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 = x 3 – 2 Find dy/dt , given that dx/dt = 2 when x = 1 Implicitly derive x and y with respect to t Now substitute in what you know 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: Given rate: Find dA/dt when r = 5 ft Substitute what you know! When radius is 5 ft, the Area is changing at rate of 10 π ft 2 /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 Given d V /dt = 5 ft 3 /min, Find d r /dt when r = 1.5 ft Now plug in the given info after a cleanup Equation:
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 Solution: Let θ be angle of elevation. Equations: Given t = 10 so s (10) = 5000 ft
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 (You might recall that a cylinder with same height and radius is 3 times as much volume as a cone.)
8. Implicitly derive with respect to time: A conical tank is 8 inches high and 8inches across the top. If water is flowing into the tank at rate of 5 in 3 /min, find the rate of change of the depth when the water is 6 inches deep. Given: What you can observe: or What you are solving for: when h = 6 in. Replace all r’s with expressions in h, and all dr/dt’s with expressions in dh/dt
9. Solution: or the height is changing inches /minute h = 6 in We know:
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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
