# Vector calculus - How to use the gradient?

• alexvenk
In summary: Distance in direction (1,2) gives an increase of .2 in T.Distance in direction (2/5,4/5) gives an increase of .12 in T.T will be stationary in direction (2/5,4/5) only.
alexvenk
Member warned about posting without the template
I have done part A (i think) really not sure where to begin with the rest of the parts, would appreciate a tip in the right direction, its revision for my first year physics exams in a few weeks.

Consider the funtion T in the plane (x,y), given by T=ln(x^2 + y^2)

at point 1,2

a) in which direction is most rapid increase in T

I did Grad(T) to get a vector which i think is in the direction of most rapid increase (2/5,4/5)

b) what distance in this direction gives an inrease of .2 in T

c) what distance in direction i + j gives and increase of .12 in T

d) in what directions will T be stationary.

I don't want solutions, just how to go about solving the problems

I have done part A (i think) really not sure where to begin with the rest of the parts, would appreciate a tip in the right direction, its revision for my first year physics exams in a few weeks.

Consider the funtion T in the plane (x,y), given by T=ln(x^2 + y^2)

at point 1,2

a) in which direction is most rapid increase in T

I did Grad(T) to get a vector which i think is in the direction of most rapid increase (2/5,4/5)

b) what distance in this direction gives an inrease of .2 in T

c) what distance in direction i + j gives and increase of .12 in T

d) in what directions will T be stationary.

I don't want solutions, just how to go about solving the problems

alexvenk said:
I have done part A (i think) really not sure where to begin with the rest of the parts, would appreciate a tip in the right direction, its revision for my first year physics exams in a few weeks.

Consider the funtion T in the plane (x,y), given by T=ln(x^2 + y^2)

at point 1,2

a) in which direction is most rapid increase in T

I did Grad(T) to get a vector which i think is in the direction of most rapid increase (2/5,4/5)

b) what distance in this direction gives an inrease of .2 in T

c) what distance in direction i + j gives and increase of .12 in T

d) in what directions will T be stationary.

I don't want solutions, just how to go about solving the problems

For (b): if you go along direction (2/5,4/5) from the point (1,2) you are looking at points of the form ##x = x(t) = 1 + (2/5)t, \:y = y(t) = 2 + (4/5)t##, where ##t > 0## is a scalar.

(DelT = delta T (change in T))

Turns out the best way to do it for those who are interested is you use DelT = GradT . r, to get the largest change in t (highest delT) r and GradT must be in the same direction. To work out how far in a certain direcction it changes by a certain amount, set delT to whatever you want the change to be (.2 for b) then set r to be a vecctor with magnitude a and direction the same as the direction it was in a, then simply solve for a. Do the same for part c, and finally for part d, set delT to 0 so GradT must be perpendicular to r, which is pretty easy to do by inspection.

For simplicity, direction of (2/5,4/5) is the same of (1,2).

## 1. What is a gradient in vector calculus?

A gradient in vector calculus is a mathematical operator that represents the rate of change or slope of a function in multiple dimensions. It is a vector that points in the direction of the greatest increase of the function at a specific point.

## 2. How is the gradient calculated?

The gradient is calculated by taking the partial derivatives of a multivariable function with respect to each of its variables and combining them into a vector. This can be represented as ∇f, where f is the function and ∇ is the gradient operator.

## 3. What is the relationship between the gradient and level curves?

The gradient is always perpendicular to the level curves of a function. This means that the gradient vector at a particular point will be tangent to the level curve at that point. In other words, the gradient points in the direction of the steepest ascent of the function at that point.

## 4. How is the gradient used in optimization?

The gradient is used in optimization to find the maximum or minimum values of a multivariable function. By taking the gradient and setting it equal to zero, we can find the critical points of the function. These points represent potential maximum or minimum values, and further analysis can determine which is the global maximum or minimum.

## 5. Can the gradient be used for vector fields?

Yes, the gradient can also be applied to vector fields in vector calculus. In this case, the gradient represents the direction and magnitude of the maximum change in the vector field at a specific point. It can be useful in analyzing the behavior of vector fields and predicting the direction of their flow.

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