Presentation on theme: "1 Derivatives: A First Look Average rate of change Instantaneous rate of change Derivative limit of difference quotients Differentiable implies continuity."— Presentation transcript:
1 Derivatives: A First Look Average rate of change Instantaneous rate of change Derivative limit of difference quotients Differentiable implies continuity Derivative notation
2 Different rates of change y=f(x) Average rate of changeAverage rate of change over [a,b]: m av = [f(b)-f(a)]/(b-a) = slope of secant line through (a, f(a)) and (b, f(b)) Instantaneous rate of changeInstantaneous rate of change at x=a: m tan =lim b a [f(b)-f(a)]/(b-a) = slope of tangent line through (a, f(a))
3 Tangent lines tangent line to the graph of f at PLet P=(x 0,y 0 ) be a point on the graph of f. Then the tangent line to the graph of f at P is the line through P with slope
4 Remarks and example Slope of tangent line important thing: m tan is (instantaneous) rate of change of f at x 0 ExampleExample: Find the equation of the tangent line to the graph of y=x 2 +3 at the point (2,7) ExampleExample: Do the same for y=2x+4 at the point (1,6)
6 Examples Compute the derivative of the constant function f(x) = c Compute the derivative of x 3 +x+13 Compute the derivative of f(x)=x 1/2 (square root of x) For last problem, what are the limits as x 0+ and x + of the derivative, and what do these limits say about graph of the square root function? Compute the derivative of f(x) = 1/x for x 0 Show that f(x) = x 1/3 is not differentiable at x = 0 Vertical tangent line for f(x) = x 1/3 at x=0
7 Differentiability f is differentiable at x.If lim h 0 [f(x+h)-f(x)]/h exists, say f is differentiable at x. Otherwise, f not differentiable there. Say f is differentiable on open interval if differentiable at every point ExampleExample: Where is f(x)=|x| differentiable?
8 Differentiability implies cont. If f is differentiable at x 0 then it’s continuous there Why? Need to see that lim h 0 f(x 0 +h)=f(x 0 ) :
10 Differential Equations: First Look Problem 1: Inflating a spherical balloon: Volume as a function of radius Consequences: When r large, V’(r) large When r small, V’(r) small
11 Two special cases Party and hot air balloon: r 0 = 1 and r 1 = 20 V(1) = 4 (1) 3 /3 4.19 cubic feet V'(1) = 4 (1) 2 12.57 cubic feet per foot V(20) = 4 (20) 3 /3 33510.32 cubic feet V'(20) = 4 (20) 2 5026.55 cubic feet per foot --- WOW!! See why balloonists use machines to inflate their balloons
12 Related calculation How much air is needed to increase radius by 3 inches for r 0 = 1 and r 1 = 20? First case: V 0 = V(r 0 + r) - V(r 0 ) = (4 (1 + 0.25) 3 /3) - (4 (1) 3 /3) = 4 ((1.25) 3 - 1 3 )/3 3.99 cubic feet Second case: V 1 =V(r 1 + r) - V(r 1 ) =(4 (20 + 0.25) 3 /3) -(4 (20) 3 /3) = 4 ((20.25) 3 - 20 3 )/3 1272.41 cubic feet
13 Average rate of changes of volumes Roughly same as instantaneous rates of change: V'(1) 12.57, and V'(20) 5026.55 Same conclusion as for the derivative of volume w.r.t. radius...when the radius is small, the rate of change (average or instantaneous) of volume with respect to radius is small. When the radius is large, so is the rate of change of volume.
14 Velocity and speed Let s(t) denote position of particle moving in straight line s/ t is the average velocity (average rate of change of s) and v(t)=ds/dt velocity (instantaneous rate of change) Suppose s(t)=2t – t 2. Graph s(t) Find a time where the velocity is zero Find an interval over which average velocity is zero Explain Difference between velocity v, and speed |v|
15 Population Growth N(t) = population at time t Instantaneous growth dN/dt Instantaneous per capita growth rate: N -1 dN/dt Assume constant per capita growth rate: dN/dt = rN Find the per capita growth rate Suppose N(0)= 347 and that r>0. Is the population size at time t=1 bigger or smaller than 347? Why?
16 Review Questions Where are the following functions differentiable?