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16.6 Applying the Derivative

16.6 Applying the Derivative. The derivative can tell us all sorts of things in regards to physical situations. One of these is the instantaneous rate of change. Def: The instantaneous rate of change of y = f ( x ) per unit change in x at x 0 is:

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16.6 Applying the Derivative

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  1. 16.6 Applying the Derivative

  2. The derivative can tell us all sorts of things in regards to physical situations. One of these is the instantaneous rate of change. Def: The instantaneous rate of change of y = f (x) per unit change in x at x0 is: (aka: the derivative evaluated at x0) Ex 1) Physics: Find the velocity and the acceleration of a moving particle with the given position function (meters) at the given time (secs). f (t) = t3 – 5t2 + t + 2 at t = 4 v(t) = f(t) = 3t2 – 10t + 1 f(4) = 3(4)2 – 10(4) + 1 = 48 – 40 + 1 = 9 m/s a(t) = v(t) = f(t) = 6t – 10 f(4) = 6(4) – 10 = 14 m/s2

  3. You don’t always have to find instantaneous rates with respect to time. Ex 2) A spherical balloon is being inflated. Let x represent the radius. Find the instantaneous rate of change of the volume when the radius is 2 ft. (Find deriv. of volume) Volume of sphere: per one foot radius

  4. We can also use the derivative to find particular maximums or minimums. (Remember the first derivative sign chart helps us determine max’s & min’s) Ex 3) An oil refiner has 100,000 gallons of gasoline that could be sold now with a profit of 25 cents/gal. For each week of delays, an additional 10,000 gal can be produced. However, for each such week, the profit decreases by 1 cent/gal. It is possible to sell all the gasoline that is on hand at any time. When should the gasoline be sold to maximize the profit? *First, we need an equation! Profit function is (gallons sold)(dollars profit per gallon) w = weeks delayed (100,000 + 10,000w)(.25 – .01w) now addtl per week now decrease per week

  5. Ex 3) cont… P(w) = (100,000 + 10,000w)(.25 – .01w) = 25,000 – 1000w + 2500w – 100w2 P(w) = – 100w2 + 1500w + 25,000 Now, the derivative to find max P (w) = – 200w + 1500 = 0 – 200w = –1500 w = 7.5 P (w) + – rising to falling is a max! 7.5 Sell right at 7 ½ weeks for max profit

  6. Homework #1606 Pg 884 #1, 7, 13, 16, 18, 20, 25, 26

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