-- square the point's x-coordinate
-- square the point's y-coordinate
-- add the two squares together
-- take the square-root of the sum
-- the answer is the distance of the point from the origin.
This works because if you draw a line down from the point to the x-axis (length is y-coordinate), then along the x-axis to the origin (length is x-coordinate), and back to the point (length is distance), you just made a right triangle.
Then you can use the Pythagorean Theorem to find the length of the long side (the distance) since you know the length of the two shorter sides.
Find the slope of the tangent to the graph at the point of interest.
If the point (x,y) is on the graph of the even function y = f(x) then so is (-x,y)
-- Take the equation -- Set either 'x' or 'y' equal to zero -- Solve the resulting equation for the remaining variable -- If the remaining variable is then also zero, then the origin is on the graph of the function If the graph is a straight line ('x' and 'y' appear in the equation only to the 1st power), then the equation has to be in the form of a simple ratio ... like (y = Kx) or (x = Ky) or (xy = K) or (x/y = K) ... in order to go through the origin.
It looks like this came from some multiple choice question, where you're given several choices. Take each choice and substitute the x and y coordinates into the equation. So for example the point is (0,3), then substitute in and get 2*0 + 3 which equals 3 and satisfies the equation, so the point is on the graph. If the point is (1,1) then 2*1 + 1 = 3 which satisfies the equation, so that point is also on the line. You want to find one where the left side does not equal 3, then that point is not on the graph of the line.
x + 5y = 0Subtract 5y from each side of the equation, just to put 'x' and 'y' on opposite sides.x = -5yThis is interesting. If EITHER 'x' OR 'y' is zero, then the other one is also zero.The only place on the graph where that is true is the origin. So the line goes through the origin,and the 'x' and 'y' intercepts are both zero.
The origin on a graph is the point (0,0).You can find the distance to a point by applying the Pythagorean theorem:Square the x coordinate and add it to the square of the y - coordinate of the point.Now take the square root of your answer.The result is the straight line distance from the origin to the point.
graph it
You can find the speed of an object from its distance-time graph by calculating the slope of the graph at a specific point. The slope represents the object's velocity at that particular moment. By determining the slope, you can find the speed of the object at that point on the graph.
It is the gradient of the straight line joining the origin to any point on the graph. Thus, if A = (p,q) is any point on the graph, the average unit rate between the origin and A is q/p (provided p is non-zero).
The distance from any point on the circle to the origin
To find the starting point of a distance vs time graph from a velocity vs time graph and a function, you would integrate the velocity function to find the displacement function. The starting point of the distance vs time graph corresponds to the initial displacement obtained from the displaced function.
A: The distance from any point inside the circle to the origin. B: The distance from any point inside the circle to the origin. C: The distance from the x-coordinate to the origin. D: The circumference.
To get speed from a distance-time graph, you would calculate the slope of the graph at a given point, as the gradient represents speed. To calculate total distance covered, you would find the total area under the graph, as this represents the total distance traveled over time.
You cannot because a dispacement-time graph is concerned only with motion in a radial direction; any motion in a transverse direction is completely ignored. For example, an object circling the origin at a fixed distance, even with a variable speed, is always at the same distance from the origin. So the displacement-time graph will be a straight line whose height is the radial distance. A straight line in the distance-time graph is to be interpreted as no motion! Really?!The average velocity in the radial direction is the final displacement minus the starting [initial] displacement, all divided by the difference in time between the two points. The instantaneous velocity in the radial direction is the slope [gradient] of the graph at the point in question.
You cannot since the graph shows displacement in the radial direction against time. Information on transverse displacement, and therefore transverse velocity, is not shown. For example, there is no difference in the graph of you're staying still and that of your running around in a circle whose centre is the origin of the graph. In both cases, your displacement from the origin does not change and so the graph is a horizontal line. In the first case the velocity is 0 and in the second it is a constantly changing vector. All that you can find is the component of the velocity in the radial direction and this is the slope of the graph at the point in question.
To determine the speed of a body from a distance-time graph when the body starts from rest, you can find the slope of the graph. The slope of a distance-time graph represents the speed of the body. A steeper slope indicates a higher speed, while a shallower slope indicates a lower speed.
Divide the ordinate (y-coord) of any point on the graph by its abscissa (x-coord).