 12 dt 4 dt π 62 dh dh 5= = 9π 4 dt dt Solution: 5 dh = 9π dt or the height is changing 5 9π inches /minute
  • 10. •Related rate problems often involve a situation in which you are asked to calculate the rate at which one quantity changes with respect to time from the rate at which a second quantity changes with respect to time. •Related rate problems can be recognized because the rate of change of one or more quantities with respect to time is given and the rate of change with respect to time of another quantity is required. http://www.mathdemos.org/mathdemos/relatedrates/relatedrates.html
  • 12. DO NOT MAKE ANY SUBSTITUTIONS before you differentiate! That creates a problem where things can’t change with time. If a rate is decreasing or getting smaller over time, it is a negative rate! Geometry formulas are in the back of your book. 2.6 p. 154 #1-8, 13-27 odd, 35, 36, 45, 46