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Lecture 9: Linear and Quadratic Approximations
Unit 2: Applications of Differentiation
Today, we’ll be using differentiation to make approximations.
Linear Approximation
y=f(x)
y y = b+a(x-x 0 )
x
b = f(x 0 ) ;
x 0 ,f(x 0 ( ))
a = f’(x 0 )
Figure 1: Tangent as a linear approximation to a curve
The tangent line approximates f (x). It gives a good approximation near the tangent point x 0.
As you move away from x 0 , however, the approximation grows less accurate.
f (x) ≈ f (x 0 ) + f
� (x 0 )(x − x 0 )
Example 1. f (x) = ln x, x 0 = 1 (basepoint)
f (1) = ln 1 = 0; f � (1) = = 1 x (^) x=
ln x
Change the basepoint:
Basepoint u 0 = x 0 − 1 = 0.
≈ f (1) + f � (1)(x − 1) = 0 + 1 · (x − 1) = x − 1
x = 1 + u =⇒ u = x − 1
ln(1 + u) ≈ u
Basic list of linear approximations
In this list, we always use base point x 0 = 0 and assume that |x| << 1.
- sin x ≈ x (if x ≈ 0) (see part a of Fig. 2)
- cos x ≈ 1 (if x ≈ 0) (see part b of Fig. 2)
x
- e ≈ 1 + x (if x ≈ 0)
- ln(1 + x) ≈ x (if x ≈ 0)
- (1 + x) r ≈ 1 + rx (if x ≈ 0)
Proofs
Proof of 1: Take f (x) = sin x, then f � (x) = cos x and f (0) = 0
f � (0) = 1 , f (x) ≈ f (0) + f � (0)(x − 0) = 0 + 1.x
So using basepoint x 0 = 0, f (x) = x. (The proofs of 2, 3 are similar. We already proved 4 above.)
Proof of 5:
f (x) = (1 + x) r ; f (0) = 1
f � (0) =
d (1 + x) r x=0 =^ r(1^ +^ x)
r− 1 x=0 =^ r dx
f (x) = f (0) + f � (0)x = 1 + rx
y = x
sin(x)
y=
cos(x)
(a) (b)
Figure 2: Linear approximation to (a) sin x (on left) and (b) cos x (on right). To find them, apply f (x) ≈ f (x 0 ) + f �(x 0 )(x − x 0 ) (x 0 = 0)
e − 2 x
Example 2. Find the linear approximation of f (x) = √ near x = 0. 1 + x
We could calculate f �(x) and find f �(0). But instead, we will do this by combining basic approxi
mations algebraically.
u e − 2 x ≈ 1 + (− 2 x) (e ≈ 1 + u, where u = − 2 x)
me
satellite
(with velocity v)
Figure 3: Illustration of Example 4: a satellite with velocity v speeding past “me” on planet Quirk.
Here, T � is the time I measure on my wristwatch, and T is the time measured onboard the satellite. � 2
2
2
v 1 v v 1 T � = T 1 − c^2
2 c^2
(1 + u) 4 ≈ 1 + ru, where u = − c^2
, r = − 2
2
If v = 4 km/s, and the speed of light (c) is 3 × 10 5 km/s,
v ≈ 10 − 10
. There’s hardly any difference c^2 between the times measured on the ground and in the satellite. Nevertheless, engineers used this very
approximation (along with several other such approximations) to calibrate the radio transmitters
on GPS satellites. (The satellites transmit at a slightly offset frequency.)
Quadratic Approximations
These are more complicated. They are only used when higher accuracy is needed.
f (x) ≈ f (x 0 ) + f � (x 0 )(x − x 0 ) +
f �� (x 0 ) (x − x 0 ) 2 (x ≈ x 0 ) 2
Geometric picture: A quadratic approximation gives a best-fit parabola to a function. For
example, let’s consider f (x) = cos(x) (see Figure 4). If x 0 = 0, then f (0) = cos(0) = 1, and
f � (x) = − sin(x) =⇒ f � (0) = − sin(0) = 0
f �� (x) = − cos(x) =⇒ f �� (0) = − cos(0) = − 1
1 1 cos(x) ≈ 1 + 0 · x − 2
x 2 = 1 − 2
x 2
You are probably wondering where that in front of the x 2 term comes from. The reason it’s 2 there is so that this approximation is exact for quadratic functions. For instance, consider
f (x) = a + bx + cx 2 ; f � (x) = b + 2cx; f �� (x) = 2c.
Set the base point x 0 = 0. Then,
f (0) = a + b 0 + c 0 2 · · =⇒ a = f (0)
f � (0) = b + 2c 0 = b = b = f � · ⇒ (0)
f �� (0) f
�� (0) = 2 c =⇒ c = 2
cos(x)
y
x
1- x
2 /
Figure 4 : Quadratic approximation to cos(x).
0.0.1 Basic Quadratic Approximations
f (x) ≈ f (0) + f � (0)x +
f ��
2
x 2 (x ≈ 0)
- sin x ≈ x (if x ≈ 0)
2 x
- cos x ≈ 1 − 2
(if x ≈ 0)
- e x ≈
1 + x + x 2 (if x ≈ 0) 2
- ln(1 + x) ≈ x −
x 2 (if x ≈ 0) 2
- (1 + x) r ≈ 1 + rx +
r(r − 1) x 2 (if x ≈ 0) 2
Proofs: The proof of these is to evaluate f (0), f � (0), f �� (0) in each case. We carry out Case 4
f (x) = ln(1 + x) = ⇒ f (0) = ln 1 = 0 1 f � (x) = [ln(1 + x)]
� = = f � (0) = 1 1 + x
f �� (x) = 1 +
� − 1
x
(1 + x)^2
=⇒ f �� (0) = − 1
Let us apply a quadratic approximation to our Planet Quirk example and see where it gives.
�
1 −
v
c^2
2
1 v
c^2
2
− 2
1 )( −
2
1 − 1)
v
c^2
2
Case 5 with x =
c
v 2
2 , r = − 2
Lecture 10: Curve Sketching
Goal: To draw the graph of f using the behavior of f �^ and f ��. We want the graph to be
qualitatively correct, but not necessarily to scale.
Typical Picture: Here, y 0 is the minimum value, and x 0 is the point where that minimum occurs.
x 0
= critical point
y
0
Figure 1: The critical point of a function
Notice that for x < x 0 , f �(x) < 0. In other words, f is decreasing to the left of the critical point.
For x > x 0 , f �(x) > 0: f is increasing to the right of the critical point.
Another typical picture: Here, y 0 is the critical (maximum) value, and x 0 is the critical point. f
is decreasing on the right side of the critical point, and increasing to the left of x 0.
x 0
= critical point
y 0
f’(x) < 0
x > x 0
Figure 2: A concave-down graph
Rubric for curve-sketching
- (Precalc skill) Plot the discontinuities of f — especially the infinite ones!
- Find the critical points. These are the points at which f � (x) = 0 (usually where the slope changes from positive to negative, or vice versa.)
- (a) Plot the critical points (and critical values), but only if it’s relatively easy to do so.
(b) Decide the sign of f �(x) in between the critical points (if it’s not already obvious).
- (Precalc skill) Find and plot the zeros of f. These are the values of x for which f (x) = 0. Only do this if it’s relatively easy.
- (Precalc skill) Determine the behavior at the endpoints (or at ±∞).
Example 1. y = 3x − x 3
- No discontinuities.
- y�^ = 3 − 3 x^2 = 3(1 − x^2 ) so, y�^ = 0 at x = ±1.
- (a) At x = 1, y = 3 − 1 = 2.
(b) At x = −1, y = −3 + 1 = −2. Mark these two points on the graph.
- Find the zeros: y = 3x − x^3 = x(3 − x^2 ) = 0 so the zeros lie at x = 0, ±
- Behavior of the function as x → ±∞. As x → ∞, the x 3 term of y dominates, so y → −∞. Likewise, as x → −∞, y → ∞.
Putting all of this information together gives us the graph as illustrated in Fig. 3)
(-√3,0)
(√3,0)
Figure 3: Sketch of the function y = 3x − x^3. Note the labeled zeros and critical points
Let us do step 3b (the sign of f � ) to double-check for consistency.
y � = 3 − 3 x 2 = 3(1 − x 2 )
y � > 0 when |x| < 1; y � < 0 when |x| > 1. Sure enough, y is increasing between x = − 1 and x = 1,
and is decreasing everywhere else.
Example 3. y = x^3 − 3 x^2 + 3x.
y � = 3x 2 − 6 x + 3 = 3(x 2 − 2 x + 1) = 3(x − 1) 2
There is a critical point at x = 1. y�^ > 0 on both sides of x = 1, so y is increasing everywhere. In
this case, the sign of y�^ doesn’t change at the critical point, but the graph does level out (see Fig. 6.
1
1
horizontal slope
(1,1)
Figure 6: Graph of y = y = x^3 − 3 x^2 + 3x
ln x Example 4. y = (Note: this function is only defined for x > 0) x
What happens as x decreases towards zero? Let x = 2 −n
. Then,
ln 2−n y = 2 −n^
= (−n ln 2) n → −∞ as n → ∞
In other words, y decreases to −∞ as x approaches zero.
Next, we want to find the critical points.
y � =
ln x
�
x( (^) x 1 ) − 1(ln x) =
1 − ln x
x x^2 x^2
y
� = 0 =⇒ 1 − ln x = 0 =⇒ ln x = 1 =⇒ x = e
In other words, the critical point is x = e (from previous page). The critical value is
ln e 1 y(x) |x=e = e
e
Next, find the zeros of this function:
y = 0 ⇔ln x = 0
So y = 0 when x = 1.
What happens as x → ∞? This time, consider x = 2 +n .
ln 2n^ n ln 2 n(0.7) y = = 2 n^2 n^
2 n
So, y → 0 as n → ∞. Putting all of this together gets us the graph in Fig. 7.
1 e
1/e
(e,1/e)
Figure 7: Graph of y = ln x^ x
Finally, let’s double-check this picture against the information we get from step 3b:
y � =
1 − ln x > 0 for 0 < x < e x^2
Sure enough, the function is increasing between 0 and the critical point.
The points where f ��^ = 0 are called inflection points. Usually, at these points the graph changes
from concave up to down, or vice versa. Refer to Fig. 10 to see how this looks on Example 1.
Inflection point (where f” = 0)
Figure 10: Inflection point: y = 3x − x^3 , y��^ = − 6 x = 0, at x = 0.
Lecture 11: Max/Min Problems
Example 1. y =
ln x (same function as in last lecture) x
x 0
=e
1/e
Figure 1: Graph of y =
ln x . x
- What is the maximum value? Answer: y =. e
- Where (or at what point) is the maximum achieved? Answer: x = e. (See Fig. 1).)
Beware: Some people will ask “What is the maximum?”. The answer is not e. You will get so used
to finding the critical point x = e, the main calculus step, that you will forget to find the maximum
1 1 value y =. Both the critical point x = e and critical value y = are important. Together, they e e 1 form the point of the graph (e, ) where it turns around. e
Example 2. Find the max and the min of the function in Fig. 2
Answer: If you’ve already graphed the function, it’s obvious where the maximum and minimum
values are. The point is to find the maximum and minimum without sketching the whole graph.
Idea: Look for the max and min among the critical points and endpoints.You can see from Fig. 2
that we only need to compare the heights or y-values corresponding to endpoints and critical points.
(Watch out for discontinuities!)
- Use the constraint equation to express everything in terms of r (and the constant V ).
h =
V
; S = πr 2
V
2 πr πr^2
- Find the critical points (solve dS/dr = 0), as well as the endpoints. S will achieve its max and min at one of these places.
dS 2 V 3 V
V
dr
= 2πr − r^2
= 0 =⇒ πr 3 − V = 0 =⇒ r = π
=⇒ r = π
We’re not done yet. We’ve still got to evaluate S at the endpoints: r = 0 and “r = ∞”.
2 V
S = πr 2
2 As r → 0, the second term, r
, goes to infinity, so S → ∞. As r → ∞, the first term πr 2 goes
to infinity, so S → ∞. Since S = +∞ at each end, the minimum is achieved at the critical point
r = (V /π) 1 / 3 , not at either endpoint.
s
r
to ∞
to ∞
Figure 4: Graph of S
We’re still not done. We want to find the minimum value of the surface area, S, and the values
of h. � � 1 / 3 � �− 2 / 3 � � 1 / 3 V V V V V V r = π
; h = πr^2
π
V
� 2 / 3 =^
π π
π π � � 2 / 3 � � 1 / 3
S = πr 2
V
= π
V
+ 2V
V
= 3π − 1 / 3 V 2 / 3 r π π
Finally, another, often better, way of answering that question is to find the proportions of the
can. In other words, what is
h
r
? Answer:
h
r
(V /π) 1 / 3
(V /π)^1 /^3
Example 4. Consider a wire of length 1, cut into two pieces. Bend each piece into a square. We
want to figure out where to cut the wire in order to enclose as much area in the two squares as
possible.
(1/4)x
(^0) x 1
(1/4)(1-x)
Figure 5: Illustration for Example 5.
x x^2 The first square will have sides of length. Its area will be. The second square will have � � 2 4 16 sides of length 1 − 4
x (^). Its area will be 1 − 4
x (^). The total area is then
x
� 2 �^
1 − x
A = +
A
�
x
− x) (−1) =
x
8
x
8
= 0 =⇒ 2 x − 1 = 0 =⇒ x =
So, one extreme value of the area is
A =
2
2
4 4 32
We’re not done yet, though. We still need to check the endpoints! At x = 0,
A = 0
2
At x = 1, � � 2 1 1 A = + 0 2 = 4 16
Lecture 12: Related Rates
Example 1. Police are 30 feet from the side of the road. Their radar sees your car approaching at
80 feet per second when your car is 50 feet away from the radar gun. The speed limit is 65 miles
per hour (which translates to 95 feet per second). Are you speeding?
First, draw a diagram of the setup (as in Fig. 1):
Road
Car
Police
30 D=
x
Figure 1: Illustration of example 1: triangle with the police, the car, the road, D and x labelled.
Next, give the variables names. The important thing to figure out is which variables are changing.
dD At D = 50, x = 40. (We know this because it’s a 3-4-5 right triangle.) In addition, = D � = dt −80. D�^ is negative because the car is moving in the −x direction. Don’t plug in the value for D
yet! D is changing, and it depends on x.
The Pythagorean theorem says 30 2
Differentiate this equation with respect to time (implicit differentiation:
d �^2 �^2 DD � 30
2
2 = 2 xx
� = 2DD
� = x
�
dt
2 x
Now, plug in the instantaneous numerical values:
50 feet x � = 40
s
This exceeds the speed limit of 95 feet per second; you are, in fact, speeding.
There is another, longer, way of solving this problem. Start with
D = 302 + x^2 = ( 2
d 1 dx D = ( 2
- x 2 ) − 1 / 2 (2x ) dt 2 dt
Plug in the values:
1 dx −80 = ( 2
- 40 2 ) − 1 / 2 (2)(40) 2 dt
and solve to find dx feet = − 100 dt s
(A third strategy is to differentiate x =
D^2 − 302 ). It is easiest to differentiate the equation in its
simplest algebraic form 302 + x^2 = D^2 , our first approach.
The general strategy for these types of problems is:
- Draw a picture. Set up variables and equations.
- Take derivatives.
- Plug in the given values. Don’t plug the values in until after taking the derivatives.
Example 2. Consider a conical tank. Its radius at the top is 4 feet, and it’s 10 feet high. It’s being
filled with water at the rate of 2 cubic feet per minute. How fast is the water level rising when it is
5 feet high?
h
r
Figure 2: Illustration of example 2: inverted cone water tank.
From Fig. 2), the volume of the tank is given by
V = πr
2 h 3
