Math 217
Midterm 1 review
02/28/ 2014
Jiyuan Han
content
1 Tricks of derivative
2 Graph and Max-Min problem
3 Optimization problem
4 Logarithm
reviewing material (with priority)
1 Practice problems (with solution) on Prof. Miller’s website;
2 Homework questions;
3 lecture notes and book;
4 quiz;
formula you might need:
Derivative formula
product rule: (f · g)0 = f 0 · g + f · g 0
quotient rule:
f 0 f 0 · g − f · g0
=
g
g2
chain rule: f (u(x))0 = f 0 (u(x)) · u0 (x)
power rule (xn )0 = n · xn−1
derivative of inverse function (f −1 (y))0 =
1
f 0 (x)
(a · f (x))0 = a · f 0 (x)
sin0 (x) = cos(x)
cos0 (x) = −sin(x)
1
x
1
tan0 (x) =
cos2 (x)
ln0 (x) =
arctan0 (x) =
1
1 + x2
(Here (y = f (x)))
trignomeric formula
sin2 (θ) + cos2 (θ) = 1
sin(2x) = 2sinx · cosx
cos(2x) = 2cos2 (x) − 1
volume and area formua
area of a disk: S = π · r2
area of a cylinder (with top and bottom disks): S = 2π · r2 + 2π · rh
volume of a cylinder: V = πr2 · h
4π 3
volume of a ball: V =
r
3
π
volume of a cone: V = r2 · h
3
perimiter of a circle: c = 2π · r
circular arc with angle θ: c = θ · r
Parametrized curves
y 0 (t)
dy
= 0
dx
x (t)
κ=
y 00 (t) · x0 (t) − x00 (t) · y 0 (t)
3
(x0 2 (t) + y 0 2 (t)) 2
q
s (t) = x0 2 (t) + y 0 2 (t) where s is the length of the curvce
0
R=
1
where R is the radius of curvature
κ
Logarithm formula
notation: log(a) = log2 (a); ln(a) = loge (a);
loga (b · c) = loga (b) + loga (c)
aloga (b) = b ; loga (ab ) = b
abc = (ab )c
ab+c = ab · ac
ln(b)
loga (b) =
ln(a)
1
loga (b) =
logb (a)
loga (bc ) = c · loga (b)
1
logac (b) = · loga (b)
c
Tricks of derivative
Derivative just tells you how fast does the function value change.
Definition: the derivative of the function f (x) at x = a is:
f (x) − f (a)
x→a
x−a
(You can interpret it in this way: consider the function f (t) is the distance, t is the time, then
f 0 (t) is just the speed.
f 0 (a) = lim
Some examples:
(1)
sin0 (x2 ) = cos(x2 ) · 2x
(2) Let sin(y) + cos(x) + x2 = 10. Find y 0 (x).
ans:
Here you have a implicit function. So we differentiate the equation directly:
(sin(y) + cos(x) + x2 )0 = (10)0
cos(y) · y 0 (x) − sinx + 2x = 0
sinx − 2x
y0 =
cosy
(3). chain rule:
√
1 + u, u = x2
√
1
x
y 0 = ( 1 + u)0 · u0 = √
· 2x = √
2 1+u
1 + x2
y=
Intermidiate Value Theorem: If f is a continuous function defined on an interval a ≤ x ≤ b,
and if y is any number between f (a) and f (b), then there is a number c with a ≤ c ≤ b such
that f (c) = y.
Exercise:
1
(sin0 (x2 ); cos0 (x2 )
2
p
( 1 − sin2 (x))0
3
4 Let y = u3 + 1, u = 3x + 7. Find
√
(tan( 1 + x2 ))0
dy dy
, .
dx du
1
5 Use the method of implicit differentiation to find f 0 (x), where f (x) = (1 − x4 ) 4 , −1 ≤
x ≤ 1.
(Check the answer on textbook page 81, example 15.3)
Graph and Max-Min problem
It’s impossible for us to draw a graph as precisely as a computer does. However, we still want
our ”coarse graph” to show some important properties of the function. For example, we want
to know if the function is increasing or decreasing; when the function attains its maxima or
minima values; if the function is concave or convex; if there exist any asymptote (vertical, horizontal, slant); the intersection points with the x, y-axis. These properties should be enough
to gurantee us to draw a pretty nice graph.
**Check those examples in your textbook in section 6. There are many good pictures.
monotunicity
The derivative tells you if the function is increasing or decreasing.
f(x) is increasing(strictly) if and only if f 0 (x) ≥ 0(f 0 (x) > 0);
f(x) is decreasing(strictly) if and only if f 0 (x) ≤ 0(f 0 (x) < 0);
the point where f 0 (x) = 0 is called the stationary point;
Mean value theorem
If f is a differentiable function on the interval a ≤ x ≤ b, then there is some number c with
a < c < b such that:
f (b) − f (a)
f 0 (c) =
b−a
Maxima and Minima
A function has a global maximum at some a in its domain if f (x) ≤ f (a) for all other x in the
domain of f . Global maxima are sometimes also called ”absolute maxima”.
A function has a global minimum at some a in its domain if f (x) ≥ f (a) for all other x
in the domain of f . Global miniima are sometimes also called ”absolute miniima”.
A function has a local maximum at some a in its domain if there is a small δ > 0 such
that f (x) ≤ f (a) for all a − δ < x < a + δ that lie in the domain of f .
A function has a local minimum at some a in its domain if there is a small δ > 0 such
that f (x) ≥ f (a) for all a − δ < x < a + δ that lie in the domain of f .
Remark:
1. A local minimum or a local maximum point is necessary to be a stationary point (where
f 0 (x) = 0); however, a stationary point is not necessary to be a local maximum point or a local
minimum point! (Think of the function y = x3 at the point x = 0). So if you want to find
local max(mini)mum points, you need to check the derivative in the left and right of that point!
2. The global mini(max)mum point is not necessary to be a stationary point. Sometimes
it could be among the boundary points. (Think of the function: y = x2 , −2 ≤ x ≤ 2).
3. Besides the way in remark 1, here is another way to check local max(mini)mum point:
Suppose c is a stationary point for a function f . If f 00 (c) < 0, then f has a local maximum at x = c, and if f 00 (c) > 0, then f has a local minimum at c.
concavity
Similiar to the monotunicity, the second derivative tells you if the function is convex or concave.
f(x) is convex if and only if f 0 (x) ≥ 0(f 00 (x) > 0);
f(x) is concave if and only if f 0 (x) ≥ 0(f 00 (x) < 0);
the point where f 00 (x) = 0 is called the inflection point;
Exercise:
1 Problem 14, page 104; Problem 20 page 104.
The answer is on my webpage: answer to HW2
2 Review Homework question: Problem 3, Page 89. It remainds you of how to find the
normal line:
Where does the normal line to the graph of y = x2 at the point (a, a2 ) intersect the
x-axis?
3 Question 40, Page 104. This question gives you experience of how to draw periodic function.
Sketch the graph of the following function as in the previous problem, but only list those
“interesting points” that lie in the interval 0 ≤ x ≤ 2π.
y = 2sinx + sin2 (x)
4 Problem 54, Page 105. Show that the equation 3y + siny = π2 has exactly one solution
y.
A similar question has shown up in our Quiz 2, problem 2. Check the answer on my
webpage.
Optimization problem
For the applied optimization problem, you will given with a real scenario, and then asked to
find max(mini)ma of a certain variable.
Here is one format to help you solve applied optimization problem by steps:
1. Object. Object is the variable you want to max(mini)mize. (e.g, if you want to find
the largest area, then the area is the object) You should find the formula you need to calculate
your object. (e.g, if the object is the volume of the cone, then you have a formula to calculate
the volume: V = π3 r2 · h)
2. Constraints. Constraints are those relations and conditions you are given. (e.g, the domain
of the variables, the equation between different variables) You need to use these relations to
simplify your object formula.
3. Differentiate your object formula. And find those stationary points.
4. Check your stationary points if they are really max(mini)ma points. Sometimes, you also
need to check the boundary points.
Here is one example:
An open-top cylindrical tank with a volume of 10 cubic feet is to be mad from a sheet of
steel. Find the dimensions of the tank that will require as little material used in the tank as
possible.
ans:
1. First we need to find what is the object. The question asks to find the minimal material that is needed. So the object is the area of the cylinder without top but with the bottom.
So we can build the formula for the area S:
Let the radius of the cylinder be r; the height of the cylinder be h;
The area of the bottom disk is: π · r2
The area of the vertical surface is: 2π · r · h
So we have the area formula: S = π · r2 + 2πr · h.
2. Now we shall find the constraints.
Of course, this is a real problem ,so r ≥ 0, h ≥ 0.
10
And we know the volume of the cylinder is 10, so we have: 10 = πr2 · h.Then we have h = πr
2.
10
5
2
2
As a result, we can simplify our area formula into: S = π · r + 2πr · π·r2 = π · r + r
5 13
3. Now let’s find the stationary point. S 0 (r) = 2π · r − r52 = 0. So r = ( 2π
) .
4. But don’t make your decision so hastily . Still, we need to check if our stationary point is
5 13
) ); 1 be the test point in
the minimum point. Let’s choose 12 be the test point between (0, ( 2π
5 13
5 13
0 1
0
(( 2π ) , ∞). And we find S ( 2 ) < 0, S (1) > 0. This would guarantee us ( 2π
) to be the global
minimum point. (Since the domain of r is (0, ∞)).
Exercise:
1 Problem 6, Page 85; Problem 16, Page 86; Problem 21, Page 86.
Ans on webpage: answer to HW1.
2 Problem 9, Page 108; Problem 11, Page 108; 12 Page 108; Problem 24, Page 109;
Ans on wepage: discussion question:applied optimization; answer to HW3; Answer to
discussion questions: Optimization&L’Hopital;
3 Problem 22, Page 109. Find the length of the shortest ladder that will reach over an 8-ft
high fence to the wall of a tall building which is 3 ft. behind the fence.
4 Prof. Miller’s practice problems: 1, 2, 5.
The answer is on his website.
parametrized curve
Not much to say. Do the calculation with formula.
Three remainders:
1. Find a line:
If you have the slope of a line is m, and the line passes through the point (x0 , y0 ), then the
line function can be set as: y − y0 = m(x − x0 )
2. The slope of the tangent line at x0 is: y 0 (x0 ). So the tangent line is horizontal means
y 0 (x0 ) = 0; the tangent line is vertical means y 0 (x0 ) = ∞.
3. Review how to find a normal line. The slope of the normal line at x0 is:
−1
.
y 0 (x0 )
Example, Problem 3, Page 115 Find the radius of curvature and the center of the osculating circle to the curve: x(t) = t, y(t) = 1t at the point t = 1.
Ans:
1. Find the curvature. x0 = 1, x00 = 0; y 0 =
Then,
−1
, y 00
t2
κ=
=
2
t3
=
2
;
t3
·1−0
3
)2 ) 2
(1 + ( −1
t2
2 · t3
3
(t4 + 1) 2
plug in t = 1
2
κ(1) = 3
22
1
=√
2
√
The radius of curvature R = κ1 = 2.
We know that, the center of the osculating circle is on the normal line. The slope of the normal
line at t = 1 is:
−1
−1
= y0 = t2 = 1
dy
( dx )
( x0 )
And the normal line also passes through the point: (x(1), y(1) = (1, 1).
So the normal line is:
y − 1 = 1 · (x − 1) ;which is: y = x
√
Since the radius
of the osculating circle is 2, the distance from the center to the point (1, 1)
√
should be 2.
Let the center point be (x0 , x0 ).
p
√
(x0 − 1)2 + (x0 − 1)2 = 2
We solve x0 = 2, 0.
However the center must be in the inner side of the curve. Since the curve y =
1
x
is convex,
(Check the second derivative by yourself!) so the center must be above the curve. Then the
only answer is x0 = 2. Thus the center point is: (2, 2).
Example Problem 5(c), Page 115
Find the point with the largest or smallest curvature.
1
y = x3
3
ans:
y 0 = x2 , y 00 = 2x.
κ=
2x
3
((x2 )2 + 1) 2
Find the stationary points of κ, (which is equivalent to find the stationary point of (κ)2 :
4x2
(κ2 )0 = ( 4
)0
3
(x + 1)
8x · (x4 + 1)3 − 4x2 · 3(x4 + 1)2 · 4x3
=
(∗)2
solve (κ2 )’ = 0
1 1
1 1
x = ( ) 4 , 0, −( ) 4
5
5
1
1
Choose −0.1 as the test point in (−( 15 ) 4 , 0), choose 0.1 as the test oint in (0, ( 15 ) 4 ); choose 2
1
1
as the test point in (( 15 ) 4 , ∞); choose −2 as the test point in (−∞, −( 15 ) 4 );
Then we find: (κ2 )0 (−2) > 0; (κ2 )0 (−0.1) < 0;(κ2 )0 (0.1) > 0;(κ2 )0 (2) < 0.
1
1
1
1
Then we know x = −( 15 ) 4 , ( 51 ) 4 are local maxima points. And κ(−( 51 ) 4 ) =
1
2( 15 ) 4
3
( 15 +1) 2
−2( 15 ) 4
3
( 15 +1) 2
1
, κ(( 15 ) 4 ) =
1
. So the largest curvature is at x = ( 15 ) 4 .
L’Hopital’s rule
f (x)
x→a g(x)
If you want to calculate the limit lim
in the form of 00 or
∞
∞
(which means lim f (x), lim g(x) =
x→a
x→a
0, or lim f (x), lim g(x) = ±∞) So before use the L’hopital’s rule, check the condition first!
x→a
x→a
Example:
x − sinx
=
x→0
x3
lim
0
, use L’Hopital’s rule
0
1 − cosx
= lim
x→0 3 · x2
0
of the form , use L’Hopital’s rule
0
sinx
= lim
x→0 6 · x
0
of the form , use L’Hopital’s rule
0
cosx
= lim
x→0 6 · 1
1
=
6
of the form
Exercise:
1 Problem 6, Page 118; Problem 2, Page 117
Answer on webpage: answer to HW4; Answer to discussion questions:Optimization&L’Hopital;
Logarithm
Logarithm is the inverse of exponential function, in the mean that, eln(x) = x, ln(ex ) = x.
The question related to logarithm is mainly about to use formula provided at the beginning
of this review sheet.
Exercise:
Review your last homework question in Precalculus textbook.