course notes2010 - MAT 1332 Calculus for Life Sciences A...

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MAT 1332: Calculus for Life Sciences A course based on the book Modeling the dynamics of life by F.R. Adler Supplementary material University of Ottawa Frithjof Lutscher, Jing Li and Robert Smith? April 13, 2010
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MAT 1332: Additional Course Notes 1 The inverse tangent function The tangent function is defined as tan( x ) = sin( x ) cos( x ) , and its derivative can be compute by the quotient rule as d dx tan( x ) = d dx sin( x ) cos( x ) - sin( x ) d dx cos( x ) cos 2 ( x ) = cos 2 ( x ) + sin 2 ( x ) cos 2 ( x ) = 1 cos 2 ( x ) . In particular, the function is defined for all x that are not odd multiples of π, and the function is monotone increasing, see Figure 1. The inverse of the tangent is denoted as arctan or tan - 1 and it is defined in the usual way as arctan(tan( x )) = x, tan(arctan( x )) = x. See Figure 1. What is its derivative? We differentiate the first equality above by the chain rule and find d dx [arctan(tan( x ))] = d dy arctan( y ) d dx tan( x ) = 1 , with y = tan( x ), since d dx x = 1 . Hence, we can divide d dy arctan( y ) = 1 d dx tan( x ) = cos 2 ( x ) cos 2 ( x ) + sin 2 ( x ) = 1 1 + sin 2 ( x ) cos 2 ( x ) = 1 1 + tan 2 ( x ) = 1 1 + y 2 . Application to integration We can now integrate the derivative of the arctan function to get Z 1 1 + x 2 dx = arctan( x ) + C. 1
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MAT 1332: Additional Course Notes 2 ! 5 ! 4 ! 3 ! 2 ! 1 0 1 2 3 4 5 ! 15 ! 10 ! 5 0 5 10 15 x y y=tan(x) x= ! ! /2 x= ! /2 ! 10 ! 5 0 5 10 ! 4 ! 3 ! 2 ! 1 0 1 2 3 4 x y y=arctan(x) y= ! /2 y= ! ! /2 Figure 1: Graphs of the tangent function and its inverse, the arctangent function 2
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MAT 1332: Additional Course Notes 3 Integrals and volumes First introductory example The E. coli bacterium has a rod-like shape with a cylinder in the middle and two half balls at the ends. What is its volume? The volume of a cylinder with radius r and height h is πr 2 h. The volume of a ball with radius r is 4 πr 3 / 3 . Now we only have to add the two volumes. For E. coli , the data are approximately r = 0 . 8 × 10 - 6 m and h = 2 × 10 - 6 m. This gives a volume of V = π (1 . 28+0 . 6825) × 10 - 18 m = 6 . 16 × 10 - 18 m. Second introductory example A termite mound is approximately cone shaped. What is its volume? If we imagine that we cut a cone into thin horizontal slices (perpendicular to its rotational axis), then each slice is approximately a cylinder. For each of these cylinders, we know how to compute the volume; and then we add the volumes. This procedure is very similar to the Riemann sums for the area under a curve, except that we are now using three-dimensional objects. But it gives us the right idea. General Idea To calculate the volume of an object with rotational symmetry, we need to know the diameter or radius at each point of its rotational axis. This gives the area of the cut surface at each point. Then we find the volume by integrating the area. See Figure 2. More precisely, assume that the rotational axis is the x -axis and the radius at each point is given by f ( x ) . Then the area of the cut surface at point x is A ( x ) = πf 2 ( x ) and the volume of the object between points a and b is given by Volume = V = Z b a A ( x ) dx = π Z b a ( f ( x )) 2 dx.
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