STEYNING & DISTRICT U3A. Discovering Mathematics Session 12. Some geometrical problems & Applications of Calculus.

Transcription

1 STEYNING & DISTRICT U3A Discovering Mathematics Session 12 Some geometrical problems & Applications of Calculus.

2 Calculate the Diagonal What is the length of the hypotenuse AC of the right angled triangle? ABCO is obviously a perfect rectangle & therefore AC=OB, But OB is the radius of the arc. Therefore, AC = 8 A B In this case OC bisects the radius, but the same result will apply to any triangle produced from a vertical line from the horizontal radius to the arc. O 4 4 C

3 Dissecting an obtuse angled triangle. Is it possible to cut an obtuse angled triangle into a number of acute angled triangles and if so what is the minimum number? If we draw a line from any vertex to the opposite side we can produce an acute angled triangle, but the second triangle will have an obtuse angle. If we divide the 2 nd triangle in the same way the 3 rd triangle will also have an obtuse angle. 1 We can continue cutting but the remaining triangle will always be obtuse. The answer to the question appears to be no, but there is a solution

4 Dissecting an obtuse triangle (cont d) The solution lies in making the first line to terminate within the triangle and from there we can draw further lines to create acute triangles. One can draw several lines from the vertex to make acute triangles but the minimum number has to be 5, otherwise the angles so formed would not be acute, eg 360/4 = 90 From this we can conclude that the minimum number of acute triangles thus inscribed is 7.

5 Differential Calculus(Recapitulation) In a study of differential calculus, we will frequently come across expressions such as dx, dy, dt & du etc. We must regard these expressions to mean minute bits of x, y, t & u etc. The mathematical term for these expressions is differentials. Because dx, say, is extremely small, the mathematical convention is to regard dx times dx, ie (dx) 2, to be of second order of smallness and therefore negligible. Obviously, 3 rd, 4 th & higher orders of smallness can also be ignored. For example, let us assume that the function x increases by a minute amount dx. We then have x+dx. If we square this we get (x+dx) 2 = x 2 + 2xdx + (dx) 2 The 2 nd term is of 1 st order smallness and is significant. However (dx) 2 is of 2 nd order of smallness and can be ignored.. If we assume that dx is 1/100 of x, then (dx) 2 is 1/10000 & is insignificent., whereas 2xdx = 2x/100 and is significent.

6 This principle can be well illustrated by a geometric representation. dx x (dx) 2 The areas total x 2 + 2xdx + (dx) 2 & one can see that the 3rd term is of little significance. If we cube the original expression, ie (x + dx) 3, it will expand to x 3 + 3x 2.dx + 3x*(dx) 2 +(dx) 3 In this example the 3 rd & 4 th factors may be ignored. x dx In differential calculus we are looking for the ratio of dy/dx for instances where y and x are variables which are explicitly related to each other. Assume y = x 2. Then y+dy = (x + dx) 2 = x 2 + 2xdx + (dx) 2 If we deduct the original equation from both sides & ignoring 2 nd order items. Then dy = 2xdx & dy/dx = 2x. This is the differential of y = x 2

7 In a further example, y = x 3 and y + dy = (x + dx) 3 = x 3 + 3x 2 dx + 3x(dx) 2 +(dx) 3 If we deduct the initial equation and ignore 2 nd & 3 rd order elements ; dy = 3x 2 dx & dy/dx = 3x 2 For the general equation, y = a*x n the differential dy/dx = anx (n-1) For an equation with a more complex relationship, the same principles apply. Eg y= 3x 4 + 2x 3 5x 2 + 2x + 5 dy/dx = 12x 3 + 6x 2 10x + 2 NB The constant 5 disappears on differentiating. Try these i) y = x 4 7x 2 + 5x + 15 dy/dx = 4x 3 14x + 5 ii) u = 2v 3 + 6v 2 10v + 12 du/dv = 6v v 10 iii) s = 2/v 3 + 6/v 2 10/v = 2v v -2 10v -1 ds/dv = -6v -4 12v v -2 = -6/v 4 12/v /v 2

8 Differentiating Products & Quotients Products For a relationship such as y = (x 3 + 3)*(5 x 2 ) we use a 2 step process. Let (x 3 + 3) = u & du/dx = 3x 2 (5 x 2 ) = v & dv/dx = -2x So y = u*v & y+dy = (u+du)*(v+dv) = u*v + v*du + u*dv + du*dv As previously, deduct the original equation & discard 2 nd order elements. Then dy = vdu + udv & dy/dx = v*du/dx + u*dv/dx So, dy/dx = (5 x 2 )*3x 2. + (x 3 + 3)*-2x Simplifying; dy/dx = 3x 2* (5 x 2 )* 2x* (x 3 + 3)

9 Quotients For an equation such as ; y = (x 3 + 3)/(5 x 2 ) we use a similar 2 step method Ie y = u/v & y + dy = (u + du)/(v + dv) The dividing process is somewhat more complicated than for a product, but after discounting higher order elements; y + dy = u/v + du/v u*dv/v 2 = u/v + (vdu udv)/v 2 After deducting the original equation & dividing both sides by dx dy/dx = (vdu/dx udv/dx)/v 2 dy/dx = [(5 x 2 )*(3x 2 ) - (x 3 + 3)*(-2x)] / (5 x 2 ) 2 dy/dx = (15x 2-3x 4 + 2x 4 + 6x) / (5 x 2 ) 2 dy/dx = (15x 2 - x 4 + 6x) / (5 x 2 ) 2

10 Integral Calculus Integral Calculus can be simply defined as a process for summation of many very small elements. Fundamentally the process of integration is the converse of differentiation. Eg differential of y = ax n is dy/dx = nax (n-1) or dy = nax (n-1).dx The integral of the same equation would reverse that process thus : y = (ax (n)* dx) = [(a*x (n+1) )/(n+1) + c] where c is a constant. The addition of a constant seems confusing but if you recall all constants disappear when we differentiate, so we must include them to integrate. Try these : y = (2x 3.dx) then y =[ x 4 /2 + c] s = (2x 2.dt) then s = [2x 3 /3 + c] Eg. The general integral of y = (bx n dx) = [ b/(n+1)*x (n+1) + c] Let s try a few practical uses of integration.

11 Laws of Motion Acceleration of a body (f) is the rate of change of velocity with respect to time. f = (v u)/t where initial vel. = u & final vel. = v after time t; or v = u + ft (Eqn. 1) But distance travelled (ds) = dv*dt Therefore s = (u + ft) dt Therefore; s = [ut + f/2 * t 2 ] or s = ut + ½*f t 2 (Eqn. 2) Vel v f You will recall that we have demonstrated this graphically previously. The distance travelled is the area of the plot of velocity against time. u The yellow area is u*t The orange area = ½*(v u)*t but v-u = f Time t Therefore; s = ut + ½ft 2

12 Maxima & Minima dy/dx = 0 dx dy y The slope of the curve at any point = dy/dx But dy/dx is clearly = 0 at the turning points of the curve. This leads to the general statement that for any case where a relationship between x & y can be established, then dy/dx = 0 for the maxima & minima points. x For the cubic curve; y = x 3-4x 2-3x + 5 (NB not the curve shown) dy/dx = 3x 2-8x - 3 dy/dx = 0 So 3x 2-8x - 3 = 0 for Max & Min Factorizing; (3x + 1)*(x - 3) = 0 x = -1/3 and x = +3 Substituting these values in the original cubic equation gives values of y 6.5 & y = 13

13 Volume of a Cone by Calculus Volume of thin slice, dv = πx 2 times dy. By similar triangles, x/y = r/h ie x = ry/h & squaring both sides x 2 = r 2 y 2 /h 2 h dy y dv = πr 2 y 2 /h 2 times dy By integration, add up all the slices from the base to the peak of the cone, ie from y=0 to y=h. Constant = c. x r h Volume V = πr 2 /h 2 y 0 2.dy V = πr 2 /h 2 [y 3 /3 + c] h 0, Substitute h & 0 for y V= πr 2 /h 2.h 3 /3 = πr 2. h/3 For any conical body, V=Area of base*h/3

14 Volume of a Sphere by Integral Calculus x Volume of slice, dv = πx 2 * dy dy y r x 2 + y 2 = r 2 or x 2 = r 2 - y 2 & dv = π(r 2 - y 2 ) * dy Integrate to summate all slices from y = 0 to y = r V= π ( r2 - y 2 ) * dy = π*[(r 2 y - y 3 /3 +c) ] r 0 r 0 Substitute r for y & also 0 for y. V= π*[(r 3 r 3 /3) +c] π*[(r /3) +c] = π* (3r 3 - r 3 )/3 So V = 2πr 3 /3 for a hemisphere. For a sphere Volume = 4πr 3 /3.

15 Applications of Calculus (1) A farmer has 1000 m. of fencing and wishes to make a paddock to graze his stock. His donkeys require twice the area of the goats and the horses three times. What are the optimum sizes of the rectangular paddocks to maximize grazing area. Horses Donkeys Goats y 3x 2x x The total length of fencing, 1000 m =12x + 4y, so 4y = x or y = 250 3x (Eqn. 1) The total area A will be 6x*y, substitute for y, A = 6x*(250 3x) = 1500x -18x 2

16 Applications of Calculus (1 cont d) A = 1500x -18x 2 (Eqn. 2) From our differentiation exercises ; da/dx = 0 for maximum & minimum values. da/dx = *18x so x = 0 or x = 1500/36 = 250/6 Substituting in the linear equation (Eqn. 1); y = 250 3*250/6 = 250/2 Therefore the total perimeter is 750 m & the intermediate fences are each 250/2 m.

17 Applications of Calculus (2) A swimming pool is being filled at the rate of 1 cubic metre per minute. The cross section of the pool is as shown and its length is 10 m. At what rate is the water rising when the depth has reached 1m? 8 We need to find a relationship between the volume of water & the depth h. x h 2 The area of the yellow triangle is xh/2 and it is similar to the white triangle. So h/2 = x/1 and yellow area = (h/2)*(h/2). 6 Therefore the total area of the trapezium below h is; 2*h 2 /4 + 6h. The volume of water (V) below h is; 10(h 2 /2 + 6h) = 5h h If we differentiate V with respect to h; dv/dh = 10h + 60 But the rate of inflow, dv/dt = 1 m 3 /minute and dv/dt = (dv/dh)*(dh/dt) or 1 = (10h + 60)* dh/dt and dh/dt = 1/(10h + 60) So when h = 1 the water is rising at the rate of one seventieth of a metre/ min.

18 A Financial Application A manufacturer has established that the demand for a product is: p = 500/ n, where p is the price and n is the market demand at that price. The cost of manufacturing the item is C = 5n + 100* n in What are the marginal cost, the marginal income and marginal profit for 100 items? The marginal cost is the rate of change of the slope of the cost curve where n = 100 C = 5n + 100* n = 5n + 100*n ½ so dc/dn = 5 + (100/2)*n -½ or / n therefore the marginal cost at n = 100 is / 100 = 10 Income equals the number of items sold times the price. Income(R) = p*n = 500n/ n Rationalize; R = 500 n The marginal income is the rate of change of this curve, where n = 100 so dr/dn = ½*500* n -½ = 250/ n = 250/ 100 = 25 Profit(P) = Income Costs, ie P = R C or 500 n (5n + 100* n + 500) P = -5n n * 500 and dp/dn = -5 + ½*400/ n = / 100 for n = 100 The Marginal Profit is = 15 Marginal profit could have been calculated by deducting Marginal Cost from Income

19 That s it Folks!