The curl of the vector field is given by

Next, we take the cross product. One can do this by multiplying across from the top left to the lower right, and continuing downward, and then subtracting the terms multiplied from top right to the bottom left. To find the given partial derivative of the function, we must treat the other variable(s) as constants.

A vector field is conservative if its curl produces the zero vector.

The curl of the vector field is given by

Next, we take the cross product. One can do this by multiplying across from the top left to the lower right, and continuing downward, and then subtracting the terms multiplied from top right to the bottom left. To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The vector field is conservative.

Solution 1)

This probably was deceptively easy and could have been very quickly solved without doing any calculations. This problem involves first the basic definition of a conservative vector field, and a useful theorem on conservative vector fields. 

1) A vector field  is conservative if there exists a scalar function  such that  is its' gradient. 

2) If a vector field is conservative, its' curl must be zero. 

 In other words, the curl of the gradient is always zero for any scalar function. 

In this problem we were given a scalar function . If we now compute the gradient, we obtain a vector field we will call   (the gradient is our vector field). Automatically we know it fits the definition of a conservative vector field because we know there is a scalar function which has  as its' gradient. That function is .  

Now we know that since the gradient is a conservative vector field, and therefore the curl must be equal to zero. 

Solution 2) 

Just for fun, let's see if it works by doing the actual calculation. 

Classify each vector field below as either conservative or non-conservative. 

This is a conservative vector field. This can be easily determined by computing the curl:

Because  the vector field  is conservative. 

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Connection to Physics! 

This vector field can also be shown to be conservative. In fact, all inverse square vector fields are conservative.

Common examples are Newton's Universal Law of Gravitation: 

 and Coulombs Law in electrostatics which gives the force exerted by on point charge  onto another point charge : 

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The easiest approach for a function defined in this way is to simply find a scalar function that has  as its' gradient. 

In general, the gradient of a vector field in terms of cylindrical coordinates  is written as: 

The vector field  in this example does not have any  or  dependency, so the first therm is the only non-zero component. Therefore, we only need to integrate the partial derivative in the  component in order to find

Now check to see if it works: 

Therefore, we have shown that there exists some scalar function  such that the gradient of  gives the vector field . Therefore,  is a conservative vector field. 

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In this case we must again conclude that vector field is conservative. All constant vector fields (vector fields for which every component is a constant) will always be conservative. To show this, start by integrating each component: 

Similarly, we can integrate the  and  components to obtain: 

Note that each expression for  is distinct despite the fact they're supposed to be the same function. The reason being is that we have no reason to anticipate that each component of the gradient will have all the information about the original function.

The differentiation with respect to  for instance to obtain the  component will "delete," all constant terms, and all terms consisting of variables being held constant. The trick is to include all unique terms from each of the three calculations and to simply add them and combine their constants of integration into one constant.  

As an exercise, you can test this to compute the gradient.  

You can also prove the vector field is conservative by showing that the curl is zero. This is obvious since each derivative in the curl will be equal to zero because all terms are constant. 

 The derivatives in the  component all vanish do due to the fact that neither partial derivative operators in the first row differentiate with respect to .

This shows that the curl is non-zero and therefore the vector field  is not conservative. 

First we need to evaluate the vector field evaluated along the curve. 

Now we need to find the derivative of 

Now we can do the product of  and .

Now we can put this into the integral and evaluate it.

The formula for work is given by

.

Writing our path in parametric equation form, we have

.

Hence

Plugging this into our work equation, we get

.

The line integral of a vector field is given by

So, we must evaluate the vector field on the curve:

Then, we take the derivative of the curve with respect to t:

Taking the dot product of these two vectors, we get

This is the integrand of our integral. Integrating, we get

The line integral of the vector field is equal to

The parameterization (using the corresponding elements of the curve) of the vector field is

The derivative of the parametric curve is

Taking the dot product of the two vectors, we get

Integrating this with respect to t on the given interval, we get

To calculate the line integral of the vector field, we must evaluate the vector field on the curve, take the derivative of the curve, and integrate the dot product on the given interval.

The vector field evaluated on the given curve is

The derivative of the curve is given by

The dot product of these is

Integrating this over our given t interval, we get