I had first class as well, and i don't see it as a big deal.
Holiness_M - my answers below:
1. Cosine Law.
The Pythagorean Theorem is just a modified version of the Cosine Law that applies only to right triangles. The Cosine Law, c^2 = a^2 + b^2 - 2ab cosC, will yield the Pythagorean Theorem, when C = 90 degrees. Since cos90 = 0, the term "2ab cosC" disappears and you're left with the Pythagorean Theorem.
2. Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus is a remarkable discovery as it shows that integration and differentiation are related; it shows that the two processes are inverses (that is, one process is the inverse of the other and vice-versa).
3. L'Hôpital's Rule
e.g. lim x -> 1 of (x - 1)/(x^2 - 1).
Instead of factoring to cancel the x - 1 term in the numerator, simply take the derivative of the numerator and the denominator, and you will get:
lim x -> 1 of 1/2x, which equals 1/2. Try it with factoring and cancelling and you will get the same answer!
L'Hôpital's Rule is generally used for more complicated functions where it is not possible to cancel terms. It also has applications to the limits of indeterminate products, differences and powers - check it out!
4. It cannot be integrated using elementary functions (numerical integraton).
The problem with integrating e^(x^2) is finding an antiderivative. Using common elementary functions, it is not possible, so mathematicians rely on analytical integration to approximate the integral of e^(x^2). Mathematicians can also use computers to compute the integrals.
There are other integrals that cannot be evaluted using elementary functions, such as (e^x)/x, sin(x^2), et cetera.
5. Squeeze, Pinching or Sandwich Theorem
Example. Say you have three functions: f(x), g(x) and h(x), where g(x) is greater than or equal to f(x) and h(x) is greater than or equal to g(x). If f(x) and h(x) approach the point 0 as x approaches 0, then by the theorem, g(x) must also approach the point 0 as x approaches 0.
6. Catenary Equation.
The catenary states that when a chain or cable is hung, it (the chain/cable) is steepest near the point(s) of suspension, because this is where most of the weight of the chain/cable is pulling down, thanks to gravity.
7. The function has a discontinuity within the given limits.
The differential is in the question: dx. It simply means integrate with respect to the variable 'x'.
The antiderivative of 1/x does in fact exist; it is ln |x| + C (where C is the integration constant).
The problem stems from the fact that in the interval [-1,1] for 1/x, there is a discontinuity at x = 0, which is in the given interval. Because there is a discontinuity in the function, its definite integral cannot be calculated. If you did calculate it (tsk, tsk), you would get an answer of ln 1 - ln |-1| = ln 1 - ln 1 = 0. However, you CANNOT do that.
On that note, when doing definite integrals, be sure to check your limits of integration to see if there are any discontinuities within the limits.
8. T(x,y) = (2x + y, x - y).
In order for a transformation to be deemed non-linear, it just has to fail one of the two main properties: additivity or homogeneity. For T(x,y) = (x + 1, y), I will show how it fails using homogeneity:
T(x,y) = (x + 1, y) - Take any vector, say u, to be u = (1,2) and a constant c = 2.
For the transformation to be linear, T(cu) = cT(u)
T(cu): cu is the same as 2(1,2) which will give a new vector, (2,4). Using x = 2 and y = 4, we plug it into the transformation equation T(x,y) which gives us (2+1, 4) = (3,4)
cT(u): instead of multiplying the vector by the constant first, we use the original vector (1,2), put it into the tranformation equation and then multiply by the constant. Doing that, we get T(x,y) = (1+1, 2) = (2,2). Now multiply by 2, and you get (4,4).
Because the two results are not equal, this system fails the property of homogeneity, and is therefore, not linear.
I encourage you to try it out with the others!
9. Inconsistent.
Having a row [0 0 0 , 0 | x ], where x is any non-zero number, means the system is inconsistent, because you're saying that nothing is equal to a specific value, which is not possible (e.g. 0x = 5; not possible to evaluate).
If you have the row [0 0 0 , 0 | 0], this is entirely possible because you're saying nothing equals nothing (e.g. 0x = 0, x = 0); having a row like this would mean that your system is consistent and will have a unique solution of infinitely many solutions (when you have one or more free variables).
10. A plane passing through the point (1,2,5), parallel to the vectors (2,0,0) and (0,1,-2)
When analyzing these types of equations, the point will never have a variable attached to it (in this case, the variables are t1 and t2).
Also, with a line, you will only be given one vector; with a plane, you will be given two or more vectors. The vectors will have a variable attached to it (in this case, t1 and t2).
Are these correct @ Holiness_M?
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