Chapter 4 Applications of the definite integral to rates, velocities and ...

Math 103 Notes Chapter 4Integrating the above, we getR(t) − R(0) =(1.5t − cos(πs/5) ) ∣ ∣∣∣t.π/50Thus, using the initial value, R(0) = r 0 (which is a constant), and evaluating the integral, leads toR(t) = r 0 + 1.5t − 5 cos(πt/5)π(The constant at the end of the expression stems from the fact that cos(0) = 1.) A graph of theradius of the tree over time (using r 0 = 1) is shown in Figure 4.5. This function is equivalent tothe area associated with the function shown in Figure 4.4. Notice that Figure 4.5 confirms that theradius keeps growing over the entire period, but that its growth rate (slope of the curve) alternatesbetween higher and lower values.+ 5 π .25201510502 4 6 8 10 12 14tFigure 4.5: Rate of change of radius, f(t) for a growing tree.After ten years we haveBut cos(2π) = 1, soR(10) = r 0 + 15 − 5 π cos(2π) + 5 π .R(10) = r 0 + 15.(b) The mass of the tree is density times volume, and since the density is taken to be 1 gm/cm 3 ,we need only obtain the initial and final volume. Taking the trunk to be cylindrical means that thevolume at a given time isV (t) = πR(t) 2 h.v.2005.1 - January 5, 2009 10