0
Find a function f such that gradient of f =F if F(x,y,z)=?
Modiegi Morudu
I tried to find f by integrating the partial derivatives, but since 1/r is multiplying the whole vector, I just took it out, I'm not sure if I can do that. Like this:
∂f∂x(x,y,z)=x
∂f∂y(x,y,z)=y
∂f∂z(x,y,z)=z
thus
f(x,y,z)=x22+g(y,z)
f(x,y,z)=y22+h(x,z)
f(x,y,z)=z22+k(x,y)
for some functions g, h, and k, so if g=y22+z22, h=x22+z22 and k=x22+y22, the function f is:
f(x,y,z)=1r(x22+y22+z22)=12r⋅r2=r2
Am I correct? If not, how can I solve this correctly, should I integrate x/r, y/r and z/r instead?
Shandy Rica Reyes Ab...
Add your answer:
Earn +20 pts
What is a gradient function?
Assume you want to know what is the formula of the gradient of the function in multivariable calculus. Let F be a scalar field function in n-dimension. Then, the gradient of a function is: ∇F = <fx1 , fx2, ... , fxn> In the 3-dimensional Cartesian space: ∇F = <fx, fy, fz>
What is the formula of gradient?
Assume you want to know what is the formula of the gradient of the function in multivariable calculus. Let F be a scalar field function in n-dimension. Then, the gradient of a function is: ∇F = <fx1 , fx2, ... , fxn> In the 3-dimensional Cartesian space: ∇F = <fx, fy, fz>
What is the formula for functions?
Assume you want to know what is the formula of the gradient of the function in multivariable calculus. Let F be a scalar field function in n-dimension. Then, the gradient of a function is: ∇F = <fx1 , fx2, ... , fxn> In the 3-dimensional Cartesian space: ∇F = <fx, fy, fz>
Why do you use gradient?
The derivative of a function, df/dx, is to single variable calculus as the gradient of a function, ∇f, is to multivariable calculus.If f is a function of three variables, x, y, and z, then the gradient of f is the vector function ∇f(x, y, z) = All of the uses of derivatives in single variable calculus are analogous to the uses of gradients in multivariable calculus:In single variable calculus the derivative tells us the instantaneous rate of change at some point, [x, f(x)]. In multivariable calculus, the gradient of a function tells us the instantaneous rate of change at some point, [x, y, f(x,y)], or if the function is of more than two variables, ∇f would tell us the instantaneous rate of change at a point [x, y, z, ….., f(x, y, z, ….)]. One Important difference in calculus of more than one variable is that a function can have many different rates of changes at one point. To understand why this is so, imagine that you are standing on a hilltop which is defined by a function of two variables f(x, y). The downward slope of the hill, the gradient, is different depending on the direction you look; to find the slope you need to specify a direction. This is why we take the 'directional derivative' which is simply the dot product of the gradient with a unit direction vector (the direction you are looking down the hill). For example suppose we want to find the instantaneous rate of change of the function f(x,y) = x2 + y2 at the point (2,1) in the direction of v = :The directional derivative in the direction of v = ∇f(x, y) ● = < ∂f/∂x(x, y, z), ∂f/∂y(x, y, z)> ● = ● = 2y evaluated at (2,1) = 2.Let's continue our comparison of derivatives and gradients. In single variable calculus a derivative of a function is equal to zero at a maximum or minimum value of the function. This fact can be used in practical applications that require maximizing and minimizing functions of one variable. The same is said of the gradient in multivariable calculus. By setting the gradient of a multivariable function equal to zero, we can solve for the point of maximum or minimum values.
Why you differentiate problems in maths?
Firstly, and most simply, it may lead to another part of a problem or question or context.Secondly, and importantlyest, (i just invented that word it means most importantly) It can be used to find the gradient of a curve.As you may know, the gradient of a straight line is constanty=mx+c the gradient is mBut for a curve, the gradient is always changing.look at a graph of y=x2 and you will see that an infinite tangents make up the curve each with a different value for m.In short, when you differentiate a function or equation, you get the gradient function, which allows you to find the gradient at any point on the graph y=f(x)differentiate y=x2 (which is the same as find dy/dx)dy/dx=2xso the gradient on the curve y=x2 always 2 times the x value in question. At x=1 the gradient is 2. At x=2 the gradient is 4.In mechanics, if you draw a graph of displacement against time for a moving object, the GRADIENT is equal to the function of velocity. Plot velocity against time and the gradient is equal to Acceleration against time.In any circumstance where a curve is involved differentiation is needed to ind a gradient.Phew, hope that helps and answers your question
What is a gradient function?
Assume you want to know what is the formula of the gradient of the function in multivariable calculus. Let F be a scalar field function in n-dimension. Then, the gradient of a function is: ∇F = <fx1 , fx2, ... , fxn> In the 3-dimensional Cartesian space: ∇F = <fx, fy, fz>
What is the formula for gradient?
Assume you want to know what is the formula of the gradient of the function in multivariable calculus. Let F be a scalar field function in n-dimension. Then, the gradient of a function is: ∇F = <fx1 , fx2, ... , fxn> In the 3-dimensional Cartesian space: ∇F = <fx, fy, fz>
What is gradient formula?
Assume you want to know what is the formula of the gradient of the function in multivariable calculus. Let F be a scalar field function in n-dimension. Then, the gradient of a function is: ∇F = <fx1 , fx2, ... , fxn> In the 3-dimensional Cartesian space: ∇F = <fx, fy, fz>
What is the formula of gradient?
Assume you want to know what is the formula of the gradient of the function in multivariable calculus. Let F be a scalar field function in n-dimension. Then, the gradient of a function is: ∇F = <fx1 , fx2, ... , fxn> In the 3-dimensional Cartesian space: ∇F = <fx, fy, fz>
What is the formula for functions?
Assume you want to know what is the formula of the gradient of the function in multivariable calculus. Let F be a scalar field function in n-dimension. Then, the gradient of a function is: ∇F = <fx1 , fx2, ... , fxn> In the 3-dimensional Cartesian space: ∇F = <fx, fy, fz>
What is a positive gradient in math?
A positive gradient is a characteristic of a function whose value increases as the value of the argument increases. So, if y is a function, f(x), of x, then an increase in the value of x is accompanied by an increase in the value of y.
Why do you use gradient?
The derivative of a function, df/dx, is to single variable calculus as the gradient of a function, ∇f, is to multivariable calculus.If f is a function of three variables, x, y, and z, then the gradient of f is the vector function ∇f(x, y, z) = All of the uses of derivatives in single variable calculus are analogous to the uses of gradients in multivariable calculus:In single variable calculus the derivative tells us the instantaneous rate of change at some point, [x, f(x)]. In multivariable calculus, the gradient of a function tells us the instantaneous rate of change at some point, [x, y, f(x,y)], or if the function is of more than two variables, ∇f would tell us the instantaneous rate of change at a point [x, y, z, ….., f(x, y, z, ….)]. One Important difference in calculus of more than one variable is that a function can have many different rates of changes at one point. To understand why this is so, imagine that you are standing on a hilltop which is defined by a function of two variables f(x, y). The downward slope of the hill, the gradient, is different depending on the direction you look; to find the slope you need to specify a direction. This is why we take the 'directional derivative' which is simply the dot product of the gradient with a unit direction vector (the direction you are looking down the hill). For example suppose we want to find the instantaneous rate of change of the function f(x,y) = x2 + y2 at the point (2,1) in the direction of v = :The directional derivative in the direction of v = ∇f(x, y) ● = < ∂f/∂x(x, y, z), ∂f/∂y(x, y, z)> ● = ● = 2y evaluated at (2,1) = 2.Let's continue our comparison of derivatives and gradients. In single variable calculus a derivative of a function is equal to zero at a maximum or minimum value of the function. This fact can be used in practical applications that require maximizing and minimizing functions of one variable. The same is said of the gradient in multivariable calculus. By setting the gradient of a multivariable function equal to zero, we can solve for the point of maximum or minimum values.
Why you differentiate problems in maths?
Firstly, and most simply, it may lead to another part of a problem or question or context.Secondly, and importantlyest, (i just invented that word it means most importantly) It can be used to find the gradient of a curve.As you may know, the gradient of a straight line is constanty=mx+c the gradient is mBut for a curve, the gradient is always changing.look at a graph of y=x2 and you will see that an infinite tangents make up the curve each with a different value for m.In short, when you differentiate a function or equation, you get the gradient function, which allows you to find the gradient at any point on the graph y=f(x)differentiate y=x2 (which is the same as find dy/dx)dy/dx=2xso the gradient on the curve y=x2 always 2 times the x value in question. At x=1 the gradient is 2. At x=2 the gradient is 4.In mechanics, if you draw a graph of displacement against time for a moving object, the GRADIENT is equal to the function of velocity. Plot velocity against time and the gradient is equal to Acceleration against time.In any circumstance where a curve is involved differentiation is needed to ind a gradient.Phew, hope that helps and answers your question
How you find fixed points of a function?
The fixed points of a function f(x) are the points where f(x)= x.
What is the equation for the gradient of a function?
The equation for the gradient of a linear function mapped in a two dimensional, Cartesian coordinate space is as follows.The easiest way is to either derive the function you use the gradient formula(y2 - y1) / (x2 - x1)were one co-ordinate is (x1, y1) and a second co-ordinate is (x2, y2)This, however, is almost always referred to as the slope of the function and is a very specific example of a gradient. When one talks about the gradient of a scalar function, they are almost always referring to the vector field that results from taking the spacial partial derivatives of a scalar function, as shown below.___________________________________________________________The equation for the gradient of a function, symbolized ∇f, depends on the coordinate system being used.For the Cartesian coordinate system:∇f(x,y,z) = ∂f/∂x i + ∂f/∂y j + ∂f/∂z k where ∂f/(∂x, ∂y, ∂z) is the partial derivative of f with respect to (x, y, z) and i, j, and k are the unit vectors in the x, y, and z directions, respectively.For the cylindrical coordinate system:∇f(ρ,θ,z) = ∂f/∂ρ iρ + (1/ρ)∂f/∂θ jθ + ∂f/∂z kz where ∂f/(∂ρ, ∂θ, ∂z) is the partial derivative of f with respect to (ρ, θ, z) and iρ, jθ, and kz are the unit vectors in the ρ, θ, and z directions, respectively.For the spherical coordinate system:∇f(r,θ,φ) = ∂f/∂r ir + (1/r)∂f/∂θ jθ + [1/(r sin(θ))]∂f/∂φ kφ where ∂f/(∂r, ∂θ, ∂φ) is the partial derivative of f with respect to (r, θ, φ) and ir, jθ, and kφare the unit vectors in the r, θ, and φ directions, respectively.Of course, the equation for ∇f can be generalized to any coordinate system in any n-dimensional space, but that is beyond the scope of this answer.
f (x) = 2x dan g(x) = 2x+¹ -3?
You want to identify the relationship between two functions, namely f(x) = 2x and g(x) = 2x + 1 - 3. Let's take a closer look: Function f(x) = 2x: This is a linear function with a slope (gradient) coefficient of 2 and no vertical shift (y-intercept at 0). Function g(x) = 2x + 1 - 3: This is also a linear function, but with a vertical shift of -2 (subtracting 3 from 1, so -2). It has the same slope (gradient) coefficient as f(x), which is 2. In this case, g(x) is f(x) that has been "shifted" downwards by 2 units. The function g(x) has a similar shape to f(x), but its position is different on the y-axis.
How do you find the equation of a tangent line in calculus?
Suppose a curve is defined by the function y = f(x) and you want the equation of the tangent at the point A.Suppose the x-coordinate at A is p. Then use y = f(p) to find the y-coordinate of A and let that be q. So, the point A is (p, q).Next, find dy/dx, the derivative of the function f, with respect to q. This will be a constant or a functio of x. In the latter case, find its value for x = p. This is m, the gradient of the tangent.So now you have the tangent line with gradient, m which passes through the point (p, q) and so its equation isy - q = m(x - p)Simplify this into the required form: y = mx + c or ax + by + c = 0.
