Gradient

We discussed the gradient of a function of two variables yesterday.

Definition:
Let f(x,y) have continuous partial derivatives. The gradient of f(x,y) is the vector:

Another notation is grad f(x,y).

Alternative Form of the Directional Derivative (Using gradient):
There is an alternate way to get the directional derivative using the gradient of a function. If f is a differentiable function of x and y, then the directional derivative of f in the direction of the unit vector u is:

Important Properties of the Gradient:

1. If the gradient of the function is 0, then the directional derivative is 0 for all u.

2. The direction of maximum increase of f is given by the gradient of the function. The maximum value of the directional derivative is the magnitude of the gradient of the function.

3. The direction of minimum increase of f is given by the negative gradient of the function. The minimum value of the directional derivative is the magnitude of the gradient of the function.

Directional Derivatives Using Angles

We continued discussing directional derivatives today. After recalling how the formula for the directional derivative is given by:

where f(x,y) is a function of two variables and vector (a,b) is a unit vector, we noticed how it is not always necessary to take the directional vector and unitize it only to plug in variable values based on what point we are taking the derivative at. We can use angles instead.

How does this work? When we originally used vectors, we had vectors such as:

unitized it into:

found the partial derivatives with respect to each variable, and multiplied it out to get the directional derivative, according to the above mentioned formula.

If we use angles, our vector now looks like:

and quite conveniently, any value of theta plugged into this 'angle vector' will also be a unit vector. This can be easily proven by finding the magnitude of u.


Another convenient property of this 'angle vector' is that no point is necessary. The angle value implies movement in a certain direction along both the x and y-axes.

Happy deriving!

Directional Derivative

The definition of the directional derivative is:

By making things simpler, we arrive at:
Let vector u = be a unit vector, then,

And if we choose to make f a function of x, y, and z, then we can write the directional derivative as:

Linear approximation & Total differentials

Back when the Ancient Greeks didn't have calculators, they relied on a method called linear approximation to help estimate the values of functions. Ever wonder how our modern calculators do it? Probably the same way! By constructing a tangent line, we can approximate the value of an unknown point when given the value of a known point. The formula we've all seen last year is:





Now when we extend this to a 3-dimensional space, we construct a plane tangent to the function. This can be summarized by:

f(x+Δx,y+Δy) = f(x,y) + fx (x,y)Δx + fy (x,y)Δy

One example we did in class is:

f(x,y) = x^2+3xy-y^2
Find: f(2.05, 2.96) Known: f(2,3)
Therefore, Δx = 0.05 and Δy = .04

fx = 2x+3y f(x+Δx,y+Δy) = f(x,y) + fx (x,y)Δx + fy (x,y)Δy
fy = 3x-2y 13+0.05(13)-0.04(0) = 13.65



The total differential of a function is the sum of all the partial derivatives. Total differentials can help estimate the maximum error (Δf) in using the differential. It can be summarized by:

z = f(x,y)
dz = fx (x,y)dx + fy (x,y)dy OR Δz = fx (x,y)Δx + fy (x,y)Δy

the curvy d vs. d

Today in class we saw how confusing and intimidating the partial derivative can look when used as a result of the chain rule.

A recap of the difference between the derivative and the partial derivative.:

If w is a function of x and y, and x and y are functions of t, then w is essentially a function if t. That is why we can find the derivative of w with respect to t.
Using the chain rule, we arrive at this formula:

However, if w is a function of x and y, and x and y are functions of s and t, we can no longer find the derivative of w with respect of t. We can find the partial derivative of w with respect to t and s.

Parametric Integration

On Friday, we finished up 2-dimensional integration by looking at how to integrate parametric functions. Noticing how the standard parametric function is denoted as:

we determined that the formula for integration:

can be rewritten as:

by simple substitution.

We proceeded to looking at some problems. One of the more interesting ones was the parametric curve:

which looks like this:

We found that the function crosses the x-axis at t = sqrt 3. After integrating using our new formula to find the area bounded by the curve, we also discovered that the integral from t = -sqrt 3 to t = sqrt 3 was negative. Not only that, but 2x the integral from t = 0 to t = sqrt (3), which we thought would yield the correct answer, was also negative.

We got these results for two reasons, one for each end.

• When we integrated from t = 0 to t = sqrt 3, we were integrating the area denoted by green. As we know, the area under the x-axis is negative, resulting in a negative area when we multiplied by 2.
• When we integrated from t = -sqrt 3 to t = sqrt 3, two things happened:

1. For the first half, from t = -sqrt 3 to 0, we were integrating the function 'backwards', denoted by the red arrow.

2. The second half, from t = 0 to t = sqrt 3, we integrated the green area, which was already established to be negative.

What we gleaned from this problem is that when integrating parametric functions, both the "direction" of integration and the location of the area we are integrating (whether above or below the x-axis) must be taken into consideration.

Happy integrating!