Bacteria Growth

Bacteria Growth In a certain culture of bacteria, the rate of growth is proportional to the number present. If it is found that the number doubles in 4 hours, how long does it take to get the population 8 fold? Solution - The equation of growth of population where the growth depends on the population at that instant is Pt = P0 e^{kt} Here let the equation be f(x) = ae^{kx} Take a = 1 and take k as a variable. Adjust k to get P_t = 2 at time t = 4 for this mark line || to y axis through x = 4 and shift curve by changing k to get intersection of curve and this line at (4, 2) Alternately find k = (1/t) ln(Pt /P0) named k1 Having obtained k find t for y = 8 it is seen to be 12 hours This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com Solution - The equation of growth of population where the growth depends on the population at that instant is Pt = P0 e^{kt} Here let the equation be f(x) = ae^{kx} Take a = 1 and take k as a variable. Adjust k to get P_t = 2 at time t = 4 for this mark line || to y axis through x = 4 and shift curve by changing k to get intersection of curve and this line at (4, 2) Alternately find k = (1/t) ln(Pt /P0) named k1 Having obtained k find t for y = 8 it is seen to be 12 hours Address your queries to vasantbarve@gmail.com Dr Barve, 12 September 2014, Created with GeoGebra

Double Cusp

Double cusp It is common experience that milk pot shows beautiful double cusp. It is formed by parallel or nearly parallel rays. These rays get reflected on the cylindrical surface. The bundled rays shows white cusp pattern on the surface of milk. In this applet the rays in the form of vectors are shown when number of rays is small. Thereafter only lines are shown. This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com Construction – This is done in GeoGebra. A semicircular arc is drawn. Hatching is drawn by taking a sequence of points at 5o interval on arc and drawing small line segments at 45o . from these points. Sequence of number of rays starting from a point on right and meeting mirror is drawn. The normal to mirror is drawn at each point. The reflected rays are now drawn. Number of rays is taken to be small to start with, for user to see and understand the reflection. Number now is increased to get beautiful cusp pattern. Sequence command is back-bone of this construction. Address your queries to vasantbarve@gmail.com Dr Vasant Barve, 11 September 2014, Created with GeoGebra

Kaleidoscope

Kaleidoscope  This is a toy based on the principle of multiple reflections in a pair of mirrors. Three rectangular glass strips aspect ratio of about 6 are arranged to form an equilateral triangle by the lengths abutting each other. One end is closed by a triangular glass piece. One more triangular glass piece is arranged at small distance and some colored glass pieces and beads are placed in the gap of two triangular glass pieces. Now the arrangement is wrapped in paper to hold tight together. Viewed from other end one observes beautiful patterns. Slight movement changes the patterns. At three corners hexagonal patterns are formed This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com  Construction – One need consider object, three primary images, six secondary images and three tertiary images. Thus Object + 12 images are drawn. Beyond these one has images but are faint and hence not drawn. Start with a triangle containing objects. GeoGebra has inbuilt function for reflections of object about a line. Complete all the reflections. The objects have been provided with a control point to move the object. By moving object one can make pattern change from one beautiful to next. Address your queries to vasantbarve@gmail.com Dr Vasant Barve, 11 September 2014, Created with GeoGebra

Cylinder in a Cone

Cyliner in Cone There is a cone of radius Rc and height Hc. One has to fit a cylinder in it of radius Rcy. Find the ratio of Rc / Rcy for a) Maximum volume of cylinder & b) For maximum curved surface area of cylinder. Note – Observe the change in volume and surface area with change in height of the cylinder by animating Rcy and one may view cone and different angles by changing α This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com Solution - Let OH be Hcy the height of cylinder. ΔAOC & ΔFHC are similar ∴ CH/CO = Rcy/RC ∴ CH = Rcy*HC/Rc But CH = HC - Hcy ∴ Hcy = HC - Rcy*HC/Rc = HC(1 - Rcy/Rc) Curved surface of cylinder S = 2 π Rcy*Hcy putting value of Hcy = 2 π Hc Rcy(1 - Rcy/Rc) Differentiating S with repect to Rcy and equating to zero 1 - 2Rcy/Rc = 0 Cylinder Curved surface is maximum when Rcy = Rc/2 Height of cylinder is half of height of cone. Volume of cylinder V = π Rcy^2*Hcy putting value of Hcy = π Hcy Rcy^2(1 - Rcy/Rc) Differentiating V wit respect to Rcy and equating to zero 2Rcy - 3(Rcy)^2/Rc = 0 Volume of cylinder is maximum when Rcy = 2Rc/3 height of cylinder is 1/3 of the height of cone. For querries vasantbarve@gmail.com Dr Vasant Barve, 25 August 2014, Created with GeoGebra

Quadratic Equation Solver

Quadratic Equation Solver Quadratic Equation Solver – User may choose the constants a, b & c for the quadratic equation to be solved. Applet will show the graph and the roots of the equation along with suitable comment. User may try different combinations of a, b & c and see the graph. This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com General form of Quadratic equation is ax^2 + bx + c = 0. Values of x satisfying the Quadratic equation are known as roots. Graphical of Geometrical method to get roots: Plot the function y = ax^2 + bx + c. Cutting points of this quadratic equation are known as zeros of the function. x coordinates of the cutting points (or zeros) are known as the roots of the equation. The plot may cut x axis at two, one or none of the points i.e. the plot may be wholly up or down of the x axis. Following cases arise. Case 1: The graph cuts x axis at P(x1,0) & Q(x2,0) then quadratic equation has two real roots R1 = x1 & R2 = x2. Case 2: The graph cuts x axis at one point P(x1,0) then quadratic equation has two real roots which have same value i.e. R1 = R2 = x1. Case 3: The doesn't cut x axis. The quadratic equation has no real roots. It has imaginary roots. There are found by algebra. Algebraic Solution: Here the roots are R1, R2 = [ -b ± √(b2 - 4ac)]/(2a) Expression under root sign D = (b2 - 4ac) decides the nature of roots and hence is known as discriminant. Following special cases arise: Case 1: D = 0, R1 = R2 = - b/(2a) two roots are identical. X2 - 2x + 1 D = 0 R1 = R2 = 1 Case 2: D is +ve b = 0 R1 = √D & R2 = - √D Case 3: D is – ve Roots are complex as √D is imaginary Then R1 = (-b + √D I)/(2a) and R2 = (-b - √D I)/(2a) There are further sub-cases with different combinations of zero & nonzero values of a, b & c. Dr Barve, 30 August 2014, Created with GeoGebra