Slopes and Rates of Change If a particle is moving in a straight line at a constant velocity, then the graph of the function of distance versus time is as follows s s = f(t) t
s s t t = average velocity s s = f(t) The slope of this line is the velocity v, the rate of change of s with time. Clearly v = s / t is also the average velocity during any interval of time, and is the same at all times. It is the slope of the line. t
If the motion does not have constant velocity, then the average velocity will be different during different intervals of time. s = average velocity over interval t t This is the slope of the secant line. s s t s = f(t) t
s s = f(t) t 0 t As t approaches t 0, the corresponding point on the curve approaches the point (t 0, s 0 ), and the slope of the secant lines (the changing average rates of change) approach what we define to be the instantaneous rate of change, or instantaneous velocity. Clearly this should also be considered the slope of the curve at the point t 0.
s P s = f(t) P 0 Tangent Line to the curve at P 0. t 0 t It is intuitively clear that the limit of the secant lines through points P and P 0, as P moves along the curve toward P 0, is the tangent line to the curve through P 0. Thus the instantaneous slope of the curve at P 0 is also the slope of the tangent line through that point. This is essentially the definition of the tangent line through the curve at P 0.
This concept can be applied to any function. If f is a function and and y are the independent variable and dependent variable resp., then when changes by an amount, y changes by a corresponding amount y. The ratio f( ) f( 0) = y 0 is called the average rate of change of y with respect to over the interval.
graph of y = f() f( ) f( 0 ) = y y 0 f( ) f( 0 ) Slope of secant line is f( ) f( 0) = y 0
Definition. Let f() be defined in an open interval, and let 0 be in that interval. Then the limit () ( 0) lim y f f = lim 0 0 0 if it eists, can be interpreted as: (a) the slope of the the graph of f at 0. (b) the slope of the the tangent line to the graph of f at 0. (c) the instantaneous rate of change of f with respect to, And it is called the derivative of f with respect to at the point 0. Eample. If the independent variable is time, t and the dependent variable is distance traveled by a moving object, then the limit given above is called instantaneous velocity.
The Black bo view of derivative = 2 f() f( ) f( ) = y 2 y is the average rate of change of y with as we move from to 2 in inputs. The instantaneous rate of change of y with respect to at is the limit of this average rate as 2 gets every closer to.
Eample. Let y= f() = 2 2 (a) Find the average rate of change of y with respect to over the interval [3, 4]. (b) Find the instantaneous rate of change of y with respect to at 3. (c) Find the instantaneous rate of change of y with respect to at a general point 0. Solution: (a) The average rate of change over the interval is f (4) 2 (3) 2 2 2 (4) f(3) = = 6 9 = 7 4 3 2 2
(b) To get the instantaneous rate of change of y at 3, we need to take the limit of the average rate of change. Thus we need. 2 9 2 2 ( 3)( 3) + 2 lim f() f(3) = lim = lim 3 3 3 3 3 3 lim ( + 3) = 3 3 2 (c) To get the instantaneous rate of change of y at an arbitrary point 0, we again need to take the limit of the average rate of change. Thus we need. 2 ( ) 2 ( )( + ) 2 2 0 2 0 0 lim f() f( 0) = lim = lim 0 0 0 0 0 0 lim ( + ) = 2 0 0 0
Eample. Let y = f() = (a) Find the average rate of change of y with respect to over the interval [2, 3]. (b) Find the instantaneous rate of change of y with respect to at 2. (c) Find the instantaneous rate of change of y with respect to at a general point 0. Solution: (a) The average rate of change over the interval is f (3) f(2) = = 3 2 3 2 6
(b) To get the instantaneous rate of change of y at 2, we need to take the limit of the average rate of change. Thus we need. () (2) lim f f = lim lim 2 2 2 2 2 = 2 2 2 = lim = 2 2 4 [ 2 ] (c) To get the instantaneous rate of change of y at an arbitrary point 0, we again need to take the limit of the average rate of change. Thus we need. 0 lim f() f( ) lim lim 0 = 0 0 = 0 0 0 0 0 0 = lim = 0 0 2 0
f() = /. a b a b = slope of f at 2 = slope of tangent line to graph of f at 2.
Eercise. Let f() be given by the formula f() = (a) Find the slope of f() at a general point 0. (b) Use the result of part (a) to find the tangent line to the graph of f at 0 = 2. Solution. (a) The slope of the secant line is Thus the slope of the tangent line is: f() f( 0) = 0 0 0 + lim = lim + 0 0 0 0 0 0 0 0 0 = lim lim 0 = = 0 + 0 0 0 2 + 0
(b) The slope at 2 is therefore 2 2 Thus the tangent line runs through the point and has slope. 2 2 (2, 2) The equation is therefore y 2 = ( 2) or y= + ( 2 ) 2 2 2 2 2
Eercise. Let f() be given by the formula (a) Find the slope of f at a general point 0. f() = (b) Use the result of part (a) to find the equation of the tangent line to the graph of f at 0 = 4. Solution. (a) The slope of the secant line is f() f( 0) 0 = 0 ( 0) = 0 ( 0) 0 0 0+ ( 0 ) = = ( 0) 0 0+ ( 0) 0( 0+
= 0 ( 0 + Thus the slope at 0 is lim 0 = lim 0 0 0 0( 0 = 20 + 0 = ( = )( ) 3 2 0 20 2 3 2
6 (b) At 0 = 4, the slope is and the value of the function is ½. Thus the tangent line is: y = ( 4) or y= + 3 2 6 6 4
s 7 6 5 4 3 2 0 2 3 4 5 6 7 8 9 t Estimate the following quantities. (a) the average velocity over the interval 0 t 4 (b) the average velocity over the interval 0 t 6 (c) the instantaneous velocity at t = 7 (d) the values of t for which the instantaneous velocity is 0.
Problem. A particle moves on a line away from its initial position so that after t hours it is s = 3t 2 + t miles from its initial position. (a) Find the average velocity of the particle over the interval [, 3] (b) Find the instantaneous velocity at t =. Solution: (a) At t =, s = 4. At t = 3, s = 30. The average velocity is s (3) s () = 30 4 = 26 = 3 2 2 2 (b) Here we need to compute 2 3( t ) + t 3( t )( t+ ) + t lim st s = lim t t t t t t 2 () () 3 + 4 = lim = lim = lim3( t+ ) += 7 t t t t t