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?
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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>
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>
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>
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.
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