Calculus 12
Calculus 12 ajlill Wed, 05/19/2010 - 13:48Rate of Change
Derivatives and their Applications
Implicit Differentiation
There will be times that you can't do normal differentiation, for example, if you are given the equation or a circle $x^2 + y^2 = r$. Sometimes, you can rearrange the equation, solve for y, and come up with a differential, but often not. Instead, we will take the derivative of both sides of the equation: $$ \frac{d}{dx}(x^2 + y^2) = \frac{d}{dx}r$$ $$\frac{d}{dx}x^2 + \frac{d}{dx}y^2 = 0$$ $$2x + \frac{d}{dx}y^2 = 0$$ So, what do we do about the $y^2$? We can't take the derivative wrt x, since there is no x. So, just how you would multiply a fraction by $\frac{a}{a}$ to get it to a common denominator, we can do the equivalent with the derivative. $$ 2x + \frac{d}{dx}y^2 = 0$$ $$2x + \frac{dy}{dx}\frac{d}{dy}y^2 = 0$$ $$2x + \frac{dy}{dx}2y = 0$$ $$\frac{dy}{dx}2y = -2x$$ $$\frac{dy}{dx} = \frac{-2x}{2y}$$ $$\frac{dy}{dx} = -\frac{x}{y}$$ Looks bizarre, treating differentiation like a fraction, but it works. Just remember that all the rules of normal differentiation apply, so if you wind up with, for example $\frac{d}{dx}xy$, you need to apply the product rule to get $y\frac{d}{dx}x + x\frac{d}{dx}y$
What's really happening
We can do this because y is really a function of x, even if it is difficult or impossible to solve for y and take a "normal" derivative. So consider replacing all instances of y with g(x), differentiate by the normal rules, and then substitute back y for g(x) and $\frac{dy}{dx}$ for g'(x) $$ \frac{d}{dx}(x^2 + (g(x))^2) = \frac{d}{dx}r$$ $$\frac{d}{dx}x^2 + \frac{d}{dx}(g(x))^2 = 0$$ $$2x + 2g(x)g'(x) = 0$$ $$2x + 2y\frac{dy}{dx} = 0 $$
Common derivatives
$$ \frac{d}{dx}x^n = nx^{n-1}$$ $$\frac{d}{dx}A^x = ln(A)A^x$$ $$\frac{d}{dx}sin(x) = cos(x)$$ $$\frac{d}{dx}cos(x_ = -sin(x)$$ $$\frac{d}{dx}tan(x) = sec^2(x)$$ $$\frac{d}{dx}(f(x) \times g(x)) = f'(x) \times g(x) + g'(x) \times f(x) $$
Lines and Planes in 3-space
Lines and Planes in 3-space ajlill Fri, 03/23/2018 - 12:42Finding the intersection of 2 planes
If you have two planes, they can either be parallel, co-incident, or intersect on a line. I'll ignore the first two, and concentrate on finding the equation of the line of intersection.
Consider the planes − 2 x + 3 y + z + 6 = 0 and 3 x − y + 2 z − 2 = 0. If this was 2 lines is 2-space we would just use substitution or elimination to find a point (this is what you would do to intersect 3 planes, btw). Unfortunately, we have more variables than equations.
We want our solution to look like $$(x,y,x) = [a_1,a_2,a_3] + t[d_1,d_2,d_3]$$, so we just pick one of our variables in the plane and set it to t, and then solve for x and y as functions of t. Lets choose z.
-2x + 3y + t + 6 = 0
3x - y + 2t - 2 = 0
we'll eliminate y by multiplying the second equation by 3 and addiong to the first
7x + 7t = 0
x = -t
sub this into the second equation and solve for y
3(-t) - y + 2t - 2 = 0
y = -t - 2
This gives us our parametric equations, which are easy to turn into a vector equation
x = -t
y = -2 - t
z = t
Geometry and Algebra of Vectors
Geometry and Algebra of Vectors ajlill Thu, 06/10/2010 - 13:32Heretofore, you have been dealing with scalar quantities. These are things like numbers, mass, length, that just have a size (magnitude). We will now formally introduce the vector. A vector is a mathematical object that has both a magnitude and a direction. Vectors are used to describe things like position, forces, weight, and hence play a big part in physics. In order to describe the direction of a vector, we need a frame of reference. In the physical world, this will quite often be horizontal and vertical axes (wrt the center of the earth). In math, we will be using the Cartesian axes, either the number line (x), x & y, or x, y, & z.
Vectors in R1
You have already been working with vectors, even though you may not have realized it. When you talk about a point on a number line, you are actually talking about a vector. If you have a number A, then it has a magnitude, which is |A|, and a direction, which is it's sign.
Vectors in R2
A vector in R2 can expressed as a magnitude and an angle from one of the axes, for example, 5N at 30 degrees up from horizontal. In R2, a vector can also be expressed in the same form as a line segment. For example, if you have two points P(1,1) and Q(2,2) you can use them to describe a vector $\vec{PQ}$. Up until now, you have been used to using an equation to describe a line in R2: Ax + By + C = 0 A vector in R2 can be also used to describe a position on a line, and so you can write an equation of a line using an initial point and a direction vector. $$\vec{v} = (x_0,y_0) + t(B,-A)$$ You should notice that the direction vector uses the same variables as the equation of the line. This is no co-incidence. Let us re-arrange the equation of the line y = (-A/B)x - C/B To go from an initial point on the line, say the y-intercept (x=0), to x=1, we move up -A and right B (remember that in this form of the equation, the number in front of the x is the slope, or rise over run). And this gives us our direction vector. Incidentally, you can also get your initial point from this equation as (0, -C/B). You could just as easily use the x-intercept, in which case it would be (-C/A, 0). This brings us to another way of describing a line in R2, using a free variable.
Vectors in R3
Vectors in R3, look and work very much like vectors in R2, just with one extra variable. If you are expressing it as an angle and magnitude, you need 3 angles to giive the direction, one from the x-axis, one from the y-axis and one from the z-axis. Theses are usually called $\alpha, \beta, and \gamma$ Lets start with equations in R3 Ax + By + Cz +D = 0 This equation describes a plane in R3. There is no way to describe a line using this form of expression.
Useful Formula
Projection of $\vec{u}$ on $\vec{v}$ is $\frac{(\vec{u}\cdotp\vec{v})\vec{v}}{|\vec{v}|^2}$ It's magnitude is $\frac{|\vec{u}\cdotp\vec{v}|}{|\vec{v}|}$ The distance from a point $P(x_1,y_1)$ to a line l: $Ax + By + C = 0$ is $d(P,l) = \frac{|Ax_1 + By_1 +C|}{\sqrt{A^2 + B^2}}$ extending to n dimensions
Worksheets
Net Force Word Problems Vector Addition and Subtraction problems Moving About Worksheets