review_2200.html

Review of Differential Calculus

As you approach the end of your course in differential calculus, you want to review the course to put into perspective what you have learned and to refresh your memory of topics you may not have thought about in a while. This project is a review of some of the main topics of the course, using Maple to check your answers and to illustrate some of the ideas you have studied. These ten topics constitute a framework of the course, but, of course, do not cover all aspects of the course, so do not count on this project to adequately prepare you for the final examination in the course. In particular, reasoning is required to set up the solutions of word problems, and Maple will not do the reasoning for you; this review will not be very useful for studying word problems, which are in many ways the heart of the course. This review is most useful for practicing routine calculus exercises.

How should you use this review? You should work each problem with pencil and paper, then use the Maple commands to check your work. If you "cheat" by entering Maple's commands before working the problems for yourself, you will destroy the value of this review, much as reading the last chapter of a mystery destroys the value of the book.

The Definition of the Derivative

This first course in calculus is a study of the derivative. You certainly should be able to state the definition of the derivative and use this definition to find the derivative of a function.

1. State the definition of the derivative of a function f with respect to x:
With pencil and paper, complete the following definition of the derivative:

Definition: The derivative of the function f with respect to x is the function f ' (x)=

Then translate the expression you wrote to the right of the equals sign into Maple notation.

Recal l how to command Maple to calculate a limit. For example to calculate , enter limit(g(x),x=a);

Type the translation of your expression in the execution cell and enter this command. If your definition is correct, the output will be, in Maple notation, f ' (x).

> restart:

> limit( );

>

2. Use the definition to find the derivative of the function f(x)=5*x^2+3*x: With pencil and paper, find the derivative of the function f(x)=5*x^2+3*x.
Check each step of your work with the following Maple commands: Define f(x), enter the quotient (f(x+h)-f(x))/h (which is called the Newton quotient of f(x)) , simplify this Newton quotient, then by inspection find the limit of this Newton quotient.

> restart:

> f:=x->5*x^2+3*x;

> (f(x+h)-f(x))/h;

The symbol % represents the preceding output. So the following command tells Maple to simplify the preceding output, which is the Newton quotient of your function f(x):

> simplify(%);

Now, by inspection, calculate your limit. Finally, check your work by asking Maple to directly compute it:

> D(f)(x);

>

3. Use the definition to find the derivative of the function f(x)=1/(5*x+3): With pencil and paper, find the derivative of the function f(x)=1/(5*x+3).

Check each step of your work with the following Maple commands: Define f(x), type the Newton quotient for f(x), simplify this Newton quotient, then by inspection find the limit of this Newton quotient.

> restart:

> f:=x->1/(5*x+3);

Type the Newton quotient in the following execution cell:

>

Simplify the Newton quotient:

> simplify(%);

Find, by hand, the limit of the Newton quotient, and check your derivative.

> D(f)(x);

>

4. Use the definition to find the derivative of the function f(x)=sqrt(5*x+3): With pencil and paper, find the derivative of the function f(x)=sqrt(5*x+3).

> restart:

> f:=x->sqrt(5*x+3);

Type the Newton quotient in the following execution cell:

>

> simplify(%);

You have just discovered that the simplify command did not adequately simplify your Newton quotient. It is difficult to force Maple to carry out the algebraic steps that you would do by hand; in this case it is easier to carry out the algebra by hand and calculate the limit by inspection. You can then use Maple to check your answer:

> D(f)(x);

>

The Tangent Line

You know that the value of the derivative of f(x) at x=a represents the slope of the line tangent to y= f(x) at the point (a,f(a)). Here are three problems concerning tangent lines.

1. Find an equation of the line tangent to a graph at a point on the graph: With pencil and paper, find an equation of the line that is tangent to the graph of f(x)=1/(5*x-3) at the point with x-coordinate x=1.
After you have found your equation, check your answer using Maple. First write your tangent line in the form y=m*(x-a)+c, and define y to be a function t(x)=m*(x-a)+c. You will, of course, have to determine the values of m, a, and c. Then plot the line and the graph of f(x) on the same coordinate system to see whether or not your answer is correct:

> restart:

> f:=x->1/(5*x-3);

> t:=x-> ;

> plot([f(x),t(x)],x=.8...1.2);

>

Does your line appear to be tangent to the graph? If not, find your error and fix it.

2. Find an equation of the line tangent to a graph, given the slope of the tangent line: With pencil and paper, find an equation of the line of slope -10 that is tangent to the graph of f(x)=1/(5*x-3).
You should have obtained two tangent lines. As in problem 1, define the first tangent line to be t1(x) and the second to be t2(x), then plot the graph of f(x) and both tangents. If your answers are correct your lines will appear to be tangent to the graph (although their slopes will not appear to be -10, since the scaling is not constrained, resulting in different units in the horizontal and vertical directions).

> restart:

> f:=x->1/(5*x-3);

> t1:=x->;

> t2:=x->;

> plot([f(x),t1(x),t2(x)],x=.4..0.8,y=-4..4,discont=true,color=[black,red,blue]);

>

3. A challenge: Find an equation of the line that passes through the origin and is tangent to the graph of f(x)=exp(5*x).

>

Then check your answer by defining your tangent line t(x) and plotting it with the graph of f(x):

> restart:

> f:=x->exp(5*x);

Enter the function whose graph is your tangent line in the following command.

> t:=

> plot([f(x),t(x)],x=-.1..0.4,y=0..4,color=[red,navy]);

>

Routine Differentiation

Find, by hand, derivatives of each of these functions, then let Maple check each one:

> restart:

1. f(x)=(x^3+5*x-1)^7

> f:=x->(x^3+5*x-1)^7;

> D(f);

>

2. f(x)=(x^2+1)/(3x^5-5x+4)

> f:=x->(x^2+1)/(3*x^5-5*x+4);

> simplify(D(f)(x));

>

3. f(x)=(x^3+5*x-1)^7

> f:=x->(x^3+5*x-1)^7;

> D(f);

>

4. f(x)=(x^2)*(sin(5x^3+7))

> f:=x->(x^2)*(sin(5*x^3+7));

> D(f);

>

5. f(x)=(sin(x^3+1))^5

> f:=x->(sin(x^3+1))^5;

> D(f);

>

6. f(x)=ln(1+x^3)

> f:=x->ln(1+x^3);

> D(f);

>

7. f(x)=sqrt(x)+1/(sqrt(x^2+1))

> f:=x->sqrt(x)+1/sqrt(x^2+1);

> D(f);

>

8. Find a function f(x) whose derivative is x^2*(x^3+5)^7. Then check it by differentiating:

> f:=x->

> D(f);

>

9. Find a function f(x) whose derivative is x*cos(x^2+7). Then check it by differentiating:

> f:=x->

> D(f);

>

10. Find a function f(x) whose derivative is x/(5x^2+9). Then check it by differentiating:

> f:=x->

> D(f);

>

Limits and Continuity

A: Evaluate each of the following limits by hand, then check allow Maple to check each one:

> restart;

> limit((x^2-2*x-3)/(x-3),x=3);

> limit(sin(3*x)/(x*cos(3*x)),x=0);

> limit((x+5)/(x^2-25),x=5);

> limit(abs(10-2*x)/(5-x),x=5,right);

> limit(abs(10-2*x)/(5-x),x=5,left);

> limit((1/x-1/5)/(x-5),x = 5);

> limit((sin(x+h)-sin(x))/h,h = 0);

What is the simple way to determine this limit? What does this limit represent?

B: Discuss the continuity of each of the following. Use Maple to check your answers.

> restart:

> f:=x->(x-1)/(x^2+4*x-5);

> plot(f(x),x=-8..2,y=-2..2,discont=true);

> f(1);

>

This lets you know that you cannot trust graphs completely! Is this function continuous or discontinuous at x=1?

2. Discuss the continuity of the following function.

> restart:

> f := x-> piecewise(x < 2,(x^2-5*x+6)/(x-2),3*x-7) ;

Now plot the graph to help you determine whether your discussion is correct.

> plot(f(x),x=0..4);

>

3. Discuss the continuity of the following function.

> restart:

> f := x-> piecewise(x < 2,(x^2-5*x+6)/(x-2),3*x^2-7) ;

Now plot the graph to help you determine whether your discussion is correct.

> plot(f(x),x=0..4,discont=true);

>

4. Determine all values of c for which this function is continuous at x=2:

> restart:

> f:=x->piecewise(x < 2,(x^2-5*x+6)/(x-2),3*x^2-c) ;

Now define c to be the value you determined, and look at the graph to help you see whether or not you are right.

> c:=

> plot(f(x),x=0..4,discont=true);

>

Maximum-Minimum Problems

Of course, the maximum-minimum word problems you worked in this course are more interesting than the routine problems presented here, but the solution of a maximum-minimum word problem requires you to first determine the quantity you want to maximize or minimize, then write that quantity as a function of a single variable. Having accomplished this, you finish the problem by solving the sort of routine problem presented here. Work these problems by hand, then look at the graphs to check your solutions.

1. Find and classify all extreme points of the function f(x)=x*exp(-x):

> f:=x->x*exp(-x);

> solve(D(f)(x)=0,x);

> plot(x*exp(-x),x=-1..5);

>

2. Find and classify all extreme points of the function x^3-3*x^2-9*x+5.

> f:=x->x^3-3*x^2-9*x+5;

> solve(D(f)(x)=0,x);

> plot(x^3-3*x^2-9*x+5,x=-2..4);

>

3. Find and classify all extreme points of f(x)=x*sqrt(1-x^2).

> f:=x->x*sqrt(1-x^2);

> solve(D(f)(x)=0,x);

> plot(f(x),x=-1..1);

>

4. Find and classify all extreme points of f(x)=x^2-1/x.

> f:=x->x^2-1/x;

> solve(D(f)(x)=0,x);

> plot(f(x),x=-2..3,y=-8..8);

>

The Derivative as a Rate of Change

You know that if f(t) is a function, then f ' (a) represents the rate of change of f with respect to t at t=a.

1. A conical tank, with its vertex down, has a top radius of 5 feet and an altitude of 7 feet. Water is leaking from the tank, so that when the water is 4 feet deep, the water level is dropping at the rate of 0.3 feet per second. Determine the rate at which water is flowing from the tank at that time. Work this problem by hand.

Solution: Use similar triangles to determine that the volume v(h) as a function of water depth h is given by:

> restart:

> v:=h->25*Pi/(3*49)*h^3;

> D(v)(h);

Now use the chain rule to determine that dv/dt = dv/dh*dh/dt = 0.3*dv/dh.

> D(v)(h)*D(h)(t);

> (D(v)(4))*0.3;

>

2. An observer is watching a weather balloon rise. The observer stands 2000 feet from a point directly below the balloon, and the angle of inclination of the observer's line of sight to the balloon is increasing at the rate of 0.01 radians per second when the balloon is 1000 feet high. How fast is the balloon rising?

First solve this problem by hand, by finding the height h of the balloon in terms of the angle z of inclination of the balloon. Then use the chain rule to find how fast the balloon is rising when it is 1000 feet high.

> restart:

> h:=z->2000*tan(z);

> D(h)(z);

Now evaluate this derivative at h=1000:

> evalf(D(h)(arctan(1000/2000)));

and use the chain rule to find dh/dt in terms of dh/dz and dz/dt:

> evalf(D(h)(arctan(.5))*.01);

>

Mean Value Theorem

State the mean value theorem, then determine the value of the number c whose existence is guaranteed for the function f(x)=x^3 on the interval [-2,0]. Then look up this theorem to make sure that you have stated it correctly, then define the number c and sketch the picture, which illustrates the mean value theorem. Your diagram should show the graph of f(x) and two line segments.

> restart:

> f:=x->x^3;

> c:=;

> plot([f(x),D(f)(c)*(x-c)+f(c),D(f)(c)*x],x=-2..0,color=[red,navy,cyan]);

>

Graphing

1. Sketch the graph of the derivative of the function f(x), then compare your sketch with Maple's graph:

> restart:

> f:=x->x^2*exp(-x);

> plot(f(x),x=-1..8);

Stop! Sketch the graph of f'(x) by hand. Then enter the following:

> plot([f(x),D(f)(x)],x=-1..8,y=-2..2);

>

2. Plot the graph of f'(x), then to sketch the graph of f(x) without Maple. Use Maple to check your graph.

> restart:

> f:=x->(2*x^2+1)/(x^2-2*x);

> plot(D(f)(x),x=-4..10,y=-1..1);

Stop! Sketch the graph of f(x) by hand. Then enter the following:

> plot(f(x),x=-4..10,y=-50..50);

>

Antidifferentiation

Recall that is the notation for the most general antiderivative of f(x); that is, the set of functions of the form F(x)+C, where F ' (x) = f(x). Evaluate the following, then check each:

> restart:

> int(sqrt(x),x)+C;

> int(x*cos(x^2+1),x)+C;

> int(1/(5*x+4)^3,x)+C;

> int(exp(-5*t),t)+C;

> int(x*sqrt(x^2+3),x)+C;

> int(x^3/(x^4+3),x)+C;

> int(x^3/((x^4+3)^2),x)+C;

> int(x^3/((x^4+3)^(-2)),x)+C;

>

Differential Equations

A: Solve each of the following differential equations. Check the purported solutions by evaluating each side of the equation separately:

1. dy/dx = sqrt(x+9).

> f:=x->(2/3)*(x+9)^(3/2);

> D(f)(x);

>

2. dy/dx = 3/(7*x+5)

> f:=x->(3/7)*ln(7*x+5);

> D(f)(x);

>

B: A particle travels along the x-axis with acceleration a(t)=5*t^2-3*t. If its initial velocity is 0 and its initial position is 5, find its position s(t) at any time. Then check your answer:

> s:=t->(5/12)*t^4-(1/2)*t^3+5;

> s(0);D(s)(0);D(D(s))(t);

>

C: A ball is dropped from a height of 400 feet. Find its impact velocity.

Solution: First find its velocity v(t) and its position s(t). Then check:

> s:=t->-16*t^2+400;

> s(0);

> fsolve(s(t)=0,t);

> D(s)(5);

>