First workshop Here are responses to questions e-mailed to me.
Problem 1
The question:
... does the answer need to be exact or can we have rounded decimals in the equation ...My answer:
The response to #1 CAN be exact (not pretty but exact). Once you have found it exactly, abbreviate it algebraically (call it w or something). Then you can write the answer in terms of w. I would prefer an exact answer if possible. There are certainly problems of this nature where exact answers can't be gotten, but here because the functions involved are just second degree polynomials, the answers will be solutions of quadratic equations, hence easily specified exactly with square roots, etc.
The question:
... i still dont understand the concept of how to get the tangent line between 2 parabolas. any hint could help ...My answer:
The problem is not horrible, but demands a slight amount of abstraction. A general comment is DON'T OVERLOAD x WITH DIFFERENT MEANINGS IN THE NOTATION.Let me try to solve an easier problem first.
Suppose I asked you to find a point on y=x^2 where the tangent line is parallel to the line y=5x+4. Well, since you are a spiffy-keen calculus student, I assume you would recognize that "parallel" is gotten easily by requiring that parallel lines have equal slopes. Therefore the problem is solved by finding a point on y=x^2 where the line tangent to the parabola has slope equal to 5. Well, we also know that one simple geometric use of the derivative is that it "encodes" the slope of the tangent line. The derivative of x^2 is 2x. Well, I want a point on the parabola. Let me be a bit careful: if x=A, then the corresponding point on y=x^2 is (A,A^2), and the slope of the tangent line to y=x^2 at (A,A^2) is 2A. I am trying NOT to make the letter x mean too many different things here. So if the tangent slope is 2A, and it is also supposed to be 5, then 2A=5 so A=5/2. And the point desired is (5/2,(5/2)^2).
The problem given on the workshop is more intricate logically, and a bit more complicated computationally. You CAN'T let x play too many different logical roles in the solution. For example, on y=x^2 a point could be (A,A^2). And on the other parabola, a point could be (B,something similar). The derivative of x^2 at x=A gives me 2A as the slope. The derivative of the other parabola's formula at x=B gives me some stuff involving B. And since the line connects the two points (A,A^2) and (B,something else) I also can compute the slope from that information (and that will involve both A and B).
There are then 3 expressions for the slope. Let me call them m_1 and m_2 and m_3 (in e-mail the underscore is frequently used to mean subscript). All three of the expressions should be equal in order to get the picture shown. So we need, say, m_1=m_2 and m_1=m_3 (this is enough information!).
Write these equations and solve them. It is slightly messy but it is two equations in two unknowns. They are not linear (degree 1) which would be nice, but they are not too difficult to solve -- you'll need to solve a quadratic equation (no, it doesn't factor: "most" quadratics with integer coefficients do not factor). And the solution(s) will give enough information to solve the problem.
Problem 2
The question:
I didn't really understand problem #2; by 'windows' do you mean different equations related to y = x^2,My answer:
In problem #2, which I really like (it was essentially invented by a colleague, not by me [I invented #4, which looks much more formidable]), the idea is to have people think about what their graphing calculator reports. One way of using such a calculator is to specify a function (here f(x)=x^2) and a "window", which is two intervals, one, a<=x<=b horizontal, and one, c<=y<=d vertical. I think it is reasonable to call a window "centered at (2,4)" if (a+b)/2=2 and (c+d)/2=4. That means 2 is in the middle of the horizontal interval and 4 is in the middle of the vertical window. Then there are (there REALLY are!) four such windows giving the four pictures shown (I have examples of such windows!) and there is a straight line picture which CANNOT be obtained through any selection of such windows. The first four windows can be gotten through a bit of thought and some experimentation, and then explaining why another picture can't be gotten needs some explanation but is not enormously difficult and needs essentially no calculus. It all depends on thinking just slightly creatively. The excuse for the problem (beside the fact that it is fun!) is that weird versions of pictures actually occur quite a bit when one is using powerful technology, and so some alertness is needed.
Problem 3
The question:
Number 3 states that y=x^2-c and y=c-x^2 are functions of c. As far as I know, that makes these "curves" lines, because x^2 is a constant and c is the variable.My answer:
NO.c is what's called a parameter. The "conventional" role for x and y are the independent and dependent variables, respectively. The question is supposed to be straightforward, and not tricky at all.
For each value of c, the curves which are the graphs of y=x^2-c and y=c-x^2 are parabolas. The first opens "up" and the second, down. For each specific value of c, these curves intersect, and there is a region for which they form a boundary. That region has an area. Find the value of c for which that region has area equal to 1.
This sort of problem actually does arise in a wide variety of applications. There are curves which model some sorts of phenomena, and these curves or functions depend on some other values. So the difficulty depends on finding the values of the parameter(s) (here only one parameter) which will best satisfy the requirements of the problem.
It really wasn't supposed to be tricky. Textbook problems are [generally] not supposed to be tricky. That's left to the evil, vicious (viscous?) instructors.
Maintained by greenfie@math.rutgers.edu and last modified 9/8/2009.