STEYNING & DISTRICT U3A. Discovering Mathematics Session 20. Geometrical Problems & Applications of Calculus

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1 STEYNING & DISTRICT U3A Discovering Mathematics Session 20 Geometrical Problems & Applications of Calculus

2 Semicircle in a Unit Square What is the area of the largest semicircle which can be inscribed in a unit square. The figure shown is one obvious solution, where r = ½ r The semi-circular area = π*r 2 /2 = π/8 = However, it is possible to improve on that value. D Consider a semi- circle with the diameter angled across the square and forming a tangent to the other 2 sides of the square. For optimal area the diameter will form an angle of 45 o to the horizontal, and hence = r 2. If the angle is C < or > than 45 o, the figure will tend towards that above. So 2 2 = r 2 and = r/ 2 Y A r 45 o r r Z But AB = 1, therefore r + r/ 2 = 1 or r*( 1 + 1/ 2) = 1 rationalize, r( 2 + 1)/ 2 = 1 or r = 2/( 2 + 1) the area A = ½*π*[ 2/( 2 + 1)] 2 A = ½*π[2/(3 + 2* 2)] = π/( ) = Some 37% larger than the earlier. B

3 Paper Folding Fold a rectangular sheet of paper by placing opposite corners together. The long side of the paper is 30 cm and is equal in length to the fold. What is the length of the short side? The folded sheet will look like the trapezium, with the base equal to the top left & right edges. X W s The unfolded sheet is shown with a diagonal of length d. The fold of length l is shown in green. Using Pythagoras; d 2 = s 2 + l 2 The portions either side of the fold are identical and the fold is bisected at right angles by the diagonal. The triangles XYZ & WVZ are similar because they share 2 angles. Therefore s/l = l/2/d/2 or l 2 = s* (s 2 + l 2 ) Squaring; l 4 = s 2 *(s 2 + l 2 ) = s 4 + s 2 *l 2 or l 4 - s 2 *l 2 - s 4 = 0 Y V l = 30 Z If we substitute u for l 2 this can be treated as a quadratic. u 2 - us 2 - s 4 =0 Using the general quadratic equation; u= [s 2 +- (s 4 + 4s 4 )]/2 = [s 2 +- (5s 4 )]/2 Rationalizing; l 2 = [s 2 +-s 2 5] /2 or l 2 /s 2 = [1 + 5]/2 & s 2 = l 2 /[1 + 5)/2 Taking square roots; s = l/ (1 + 5)/2 = l/1.272 l = 30 cm. therefore s = 30/1.272 = cm.

4 Differential & Integral Calculus You will recall that, put very simply, we may say that : Differential Calculus consists of finding the value of very small elements, And that Integral Calculus is a means of summating the small elements into a whole. y dy/d = 0 In simple terms integration is the converse of differentiation. In the diagram the curve is an -y relationship. The slope of the curve at any point is dy/d & when dy = 0, then dy/d = 0. Thus we can make the general conclusion that the differential of any -y type relationship = 0 for Maimum and Minimum points of the relationship dy/dc = 0

5 Maimum Volume of a Bo What is the maimum volume of a square bo to be constructed from a 1 metre square card? If we assume that squares of side are cut from the sheet, the sides of the bo will be 1 2 metres. The volume V will be (1-2)*(1-2)*. 1 m 1 m Epanding this we get [ ]* or V = Differentiating; dv/d = But dv/d = 0 for Ma. & Minimum. So for maimum volume; = 0, 0r using the general quadratic equation. = [8 +- (64-4*12*1)]/24 = [ ]/24 = [8 +4]/24 or [8 4]/24 = 12/24 or 4/24 The results are ½ or 1/6, but cannot be ½ because this would give a bo with no base. Therefore the only feasible result is = 1/6 and side = 2/3 m So volume is 2/3 * 2/3 * 1/6 = 2/27 m 3

6 Add a Lid to the Bo If we add a tight fitting lid of 1 cm depth, What size of card is required? The length & width of the bo = 2/3 m Length of card side = 2/3 + 1/10 + 1/10 m. =2/3 + 1/5 = (10 + 3)/15 Thus card size = 13/15 metres square. 2/3 m 0.1 m

7 Relative Speed of Two Cars Two cars are travelling; one due North from a crossroad and one due West towards the crossroad. The first is travelling at 50 kmph and the second at 40 kmph. How fast is the distance between them changing when the first car is 0.3 km from the crossroad and the second car is 0.4 km from it. y kmph 40 kmph 0.4 z The speeds can be represented as differentials; dy/dt = 50 d/dt = -40 (Why negative?) dz/dt is unknown Apply Pythagoras to the right angle triangle; 2 + y 2 = z 2 Differentiate this with respect to time (t) 2*d/dt + 2y*dy/dt = 2z*dz/dt (Eqn. 1) Applying Pythagoras again; z 2 = = so z = 0.25 = The negative value may be discarded. Substitute these values into Eqn. 1 2*0.4*(-40) + 2*0.3*50 = 2*0.5*dz/dt = dz/dt or dz/dt = -2 Therefore at the instant described the distance between the cars is decreasing at 2 kmph.

8 Particle Velocity A particle accelerates through an electrical field such that; acceleration f = -20/(1 + 2t) 2 where t is time in secs. & f is in cm/sec 2 What is the equation for velocity with respect to time if v = 30 cm/sec when time t = 0 Acceleration is rate of change of velocity. Ie. f = dv/dt. therefore velocity v = f*dt, so v = -20* (1/(1 + 2t) 2 dt Substitute u for (1 + 2t) then du = 2*dt & dt = du/2. then v = -20* u -2 *du/2 So v = -10[-u -1 + C] = 10/(1 + 2t) + Constant When t = 0 v = 30, so 30 = 10 + C, so C = 20 Therefore the equation for velocity is v = [10/(1 + 2t) + 20] cm/sec.

9 A Vertical Projectile A projectile is shot vertically upwards at 30 m/s. What is its height after 5 secs. The downward force acting on the object is gravity with a value of 9.8 m/sec 2 The velocity, v = f*dt = -9.8*dt = -[9.8t + C]. At time = 0 v = 30, therefore C = 30 So the equation for velocity is v = -9.8*t +30 But distance travelled is s = v*dt = (-9.8t + 30)*dt = [- 4.9t t +C 1 ] However, when t = 0, s = 0, therefore C 1 = 0 so s = [- 4.9t t] Therefore at time t = 5 secs, s = -4.9* *5 = 27.5 m.

10 Pressure in Liquids If a body of area A is immersed in a liquid with density ρ (rho) to a depth y, the force on the area is F = da* ρ*y. where da = *dy. The force on the body will increase with depth, so we have to allow for this and the shape of the immersed body. y1 surface In the case shown; F= ρ y1 y2 (*y*dy) Eqn. 1 y2 dy y The base of the triangle is 1 m. the height is 2 m & y1 = 0.5 m. When = 0, y = y1 = 0.5 & when = 1, y = y2 = 2.5 m. to find the relationship between & y, assume y = m + c so 0.5 = m*0 + c & c = 0.5 also 2.5 = m* , so m = 2.5/0.5 = 5 therefore the equation is y = or = (y 0.5)/5 Substitute this value in eqn. 1; F = ρ/5 (y 0.5)*y*dy = ρ/5 (y 2 0.5*y)*dy Integrating; F = ρ/5[(y 3 /3 0.5*y 2 /2 + K] between the limits of y1 & y2 For water, ρ = 9800 N/m 3 ; F = 1960[(2.5 3 / /4) (0.5 3 / /4)] F = 1960[ ] = N.

11 Spring Compression A helical spring has a spring constant (k) of 15 N/m So to compress or stretch the spring by 5 cm would require a force F = k* = 15*5/10 = 7.5N Assuming a natural length of spring of 1 m. What is the energy required to compress it to 0.75 m Now Work done = Force * distance moved. Work done = k**d between the limits of 0 and 0.25 m. NB. The limits are measured from the natural length of the spring. Work done = 15*d = [15* 2 /2] = [7.5*(0.25) 2 7.5*(0) 2 ] = 7.5/16 Nm What additional work is required to compress it a further 0.2 m Work done = 15*d between the linits 0.25 & 0.45 = [7.5*(0.45) 2 7.5*(0.25) 2 ] but remember that a 2 b 2 = (a+b)*(a-b) so 7.5[( )*( )] = 7.5*0.2*0.7 = 1.05 Nm

12 Compound Interest A sum of 5000 (P) is invested at a rate of 6% compounded daily. How many years will it take to double the initial sum. n = Investment period in years, I = interest rate of 6% per annum The relevant relationship is; F = P[1 + i] n 10,000 = 5000*[ /365] n*365 2 = [( )/365] n*365 = [ ] n*365 Taking logarithms of both sides; log2 = 365n*log[ ] so 365n = log2/log[ ] = days and n = years

13 That s it folks!

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

STEYNING & DISTRICT U3A. Discovering Mathematics Session 12. Some geometrical problems & Applications of Calculus. STEYNING & DISTRICT U3A Discovering Mathematics Session 12 Some geometrical problems & Applications of Calculus. Calculate the Diagonal What is the length of the hypotenuse AC of the right angled triangle?