Page 1 | © 2012 by Janice L. Epstein 2.7 Tangents Page 2 | © 2012 by Janice L. Epstein 2.7 Tangents Tangents, Velocities, & Other Rates of Change (2.7)
EXAMPLE 2
Find the equation of the tangent line to the curve at the given point.
1
, (1,1)
a) y =
x
4
Q(x,f(x))
Slope of the secant line PQ?
3
2
P(a, f( a))
1
x
1
2
3
4
5
The tangent line to the curve y = f ( x) at the point P (a, f ( a )) is
the line through P with slope
m = lim
x a
f ( x) - f (a)
f ( a + h) - f ( a )
= lim
h

0
x-a
h
EXAMPLE 1
a) Find the slope of the tangent line to the curve y = x 3 at (-1, -1)
using both definitions given above.
b) Find the equation of the tangent line.
c) Graph the curve and the tangent line
b) y =
x
, (0,0)
1- x
Page 3 | © 2012 by Janice L. Epstein 2.7 Tangents Page 4 | © 2012 by Janice L. Epstein 2.7 Tangents Let C be the graph of the vector function r (t ) = x (t ), y (t ) . The
tangent vector to the curve C at the point P corresponding to r (a)
is given by
1
1
r (t ) = lim
r (t ) - r ( a ) ] = lim [ r ( a + h) - r ( a ) ]
[
ta t - a
h 0 h
If the position of an object at time t is given by the function
s = f (t ) , then the velocity of the function at time t = a is
f ( a + h) - f ( a )
v ( a ) = lim
h 0
h
EXAMPLE 3
Let r (t ) = (t 2 - t - 2) i + 4t 2 j
a) Find a vector tangent to the curve given by the graph of r (t ) at
the point (-2,4) .
b) Find parametric equations for the tangent line to the curve at
the point (-2,4) .
EXAMPLE 4
The displacement (in meters) of a particle moving in a straight line
is given by s = t 2 - 8t +18 where t is measured in seconds.
a) Find the average velocity over the following time intervals
[3,4]
[3.5,4]
[4,5]
[4,4.5]
b) Find the instantaneous velocity when t = 4
c) Draw the graph of s as a function of t and draw the secant and
tangent lines from parts (a) and (b).
Page 5 | © 2012 by Janice L. Epstein 2.7 Tangents Page 6 | © 2012 by Janice L. Epstein 2.7 Tangents EXAMPLE 5
If an arrow is shot upward on the moon with a velocity of 58 m/s,
its height (in meters) after t seconds is given by
H (t ) = 58t - 0.83t 2
(a) Find the velocity of the arrow after 1 s.
(b) Find the velocity of the arrow when t = a
(c) When will the arrow hit the moon?
(d) With what velocity will the arrow hit the moon?
We definite the instantaneous velocity v ( a ) at a time t = a to be
1
1
v (a) = lim
r (t ) - r (a)] = lim [ r (a + h) - r (a)]
[
t a t - a
h0 h
EXAMPLE 6
If an arrow is shot upward on the moon with a velocity of
15i + 40 j m/s, its position after t seconds is given by
r (t ) = 15t i + (40t - 0.83t 2 ) j
(a) Find the velocity of the arrow after 1 s.
(b) Find the velocity of the arrow when t = a
(c) When will the arrow hit the moon?
(d) With what velocity will the arrow hit the moon?
Page 7 | © 2012 by Janice L. Epstein 2.7 Tangents Page 8 | © 2012 by Janice L. Epstein 2.7 Tangents The average rate of change of y = f ( x ) with respect to x is
f ( x2 ) - f ( x1 )
Dy
=
Dx
x2 - x1
EXAMPLE 8
The population P (in thousands) of a city from 1990 to 1996 is
given in the following table:
year 1990 1991 1992 1993 1994 1995 1996
105 110 117 126 137 150 164
P
The instantaneous rate of change of y = f ( x) with respect to x at
the point x = x1 , which is interpreted as the slope of the tangent to
the curve at the point P ( x1 , f ( x1 )) is
(a) Find the average rate of growth from
(i) 1992 to 1996
f ( x2 ) - f ( x1 )
Dx
= lim
Dx0 Dy
x2  x1
x2 - x1
lim
EXAMPLE 7
If a cylindrical tank holds 100,000 gallons of water, which can be
drained from the bottom of the tank in 1 hour, then Torricelli’s
Law gives the volume V of water remaining in the tank after t
minutes as
2
æ
t ö÷
ç
V (t ) = 100,000ç1- ÷÷ 0 £ t £ 60
çè 60 ø
Find the rate at which the water is flowing out of the tank after 20
minutes.
(ii) 1992 to 1995
(iii) 1992 to 1994
(iv) 1992 to 1993
(b) Estimate the instantaneous rate of growth in 1992 by measuring
the slope of a tangent.