what is the relationship between gradients and contour curves?
152816
Contour curves connect points where a function of 2 variables has the same particular value. Collection of all such curves is a Contour Map.\n\nA gradient is always perpendicular to the contour curve.\n\nFor more info: \nhttp://en.wikipedia.org/wiki/Contour_line
873172
u can get them with the help of second and higher derivatives.\ninflection point is there where the second derivative is zero.
184296
Differentiation of a function mapping R into R considers the limiting case of a secant to the curve to get the unique tangent line to the curve at each point. Ie., if we have a graph y=f(x), then the slope of a secant line between two points (x,f(x)) and (x+h, f(x+h)) on the curve is [f(x+h) - f(x)]/[x+h - x] = [f(x+h) - f(x)]/h. The tangent line to the curve at the point (x, f(x)) is hypothesized to have the slope given by the limit of the secant's slope as we rotate the secant so that the second intersection point moves closer to (x,f(x)). Algebraically, we want to study the limit of the above quotient as h approaches 0. The usefulness of the tangent at each point is that such a line gives a linear approximation to whatever f(x) stands for at each point. The tangent line also tells you when there is a maximum or minimum, as the line will be horizontal there (for certain well-behaved curves).\nThe first half of your study of calculus will involve justifying this concept and making concrete and mathematically precise what we mean when we say we are going to take the limit of an expression.\nThe second half of your calculus course will concetrate on integration, which studies the problem of calculating the area of arbitrarily shaped regions (instead of just the simple geometrical shapes you've studied before). Given an arbitrary curve y=f(x) or a bunch of curves enclosing a region, you will be able to calculate the area contained in the region. The motivation for the integral also involves limiting behavior. In the case of the Riemann integral, the hypothesis is that you can cover the area between a curve and the x-axis by a finite amount of rectangles, each with a finite area, and as the width of the rectangles get smaller, the difference between their total area and the true area under the curve goes to 0. The rigorous proof of this concept will be given by your textbook or teacher.\nYou will also study the Fundamental Theorem of Calculus, which connects the behavior of integrals to the behavior of derivatives in such a fundamental way that integrals are also known as antiderivatives. This theorem makes evaluating integrals a lot easier.\nAs you study further in higher-dimensional calculus and manifold theory, you will learn more general ways of thinking about derivatives and integrals, in that you will simply view them as forms over fields. We know all about linear algebra; we have to use derivatives to reduce the study of nonlinear forms to our familiar territory of linear algebra. This view is very useful in theoretical physics and differential geometry.\n\nAs some examples, let's study a curve we know well, the parabola y=x^2. To spice this up a little, let the parabola be the path of a bullet fired at some (horizontal) muzzle velocity v. Then from physics we know y(t) = -4.5*t^2 + 1 and horizontally x(t)=vt, so the path of the bullet is given by y(x) = (-4.5/v^2)x^2 + 1. We want to know how hard the bullet will hit the ground, so we need to know the instantaneous velocity of the bullet when y=0, which is when x=v/Sqrt[4.5] and thus t=Sqrt[4.5]. The velocity consists of the speed and direction of the bullet. First we get the direction, which is the slope of y(x) when y=0. Let -4.5/v^2 = k. Thus we want to study the quotient [f(x+h) - f(x)]/h = [ k(x+h)^2 + 1 - kx^2 - 1]/h = k*[x^2 + 2xh + h^2 - x^2]/h = k*[2xh + h^2]/h = k*[2x + h] as h approaches zero, which is just 2kx, so we have 2*(-4.5/v^2)*(v/Sqrt[4.5]) = -2*Sqrt[4.5]/v so the final slope decreases in absolute magnitude as muzzle velocity increases, as one would intuitively expect. The speed is a little more tedious, but if you've followed the above, you can do it yourself in no time.\nIntegration at this level requires knowledge of series and limits; if you're interested, pick up a good calculus text.
766685
You need to differentiate the equation y=2x^2. \n\ndy/dx = 4x therfore the gradient at x=3 = 4*x = 12
770231
Nice.  Now, to answer your question...\n\nThe previous post pointed out that there will be more lines in a mountainous area than on a plain, IF the contour intervals are the same.\n\nSince the lines are further apart on a plain, the mapmakers have the opportunity to decrease the contour interval to provide more detail.  In a mountainous area, decreasing the contour interval would make the map unreadable, since there would be too many lines.
900462
It really depends on how the equation of the path/curve is originally specified.\n\nIf it's specified in terms of x and y, an easy parameterization is:\n\nx = t\ny = f(x) = f(t)\n\nBut there are many other ways to do it.
416901
A-a??\n\nA gradient is a vector field consisting of derivative vectors of a source of data (like a moving ocean)\nYou could use a gradient to show how fast the water is moving across the ocean surface. The derivitaive vectors that make up the gradient are instantaneous calculations of the rate of flow at several points on the surface of the ocean)
826403
hmmm taking me way back here.  when you break the area under a curve into rectangles of differential width,  either the left top corner will lie on the curve or the right top corner will.  \n\nas you decrease the differential width the area will become more exact until bada bing you are doing calculus.  i think.
270608
Everything you wanted to know about a transformation curve. [please see link]
751800
The 1st derivative is the slope of the original function.  If it is positive, you know the original function was rising (from left to right).  If it's positive but leveling off, the original still rising but not as much.  If it hits the x axis, there is a critical point on the original function (relative max/min or inflection point).
646764
Differentiation is a branch of calculus which determains the gradiant at a specific point on a graph. A straight line graph is easy enough, because between two points on the line, you can divide the distance vertically (y direction) by the distance horizontally (x direction). For example, a graph that has a straight line that is 45 degrees from the x-axis (which, incidently would be 45 degrees from the y-axis as well) would have a gradiant of 1.\n\nIf you wanted to determain the gradiant at a point on a graph which is curved though, you cannot just simply divide the x distance by the y distance, because it is a curve. This is where differentiation comes in. Imagine two points on the curved line, determain the gradiant of that. Then bring the distance of the two points closer together and determain the gradiant of that. As the distance of the two points tends to zero, you would get closer and closer to the gradient of the first point.\n\nAs a generalised equation, you could determain the x and y coordinates of the first point, and the coordinates of the other point by the coordinates of the first point plus the distance between them (e.g. x+dx, x+dy). y would be a function of x in a general equation so you would have the first point coordinates as x, f(x).\n\nNot easy to explain this without diagrams, but I have added a link which you can visually see this.
837855
if it is a straight line evolved from a curve, it works much better,but if is is  curve evolved from a straight line then you WILL have trouble with it\nand the subject will pay the ultimate price for this mistake\nso please pay attention tot he variables in your equations of your control functions, because if you do not, the manipulative's will control it all\nand I think you know what happens when that happens....\n\ngood luck\nsmile
807353
Just an idea:\n\nthe two tangency points are related by the fact that the slope from point Pf to Pg is the slope of the line AND the derivatives at those two points. \n\nYou have 2 free variables: the x coords of Pf and Pg.  From these the y coords are given by the function values and the slopes are given by the derivatives at those values.  There should be enough relationships to allow solution.
839609
Curl, divergence, gradient.
502267
Transpiration and concentration gradients
1050238
Here's a good statement:\n"Every continuous curve in the plane has zero area".\nIt looks like it should be true. But it is FALSE. In fact,there is a continuous curve that fills a whole square! This curve 'intersects itself' infinitely often, so we might then conjecture\n"Every arc (a one-to-one curve) has zero area".\nBut this is FALSE also! \nFinding an example of an arc with strictly positive area is not easy, but it can be done (the usual examples require the use of Cantor sets).\n\nDepending on the level, you might be able to say:\n"Every continuous function is differentiable except at finitely many points".\nThis is also FALSE. There are continuous functions that are differentiable NOWHERE.\nOr you might try:\n"Every differentiable function is monotone on some interval".\nAgain, this is FALSE.\nOr another:\n"A function with derivative zero almost everywhere is constant".\nThe 'Devil's staircase' is a counterexample.
355882
The relationship is defined in one dimension as f'(x) being the slope of the tangent line to the graph of f(x) at the point x. In symbols, f'(x) is the limit as h approaches 0 of the slope of a secant to the graph passing through x: [f(x + h) - f(x)]/h.\nf'(x) being the slope of the tangent line to the graph of f at x has its advantages. When f'(x) is 0, we know the tangent line is horizontal there, meaning the graph is either at a minimum, a maximum, or a temporary flat area. Higher derivatives (treating the derivative as a function itself) can tell us which of the above it is.\nDerivatives are most concrete in physical examples. If x(t) is the position of an object at time t, then the velocity of the object at time t can be idealized as the derivative of x(t) at t (instead of the average velocity, which is the slope of a secant line to the graph of x(t)). Similarly, the acceleration at time t is given by the second derivative.\nIn higher dimensions, where x = (x1, x2, ..., xn) is a point and f(x) = (f1(x), f2(x), ..., fn(x)) is also a point, this is generalized to f'(x) being the unique linear function that vanishes with the change in f at x. In symbols, the limit as ||h|| vanishes of [f(x + h) - f(x) - f'(h)]/||h|| vanishes. This coincides with the definition in 1 dimension above. In higher dimensions, f'(x) corresponds to the basis of a tangent space at x, as it did in 1 dimension.\nIf f is a map from R^2 into R, f'(x) is a linear map from R^2 into R, which is a plane. This plane is a tangent space to the graph of f at x. Higher-dimensional tangent spaces are not as visualizable.