Directional Derivatives

Introduction

The term directional derivatives seem scary when we hear it for the first time. We have made efforts to make it simple and easy to understand. Directional derivatives come mostly under the topic of vector algebra, although it is applied widely in science too to solve problems of various other domains. We will discuss in detail about directional derivative and the derivation of its formulas in easiest possible ways. Moreover, we will take a glance at the concept of gradient descent(along with a simple problem explained for better understanding) and the maximum value of directional derivative. We would also take up some sample problems for the better understanding of the topic to the students.

What is directional derivative?

There shall be no simpler terms to define directional derivative but as to, it is the rate at which the direction of fX,Y changes in the direction of the unit vector u=⟨a,b⟩ .

In calculus the directional derivative is denoted as Dufx,y.

The exact definition of the directional derivative could be interpreted by the formula, 

Dufx,y=h→0fx+ah,y+bh-fx,yh

Dufx,y= fx(x, y)a+fy(x,y)b

Duf (x,y,z)=fx(x,y,z)a+fy(x,y,z)b+fz(x,y,z)c

Many times the definition of directional derivative sounds much more like a partial derivative.

What is gradient?

The gradient/gradient vector of ‘f’ could be defined as,

∇f =⟨fx, fy, fz⟩ or ∇f = ⟨fx, fy⟩

Gradient, when we use the standard basis vectors will be,

∇f =fx→i+fy→j or ∇f= fx→i + fy→j + fz→k

The derivation of directional derivative formula

Actually, it’s not a cakewalk to derive an equivalent formula to take up directional derivatives.

To make it simple let’s introduce a new function of single variable,

kz=fx0+az,yO+bz

Where x0, yO, a and b are fixed numbers.

This is a single variable function and z is the only variable.

By applying single variable function derivative, we get,

k’z=h→0kz+h-kzh

At z=0 the derivative turns out to be,

k’0=h→0kh-k0h

Now, substituting k(z) we get,

k'(0)=h→0f x0+ah,yO+bh-f x0, yOh

So, we get the following resulting relationship.

k′(0)=D_u ⃗ f(x0,y0)

Let’s rewrite k(z) as,

k(z) = f(x, y) where x=x0+az and y=y0+bz

Now, applying chain rule we get,

k’(z)=dg/dz= ∂f/∂x.dx/dz + ∂f/∂y. dy/dz = fx(x, y)a+ fy(x, y)

k’(z)= fx(x, y)a + fy(x, y)b                                                                      …………(2)

now taking z=0 we get that x=x0 and y=y0 (same as we defined x and y above) and put these into (2) in order to get,

k’(0)= fx(x0, y0)a + fy(x0, y0)b                     …..…(3)

Now, equating (1) and (3), we reach to the following resulted equation,

Duf (x0, y0)=g′(0)=fx(x0,y0)a+fy(x0,y0)b

When we allow x and y to be any number we reach to the formula of directional derivatives.

Duf (x,y)=fx(x,y)a+fy(x,y)b

Definition for more than two variables’ functions such as f(x, y, z), which is in the direction of the unit vector →u=⟨a,  b,  c⟩ is given by,

Duf (x, y, z) = fx(x,y,z)a + fy(x,y,z)b + fz(x,y,z)c

Maximum directional derivative

The maximum value of directional derivative and also the maximum rate of change of the function g(x) is given by ∥∇g(x)∥ and will occur in the direction given by ∇g(x).

Sample problem: Consider that the height of a feature above sea level is given by z=1000−0.01×2−0.02y2. If you are at the point (60, 100) in what direction the fastest elevation change is taking place? find the highest rate of change in elevation at this said point.

Solution. 

It is an elliptic paraboloid opening downwards. So, it has to be somewhat hilly.

Now, gradient could be defined as.

∇f(x→)=⟨−0.02x,−0.04y⟩

The direction of maximum rate of change of the elevation will be in the direction of ∇f(60,100)=⟨−1.2,−4⟩

The maximum rate of change of the elevation at this point the maximum rate of change could be determined as,

∥∇f(60,100)∥=√(−1.2)2+(−4)2=√17.44=4.176

The direction of maximum rate of change would be pointing up the hill towards the centre rather than away from the hill.

Directional derivative: sample problems

 Find each of the directional derivatives.

1Duf (→x) for f(x,y)=xcos(y) in the direction of →v=⟨2,1⟩.

Solution 

Duf (→x) for f(x,y)=xcos(y) in the direction of →v=⟨2,1⟩

Let’s first compute the gradient for this function.

∇f=⟨cos(y),−xsin(y)⟩

Also, as we saw earlier in this section the unit vector for this direction is,

→u=⟨2/√5,1/√5⟩

The directional derivative is then,

Duf (→x)=⟨cos(y),−xsin(y)⟩⋅⟨2/√5,1/√5⟩=1/√5(2cos(y)−xsin(y))

2. Duf (→x)  for f(x,y,z)= sin(yz) + ln(x2) at (1,1,π) in the direction of →v=⟨1,1,−1⟩

Solution. 

 Duf (→x) for f(x,y,z)=sin(yz)+ln(x2) at (1,1,π) in the direction of →v=⟨1,1,−1⟩. 

In such cases where we are asked for the directional derivative at a given point. Before moving further, we first need to find out the gradient at the given point, and then proceed to the dot product. So, let’s first find the gradient.

∇f(x,y,z) = ⟨2/x,zcos(yz), ycos(yz)⟩

∇f(1,1,π) = ⟨21,πcos(π),cos(π)⟩ = ⟨2,−π,−1⟩

Next is the requirement of unit vector for the direction,

∥→v∥=√3

→u=⟨1/√3,1/√3,−1/3⟩

Finally, the directional derivative at the point in question is,

D→uf(1,1,π)=⟨2,−π,−1⟩⋅⟨1/√3,1/√3,−1/√3⟩

                      =1√3(2−π+1)

                      =(3−π)/√3

Conclusion

The topic of directional derivatives has looked into what directional derivatives in simple words and gradient meaning. Moreover, it has also looked into the derivation of the formula for directional derivatives in different cases, in great detail. The various directional derivatives sample problems have been discussed under the topic. The FAQs section attempts to address the most probable queries that might arise. The FAQs section provides additional information which will aid a better understanding of the topic.