Real-World Applications of Differentiation PDF

 Differentiation is very important in many fields.  It has many real-world applications from sports to engineering to astronomy to space travel.  Functions and derivatives in calculus are relate to several concepts in economics.  Economic research uses calculus functional relationships,...

 Differentiation is very important in many fields.  It has many real-world applications from sports to engineering to astronomy to space travel.  Functions and derivatives in calculus are relate to several concepts in economics.  Economic research uses calculus functional relationships, relation of income, market prediction etc.  Isaac Newton developed the use of calculus in his laws of motion and gravitation.  Astronomical science deeply depends on calculus.  Calculus is used to build tracks…  Calculus was initially developed for better navigation system.  Engineers use calculus for building skyscrapers , bridges.  In robotics, calculus is used to model how robotic parts will work given commands.  Calculus is used for measuring growth rate of bacteria and certain species.  The concentration of drugs in a living organism can be modelled with calculus.  In this section, we bring the power of the calculus to bear on a number of applied problems involving finding a maximum or a minimum. We start by giving a few general guidelines.  If you need to draw a diagram to draw, draw it! Don’t try to visualize how things look in your head. Put a diagram down on paper and label it.  Determine what the variables are and how they are related.  Decide what quantity needs to be maximized or minimized.  Write an expression for the quantity to be maximized or minimized in terms of only one variable. To do this, you may need to solve for any other variables in terms of this one variable.  Determine the minimum and maximum allowable values (if any) of the variable you’re using.  Solve the problem and be sure to answer the question that is asked.  You have 40 (linear) feet of fencing with which to enclose a rectangular space for a garden. Find the largest area that can be enclosed with this much fencing and the dimensions of the corresponding garden. The only critical number is x = 10 and this is in the interval under consideration. We see the maximum value of the function, maximum The rectangle of perimeter 40 with maximum area is a square 10 ft on a side.  Minimizing the Cost of Highway Construction  The state is building a newhighway to link an existing bridge with a turnpike interchange, located 8 miles to the east and 8 miles to the south of the bridge.  There is a 5-mile-wide stretchof marshland adjacent to the bridge that must be crossed.  The highway costs $10 million per mile to build over the marsh and $7 million per mile to build over dry land.  How far to the east of the bridge should the highway be when it crosses out of the marsh?  Minimizing the Cost of Highway Construction  Note that the only critical numbers are where C’(x) = 0.  Minimizing the Cost of Highway Construction  Compare the value of C(x) at the endpoints and at this one critical number: The highway should be about 3.56 miles to the east of the bridge when it crosses over the marsh.  In economics, the term marginal is used to indicate a rate.  Thus, marginal cost is the derivative of the cost function, marginal profit is the derivative of the profit function and so on.  Total Costs, TC = FC + VC  Total Revenue, TR = P * Q  Profit = TR – TC  Break even: Profit = 0, or TR = TC  Profit Maximisation: MR = MC  Calculating Marginal Functions d (TR ) d (TC ) MR = MC = dQ dQ  A firm faces the demand curve P=17-3Q (i) Find an expression for TR in terms of Q TR = P.Q = 17Q – 3Q2  (ii) Find an expression for MR in terms of Q d (TR ) MR = = 17 − 6Q dQ 11 A firms total cost curve is given by TC=Q3- 4Q2+12Q (i) Find an expression for AC in terms of Q TC = Q3 – 4Q2 + 12Q Then, AC = TC/Q = Q2 – 4Q + 12 (ii) Find an expression for MC in terms of Q: MC = 3Q2 – 8Q + 12 (iii) When does AC=MC? Q2 – 4Q + 12 = 3Q2 – 8Q + 12 ⇒Q =2 Thus, AC = MC when Q = 2 (iv) When does the slope of AC=0? 2Q – 4 = 0, Q = 2 when slope AC = 0 12  Suppose that C(x) = 0.02x2 + 2x + 4000 is the total cost (in dollars) for a company to produce x units of a certain product.  Compute the marginal cost at x = 100 and compare this to the actual cost of producing the 100th unit. The marginal cost at x = 100 is C’(100) = 4 + 2 = 6 dollars per unit. The actual cost of producing item number 100 is:  Another quantity that businesses use to analyze production is average cost.  Suppose that C(x) = 0.02x2 + 2x + 4000 is the total cost (in dollars) for a company to produce x units of a certain product.  Find the production level x that minimizes the average cost. The only relevant critical number is at approximately x = 447. Further, so this critical number is the location of the absolute minimum on the domain x > 0.  Suppose that the demand x for an item is a function of its price p. That is, x = f (p).  If the price changes by a small amount p, then the relative change in price equals  However, a change in price creates a change in demand x, with a relative change in demand of  Economists define the elasticity of demand at price p to be the relative change in demand divided by the relative change in price for very small changes in price.  Suppose that f (p) = 400(20 − p) is the demand for an item at price p (in dollars) with p < 20.  (a) Find the elasticity of demand.  (b) Find the range of prices for which E < −1. Compare this to the range of prices for which revenue is a decreasing function of p. R’(p) = 0 if p = 10 and R’(p) < 0 if p > 10. The revenue decreases if the price exceeds 10.  Population dynamics is an area of biology that makes extensive use of calculus.  We now explore one aspect of a basic model of population growth called the logistic equation. This states that if p(t) represents population (measured as a fraction of the maximum sustainable population), then the rate of change of the population satisfies the equation p’(t) = rp(t)[1 − p(t)], for some constant r. p(t), r = 1 and p(0) = 0.05  Suppose that a population grows according to the equation p’(t) = 2p(t)[1 − p(t)] (the logistic equation with r =2).  Find the population for which the growth rate is a maximum. Interpret this point graphically.  The only critical number is p = 1/2.  The graph of y = f (p) is a parabola opening downward and hence, the critical number must correspond to the absolute maximum.