Derivative = Slope Since the first derivative of a continuous function y = f(x) represents the slope of the tangent to the curve, and, since a slope defines a change in the dependent variable (y) corresponding to a change in the independent variable ( x), the derivative is a RATE OF CHANGE. That's why the best notation for those of us not too lazy to use it is dy/dx or ds/dt etc. since this notation indicates a ratio of change in one variable to change in another. Now we consider 2 specific rates of change: 1. Change in position to change in time --velocity 2. Change in velocity to change in time --acceleration. Velocity and Acceleration If s(t) defines the position of a point P with respect to time t, then the first derivative ds/dt represents the change in position with respect to a change in time or VELOCITY. The second derivative ( d²s /dt² ) represents the change in VELOCITY with respect to a change in time or ACCELERATION Let's investigate the units involved in both derivatives. If we're measuring position in meters and time in seconds, The 1st derivative units will be meters /second which is a speed or velocity. The 2nd derivative units will be meters /second 2 which is acceleration. .Because the derivative is found by taking the limit as either h or delta t approach 0, when we use the first derivative of a position function to find velocity, we find the instantaneous velocity at the specified point in time. When we need average velocity over an interval of time, we must find the actual change in position rather than an approximation of it --(see Increments and Differentials). average velocity over interval (t , t +t ) = f (t + t ) − f (t) t .. The Derivative as a Rate of ChangeExample 1 The position function s of a point P on a coordinate line is: s(t) = t 3 – 12t² + 36t – 20 , t in seconds, s(t) in cm. Describe the motion of P during the time interval [–1, 9]. Since s'(t) = v(t) = the velocity function, and s''(t) = a (t) = the acceleration function, The signs of the 2 derivatives will tell us if the point a) is moving to the left (v (t) < 0 ), or to the right, (v (t) > 0 ) b) where velocity is decreasing (a (t) < 0 ), and where it is increasing(a (t) > 0 ). s'(t) = v (t) = 3t 2 – 24t + 36 = 3(t – 2)(t – 6) s''(t) = a (t) = 6t – 24 = 6(t – 4) -4 9 -60 -1 60 0 a(t) is v(t) is v(t) > 0 when t < 2, and when t > 6 t = + -0 + + a(t) < 0 when t < 4, a(t) > 0 when t > 4 v(t) = 3(t -2)(t -6) 2 v(t) = 0 at t = 2, t = 6 -70 -50 -40 -30 -20 -10 0 10 20 30 40 50 60 v(t) < 0 on 2 < t < 6, a(t) = 6(t -4) . t = 2 t = 4 t = -1 t = 9 . . . . a(t) = 0, t = 4, P moves right between t = -1 and t = 2 P moves left between t = 2 and t = 6 The red line shows tsh(e position of P at time t P moves right again until t = 9. t = 6 s''(t) = a (t) = –30 –12 0 12 30 s'(t) = v (t) = 63 0 –12 0 63 s (t)= –69 12 –4 –20 61 t = –1 2 4 6 9 So, P is at (–1, –69) to begin, it moves right until (2, 12) where v (t) = 0. From (2, 12), P moves left to (4, –4) where the acceleration a (t) = 0. It continues left to (6, –20) where v (t) = 0 Then it begins to move right until (9, 61). The velocity and acceleration at time t are as shown in the table. The Derivative as a Rate of ChangeTemperature Change over Time Example 2 We've created a new synthetic material called Goodstuff , so we're testing it's reactions to temperature changes. The mad scientists on our team tell us that if we heat it for t minutes, the temperature in degrees Celsius will be g(t) = 30t + 6 t + 8, for 0 [ t [ 5 They warn us that we can heat it for up to 5 minutes, past that --no one knows what happens. We need to know: 1) the average rate of temperature change over the interval between 4 and 4.41 minutes. and 2) the temperature change at exactly 4 minutes. 1) We want the average rate of change, so we'll set t = 4 min and t = 0.41 min. Then, g (t + t ) − g (t) t = g(4.41) − g(4) 0.41 = 12.9 0.41 l 31.46 ) C/min .2) We want the instantaneous rate of change, so we'll find g'(t) at t = 4. gΠ(t) = 30 + 3t , gΠ(4) = 30 + 34 = 31.5)C/min . . Derivative = slope velocity and acceleration Commerce applications practice solutions . The Derivative as a Rate of ChangeCommerce Applications The commerce applications of this theory involve 4 functions: 1. C(x) --the Cost function --usually a polynomial in x, the number of items produced --represents the cost of materials and labour needed to make the product. 2. p --the demand function --is the unit selling price (ususally in $) of the product. 3. R(x) = px --the Revenue function --income from selling x units at $p each. 4. P(x) --the Profit function = R(x) -C(x) --(money in) -(money out) C(x) is the cost of producing x units of product, so is the average Cost function or the C(x) x production cost per unit. The average Cost function is denoted c(x) or said C bar, or C hat. __ C (x) Generally, C(x) will have a constant term which represents the fixed costs of running a business such as heat, electricity, rent, etc. Since this fixed cost doesn't affect the cost of producing n more units, we consider the change in cost per unit produced called marginal cost = C'(x). Just as we consider average and marginal Cost function, we also consider the average and marginal Revenue and Profit functions denoted R'(x) and P '(x) respectively. __ R (x), __ P (x), And there's also the average marginal Cost, Revenue and Profit functions denoted respectively. __ C Π(x), __ R Π(x), and __ P Π(x) Just as we use dy/dx or f '(x) to approximate changes to the function when the change in x is small, we use C '(x) --the marginal cost function to approximate the cost of producing the (x + 1)th unit when x is large to begin with. The logic being that a change of 1 unit when we started at 550 or 1000 units is small enough to allow the approximation to be within acceptable norms. So, if asked to find the cost of producing the 501st thingy we'll evaluate C '(500). (see example 4 part b). To minimize the average cost, set marginal cost = average cost. C'(x) = c(x) To maximize revenue, set marginal revenue = marginal cost. R'(x) = C'(x) . The Derivative as a Rate of ChangeExample 3 A widget manufacturer has a monthly fixed cost of $10,000, a production cost of $12 per widget and he sells the widgets at $20 each. Find: a) C(x) cost function, c(x) avg. cost function, R(x) revenue function , and P(x) profit function. b) Find values for C(x), c(x), R(x), and P(x) if x = 1000. c) How many widgets must be sold to break even? (no profit, no loss) Solution c) set profit P(x) = 0, so 8x – 10,000 = 0 which means x = 1250 to break even. b) C(1000) = 22,000 c(1000) = 22 R(1000) = 20,000 P(1000) = –2000 a) C(x) = 12x + 10,000 c(x) = 12 + 10,000/x R(x) = 20x P(x) = 8x – 10,000 .Example 4 A paint manufacturer determines that the total cost in dollars of producing x gallons of paint per day is: C(x) = 5000 + x + 0.001x² We want to: a) Find the marginal cost of producing 500 gallons per day. b) Use this marginal cost to approximate the cost of producing the 501st gallon of paint. c) Find the exact cost of producing the 501st gallon of paint. Solution a) The marginal cost is C ' (x) = 1 + 0.002x, so, C ' (500 ) = 1 + 0.002(500) = 2. b) Since C '(500 ) = 2, the approximate cost of producing the 501st gallon of paint is $2.00. c) The exact cost of producing the 501st gallon of paint is C(501 ) – C(500 ) = {5000 + 501 + 0.001(501)²} -{5000 + 500 + 0.001(500)²} = 2.001 . The Derivative as a Rate of ChangeDerivative = slope velocity and acceleration Commerce applications practice solutions .Practice 1) Our company estimates that the cost in dollars of producing x boblets per day is: C (x) = 2600 + 2x + 0.001x² a) Find the cost, average cost and marginal cost of producing 1000 boblets, and 2000 boblets. b) What production level (x) will minimize the average cost? c) What is the minimum average cost per boblet? .2) The average cost of producing x units of a commodity is: c (x) = 21.4 – 0.002x Find the marginal cost function and evaluate it at a production level of 1000 units. .3) The cost in dollars of producing x meters of fine silk fabric is: C (x) = 1200 + 12x – 0.1x² + 0.0005x³ The demand function p --the selling price in dollars per meter --is: p (x) = 29 – 0.00021x What production level will maximize the profits? .4) The position function s of a point P on a coordinate line is: s (t) = t³ – 9t + 1 , t in seconds, s (t) in cm. a) Find the velocity and acceleration at time t. b) Describe the motion of P during the time interval [–3, 3]. c) Draw a diagram of the motion of P during the time interval [–3, 3]. .5) Show that the rate of change of the volume of a sphere with respect to its radius is numerically equal to the surface area of the sphere. . Derivative = slope velocity and acceleration Commerce applications practice solutions . The Derivative as a Rate of ChangeSolutions 1) Our company estimates that the cost in dollars of producing x boblets per day is: C(x) = 2600 + 2x + 0.001x 2 a) Find the cost, average cost and marginal cost of producing 1000 boblets, and 2000 boblets. c(x) = C(x)/x = 2600 /x + 2 + 0.001x (average cost function) C '(x) = 2 + 0.002x (marginal cost function) 2,000 10,600,000 5.30 6.00 1,000 5,600,000 5.60 4.00 x C(x) c(x) C '(x) .b) What production level (x) will minimize the average cost? Set marginal cost C '(x) = average cost c(x) to minimize average cost. 2 + 0.002x = 2600 /x + 2 + 0.001x 0.001x = 2600 x ux2 = 2600 0.001 = 2, 600, 000 x = 2, 600, 000 l 1612 Producing 1612 boblets per day will minimize the average cost. .c) What is the minimum average cost per boblet? c(1612) = $5.22 per boblet. .2) The average cost of producing x units of a commodity is: c(x) = 21.4 -0.002x Find the marginal cost function and evaluate it at a production level of 1000 units. Since c(x) = C(x)/x, we will multiply c(x) by x to get C(x). C(x) = 21.4x -0.002x 2 Now, marginal cost function = C '(x) = 21.4 -0.004x C '(1000) = 21.4 -0.004(1000) = 17.40 The marginal cost of 1000 units is $17.40. . The Derivative as a Rate of Change3) The cost in dollars of producing x meters of fine silk fabric is: C(x) = 1200 + 12x -0.1x 2 + 0.0005x 3 The demand function p --(the selling price in dollars per meter) --is: p(x) = 29 -0.00021x What production level will maximize the profits? The Profit function P(x) is R(x) -C(x) , so we must find R(x). R(x) = px = 29x -0.00021x 2 So, P(x) = (29x -0.00021x 2) -(1200 + 12x -0.1x 2 + 0.0005x 3) P(x) = -0.0005x 3 + 0.09979x 2 + 17x -1200 . To maximize this function, we set P'(x) = 0 P'(x) = -0.0015x 2 + 0.19958x + 17 = 0 The zeros are at -59 and 192 so the answer is 192 meters .4) The position function s of a point P on a coordinate line is: s(t) = t 3 -9t + 1 , t in seconds, s(t) in cm. a) Find the velocity and acceleration at time t. v(t) = s'(t) = 3t 2 -9 a(t) = s''(t) = 6t b) Describe the motion of P during the time interval [–3, 3]. v(t)= 3t 2 – 9 = 3(t 2 – 3) = 3(t + 3 )(t − 3 ) v(t) = 0 at t = − 3 , and t = 3 v(t) < 0 so P moves left on -3 [ t [ 3 v(t) > 0 so P moves right on−3 [ t [ − 3 , and on 3 [ t [ 3 a(t) = 0 when t = 0 a(t) < 0 when t < 0, and a(t) > 0 when t > 0 s''(t) = a(t) = –18 −6 3 0 6 3 18 s'(t) = v(t) = 18 0 –9 0 18 s(t) = 1 11.39 1 –9.39 1t = –3 − 3 0 3 3 The Derivative as a Rate of Changec) Draw a diagram of the motion of P during the time interval [ – 3, 3]. -10 -8 -6 -4 -2 0 2 4 6 8 10 t = 3 t = 0 t = -3 t = -3 t = 3 .5) Show that the rate of change of the volume of a sphere with respect to its radius is numerically equal to the surface area of the sphere. V = 43 r 3 u dV dr = 4r 2 4r 2 is the surface area of the sphere. . Derivative = slope velocity and acceleration Commerce applications practice solutions .. Back to Cal I MathRoom Index . (all content © MathRoom Learning Service; 2004 -). The Derivative as a Rate of Change