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

Cyclohexane Chair & Boat

Cyclohexane The applet shows boat & chair conformations of cyclohexane. Radio buttons are provided to switch between two conformations. Angle θ provides facility to rotate molecule about y axis and α is about x axis thus one can study full details of geometry. The angles between the bonds of cyclohexane are close to 109o 28', hence it is quite stable and nonreactive. Two most most important conformations for cyclohexane are chair & boat forms. Out of these two chair form is more stable than boat. The positions of Hydrogen atoms in chair form are at a greater distance than those in boat form. The flag pole interaction (repulsion) between the Hydrogen atoms in boat form leads to changing boat to chair form. Therefore cyclohexane exists mainly in chair form conformation. 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 Coordinate system – x axis is taken to right, y axis up and z coming out of screen towards viewer. Thus it is right hand coordinate system. Rotations about two axes are sufficient to view all the details of the 3D body. α is rotation about x axis such that y axis move towards z axis. And θ is rotation about y axis so that z moves towards x axis. Thus rotations follow right hand rule. Let there be a point P(X, Y, Z) in 3D. If this is viewed in 2D with both alpha & Theta to be zero, then point will appear at (X, Y) Due to θ rotation about y axis x = X*cos(θ) + Z*sin(θ) and y = Y, And then by θ & α rotations together x = X*cos(θ) +Z*sin(θ) & y = Y*cos(α) - Z*cos(θ)* sin(α). Let us plot the three +Ve axes. Let L be the length of the vector of axes. The Tip of x axis XT(L, 0, 0) YT(0, L, 0) & ZT = (0, 0, L) then xT is (L*cos(θ), 0) & yT is (0, L*cos(α) and zT is (L*sin(θ), - L*cos(θ)*sin(α). For the tetrahedron formed by Methane molecule: let bond length be B, Then the height of the tetrahedron H = 4*B/3. The angle made by Height with two sides appearing to be coincident in elevation γ = asin(1/3) = 19.47° and median of triangle in plan is M = (3/2)B*cos(γ) = 1.41 B Side of tetrahedron S = M/cos(30°) = 1.63 B. The angle made by top vertex with the vertex at right at base with center is obtained as δ = 2*asin(S/(2B)) = 109.47°. The angle at top vertex made by other two vertexes coinciding with each other in elevation and center is ϕ = (360-2δ)/2 = 125.26° Let rectangle formed by C2, C3, C5 & C6 lie in xz plane with origin at its center. Then x of C2 & C6 will be – B/2 and x of C3 & C4 will be B/2. C2, C3, C5 & C6 will have y to be zero. C2 & C3 will have z = -S/2 and C5 & C6 will have z = S/2. C1 & C4 have z = 0. Now to get x & y coordinates of C1, consider a tetrahedron with C2 at center and C3 other vertex to right in in elevation with C1 as one of the two vertexes coincident with each other. Hence C1 will have X = - B/2 - H/4, Y = M*sin(γ) & Z = 0. C4 is mirror reflection of C1.Thus matrix C(3 x 6) is generated. These are 3D coordinates of 6 carbon atoms. Now to get 2D coordinates one has to multiply T & C(Transpose). Hydrogen atoms H1 to H4 lie in x-y plane and hence their locations are found by considering a tetrahedron with C1 at center and C2, C6, H1 & H2 at vertexes. XH1 = xC1+ B cos(2ϕ – π) & yH1= yC1 + B sin(2ϕ – π). zH1 is obviously zero. H2 is aligned with x axis so xH2 = xC1 – B & yH2 = yC1. Obtaining coordinates of other Hydrogen atoms is bit complicated and is achieved as follows: Pass a plane through three known atoms draw two vectors of bond length from the central atom. The Hydrogen atoms lie in a plane which bisects this plane. The bisecting plane is located with the help of unit vector along two bond. Their resultant lies in plane and cross product lies ⊥ to plane. Now the Hydrogen atoms are at angle ϕ = 125.26° in bisecting plane. Dr Vasant Barve, 24 August 2014, Created with GeoGebra

Quadrilateral Family

Quadrilateral Family This applet shows the family tree of the quadrilaterals. Clicking the check boxes will display the properties. The vertexes marked red can be moved to see the variation of shape and size of the a quadrilateral. 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 Exclusive check boxes are used to display the description of properties. Red vertexes are independent object and hence can be moved to see the change in shape, size & orientation of a quadrilateral. For any queries write to vasantbarve@gmail.com Dr Vasant Barve, 20 April 2014, Created with GeoGebra

Straight Line in 3D

Straight Line in 3D P(x, y, z) & Q(x, y, z) are two points given is space. X, Y & Z axes are drawn. θ represents rotation of system about Y axis from X to Z. α represents rotation of of system about X axis from Y to Z. So if θ = α = 0 one sees X & Y ⊥ to each other and Z as a point. If θ = 90° and α = 0 then X & Z axes are seen with Y as point. If θ = 0 & α = 90° then one sees X & Z axes. A slider L is provided to change lengths of axes. A rectangular parallelepiped is constructed with P & Q as opposite vertexes. So PCFE & QBAD are faces ⊥ to X axis. large number of problems may be tackled by choice of n by a slider. α & θ are changed by sliders. 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 An array of coordinates Cor of set of points P & Q is written with 1st 3 columns for P & next 3 for Q. Axes tips and vertexes of rectangular parallelepiped are now defined in 3D as X3D, Y3D, P3D etc,  All Points are transformed to 2D coordinated    x = X cos(θ) - Z sin(θ) & y = Y cos(α) - Z sin(α) cos(θ) - X sin(α) sin(θ) These points are plotted and joined by relevant segments. Some faces are shown as colored polygons.   Required distances are calculated by    dPQ = sqrt[ (xQ - xP)² + (yQ - yP)² + (zQ - zP)² ] Dr Vasant Barve, 10 August 2014, Created with GeoGebra