Pg 869 #1, 5, 9, 12, 13, 15, 16, 18, 20, 21, 23, 25, 30, 32, 34, 35, 37,
40, 42
16.3 Tangent to a Curve
What if you were asked to find the slope of a curve?
Could you do this? Does it make sense?
(No, not really, slopes are for lines, they are straight, curves
might not be straight)
So, what if I told you this is exactly what we are going to do!
We will be utilizing limits!!
Let’s think about geometry for a second.
A secant was a line that intersected a circle at two points.
A tangent was a line that intersected a circle at just one point.
Let’s extend this to a curve – any curve!
Secant Line to a Curve on Desmos
(Slide the dot on the right slowly towards the dot on the left)
The secant line becomes a tangent line!
So, we want to make the secant become the tangent.
Q
(x, f (x)) What is the slope of
(c, f (c))
P
c
x
f ( x )  f (c )
PQ 
xc
We want Q to get closer to P
So x needs to get closer to c
f ( x )  f (c )
lim
xc
xc
this is the
difference quotient!!
Ex 1) Find the slope of the line tangent to the curve f ( x) 
at P(5, 3)
any other
Draw your own sketch
(5, 3)
point
(x, f (x))
slope
f ( x)  f (5)
x  4 3  x  4 3
m  lim
 lim
x5
x5
x 5
x 5
 x4 3

x  4   32
x  49
 lim
 lim
x5 
x  5   x  4  3 x5  x  5   x  4  3
 lim
x5
2
x 5
 x  5 
x4
We can’t
substitute 5 in,
so algebra to
work!
1
1
1
1
 lim



9 3 33 6
x  4  3 x5 x  4  3
*now we plug in 5*
If we know the slope of the tangent line, we can write the equation of the tangent line.
Ex 2) Find equation of tangent line to f (x) = x3 – 2x2 + 3 at P(1, 2).
*to find equation of a line, we need two things: (1) slope (2) point
f ( x)  f (1)
x3  2 x 2  3  2
x3  2 x 2  1
m  lim
 lim
 lim
x1
x

1
x1
x 1
x 1
x 1
( x  1)( x 2  x  1)
 lim
 lim( x 2  x  1)  1  1  1  1
x1
x1
x 1
m = –1
P(1, 2)
y – 2 = –1(x – 1)
*you can leave like this –
that is what calculus does
So back to an original question – how to find slope of a curve…
the slope of a curve at point P
the slope of the tangent at point P
The slope of a curve might vary from point to point, so it is helpful to be
able to represent it in generic form using an arbitrary point. Then we can
use it with specific slope values.
Ex 3) Find equation of line with slope 5 tangent to the graph of
f (x) = x2 + 3x – 1
This time we have slope, but not the point
general terms: f (x) and f (c)
 x 2  3x  1   c 2  3c  1
f ( x )  f (c )
m  lim
 lim
x c
x c
xc
xc
x 2  3 x  c 2  3c
x 2  c 2  3x  3c
 lim
 lim
x c
x c
xc
xc
( x  c)( x  c)  3( x  c)
( x  c)  x  c  3
 lim
 lim
x c
x c
xc
xc
 lim( x  c  3)  c  c  3  2c  3  m
x c
want 2c + 3 = 5
2c = 2
c=1
point (1, ? )
f (1) = 1 + 3 – 1 = 3
(1, 3)
y – 3 = 5(x – 1)
Physical quantities can also be found using the idea of a secant becoming a tangent.
Average rates are similar to secants (slope of line)
Instantaneous rates are similar to tangents (limit of slope of line)
Let’s look at velocity (a rate!)
A position function f (t) describes the path something takes.
Ex 4) The motion of an object is given by the function f (t) = t2 – 3t + 5
where f (t) is height of object in feet at time t seconds.
a) What is the average velocity of the object between t = 2 s and t = 4 s?
f (4)  f (2) (16  12  5)  (4  6  5) 6
slope of

  3 ft/s
secant
42
2
2
b) What is the instantaneous velocity of the object at time t = 2 s?
2
2
f
(
t
)

f
(2)
t

3
t

5

3
t
 3t  2
slope of
m  lim
 lim
 lim
t

2
t

2
t 2
t 2
t 2
t 2
tangent
f (t) & f (2)
(t  2)(t  1)
 lim
 lim(t  1)  2  1  1 ft/s
t 2
t 2
t 2
Homework
Pg 869 #1, 5, 9, 12, 13, 15, 16, 18, 20, 21, 23, 25, 30, 32,
34, 35, 37, 40, 42