y = x - 4/x
so gradient = dy/dx = 1 + 4/x2
When y = 3, x - 4/x = 3
x2 - 3x - 4 = 0 so x = -1 or x = 4
When x = -1, gradient = 1 + 4/(-1)2 = 1 + 4/1 = 1+4 = 5
When x = 4, gradient = 1 + 4/(4)2 = 1 + 4/16 = 1+1/4 = 1.25
You find the gradient of the curve using differentiation. The answer is 0.07111... (repeating).
The line described by the equation is a hyperbola and its gradient is different along its length. Its gradient is -10/(5x-3)2 except where x = 3/5 and the curve is not defined. The general form of a parallel curve is y = a + 2/[5(x + b) - 3] where a and b are constants.
The gradient (slope) of the curve at the midpoint at which the changes are calculated or the derivative of the curve at that point. These are only crude approximations unless the function is linear or if the changes are taken over very very small intervals. Equivalently, it is the slope of the tangent to the curve.
If you mean a line of y = 2x+5/4 and a curve of y^2 = 10x Then it works out that the line touches the curve at: (5/8, 5/2)
Ah, gradients are like the gentle slopes in our paintings, they show us how things change over distance. To find the gradient of 2.8km, we need to know how much something changes over that distance. If you have that information, we can calculate the gradient together and create a beautiful landscape of understanding.
You find the gradient of the curve using differentiation. The answer is 0.07111... (repeating).
The line described by the equation is a hyperbola and its gradient is different along its length. Its gradient is -10/(5x-3)2 except where x = 3/5 and the curve is not defined. The general form of a parallel curve is y = a + 2/[5(x + b) - 3] where a and b are constants.
The gradient (slope) of the curve at the midpoint at which the changes are calculated or the derivative of the curve at that point. These are only crude approximations unless the function is linear or if the changes are taken over very very small intervals. Equivalently, it is the slope of the tangent to the curve.
The period is the length of x over which the equation repeats itself. In this case, y=sin x delivers y=0 at x=0 at a gradient of 1. y next equals 0 when x equals pi, but at this point the gradient is minus 1. y next equals 0 when x equals 2pi, and at this point the gradient is 1 again. Therefore the period of y=sinx is 2pi.
If you mean a line of y = 2x+5/4 and a curve of y^2 = 10x Then it works out that the line touches the curve at: (5/8, 5/2)
rate is amount of product formed over time so for a plot of rate vs. time (reaction rate plot) a higher rate (M/hr), i.e. steeper curve/gradient corresponds to a higher amount of product formed over time
To find the maximum speed in a time-position graph, you would need to locate the steepest slope or the point with the highest gradient on the graph. This slope represents the highest rate of change in position over time, which corresponds to the maximum speed.
Ah, gradients are like the gentle slopes in our paintings, they show us how things change over distance. To find the gradient of 2.8km, we need to know how much something changes over that distance. If you have that information, we can calculate the gradient together and create a beautiful landscape of understanding.
The vertical drop of a stream channel over distance is known as the Gradient.
The amplitude of a sine (or cosine) curve is the difference between the maximum and minimum values of the curve, measured over a whole cycle.
To find the temperature gradient between two thermometers, subtract the temperature reading of the first thermometer from the temperature reading of the second thermometer. This difference represents the temperature change over the distance between the two thermometers, indicating the temperature gradient.
y = 2(x) - (pi/3) + (sqrt(3)/2)