Maths Group Presentation for Engineering

1. TEAM-2
DIRECTIONAL DERIVATIVE & GRADIENT

2. I would like to express my special thanks of gratitude to my faculty
Dr. Bhumika Choksi, who gave us the golden opportunity to do this
wonderful project of MATHS GROUP ACTIVITY on the topic of
"DIRECTIONAL DERIVATIVE & GRADIENT ”, . We came to learn about so
many new things we are really thankful to them. Secondly we would also
like to thank the group members for the cooperation and who helped and
gave their efforts to finalizing this project within the limited time frame.
ACKNOWLEDGMENT

3. INDEX
● INTRODUCTION
● Definition Of Directional Derivatives
● Directional Derivative example
● Application of Directional Derivative: Slope
● Gradient
● Definition of gradient
● Estimating Directional Derivative from level curve
● Example of level curve
● Function of 3 variable
● Properties of gradient
● The direction of gradient
● Application of gradient
● Application of gradient 2
● Gradient in real life
● Example of gradient
● Summary of ideas : Directional Derivative and the gradient vector

4. In the introduction to the directional derivative and the gradient, we illustrated the concepts behind the
directional derivative. The main points were that, given a multivariable scalar-valued function
1. the directional derivative Duf is a generalization of the partial derivative to the slope of f in a
direction of an arbitrary unit vector u,
2. the gradient ∇f is a vector that points in the direction of the greatest upward slope whose length is
the directional derivative in that direction, and
3. the directional derivative is the dot product between the gradient and the unit vector:
➼ Duf = ∇f ⋅ u
This introduction is missing one important piece of information: what exactly is the gradient? How can
we calculate it from f? It's actually pretty simple to calculate an expression for the gradient, if you can
remember what it means for a function to be differentiable
INTRODUCTION
DERIVATION OF DIRECTIONAL DERIVATIVE & GRADIENT

5. What does it mean for a function f(x) to be differentiable at the point x=a? The
function must be locally be essentially linear, i.e., there must be a linear approximation
L(x)=f(a)+Df(a)(x−a)
that is very close to f(x)for all x near a.The definition of differentiability means that, for
all directions emanating out of a,f(x) and L(x) have the same slope. We can therefore
calculate the directional derivatives of f at a in the direction of u:
Since Df(x) is a1×nrow vector and u is an n×1 column vector, the matrix-vector product is a
scalar. We could rewrite this product as a dot-product between two vectors, by reforming the
1×n matrix of partial derivatives into a vector. We denote the vector by ∇f and we call it the
gradient. We obtain that the directional derivative is as promised.
Duf(a)=∇f(a)⋅u

6. DIRECTIONAL
DERIVATIVE

7. Definition of Directional Derivative ?
➼The directional derivative of a multivariate
differentiable (scalar) function along a given
vector v at a given point x intuitively
represents
the instantaneous rate of change of the
function,
moving through x with a velocity specified by
v.

8. ➼The directional derivative of a scalar function f with respect to a
vector v at a point (e.g., position) x may be denoted by any of the
following:
➼It therefore generalizes the notion of a partial derivative, in which the
rate of change is taken along one of the curvilinear coordinate curves,
all other coordinates being constant.

9. COMPUTING THE DIRECTIONAL DERIVATIVE
Suppose we consider the situation where we are interested in the instantaneous rate
of change of f at a point (x_0,y_0) in the direction u=(u1,u2) where u is a unit vector.
The variables x and y are therefore changing according to the parameterization
x = x_0 + u_1t and y = y_0 + u_2t.
Observe that dx/dt = u_1 and dy/dt = u_2 for all values of t. Since u is a unit vector,
it follows that a point moving along this line moves one unit of distance per one unit of
time; that is, each single unit of time corresponds to movement of a single unit of
distance in that direction.

10. This observation allows us to use the Chain Rule to calculate the directional derivative,
which measures the instantaneous rate of change of f with respect to change in the direction
u, In particular, by the Chain Rule, it follows that
This now allows us to compute the directional derivative at an arbitrary point
according to the following formula.

11. ➼Use this weather map to estimate
the value of directional derivative of
the temperature function at reno in
south easterly direction
➼The unit vector directed toward the
sound east is
u=(i-j)/√2
➼We start by drawing a line through
reno towards south east.
➼We approximate the directional
derivative DuT by:
DIRECTIONAL DERIVATIVE EXAMPLE

12. ➼The average rate of change of the temperature
between the points where the line intersects the
isothermals
T=50 and T=60
➼The temperature at the point southeast of Reno is
T=60 degree F.
➼The temperature at the point northwest of Reno is
T=50 degree F.
➼The distance between these points looks to about 75
miles
➼So the rate of change of temp is in the south easterly
direction is:
➡Dut=60-50/75
➡Dut=10/75
➡Dut=0.13 degree F/mi.

13. APPLICATION OF DIRECTIONAL DERIVATIVE:SLOPE
We know that for a function f of two variables x and y, the partial derivatives ∂f/∂x and ∂f/∂y give
the slope of the tangent to the trace, or curve of intersection of the surface defined by
z = and vertical planes that are, respectively, parallel to the x- and y-coordinates axes.
Partial derivatives gives us an
Understanding of how a
Surface changes when we
Move in x or y direction.
The partial derivatives
∂f/∂x and ∂f/∂y, are restricted
To or just defined by a
Certain limit.

14. Therefore the partial derivatives give the slope and the curve of intersection that are, respectively,
parallel to the x- and y-coordinates axes.
But then what about other directions?
That’s where the directional derivative comes to play. We can generalize the partial derivatives to
calculate slope in any direction and that result is called directional derivative.
The first step in taking a directional derivative is specifying the direction.
So let’s take the direction u = <u1,u2>, where |u| = 1
x(s) = x_0 + u_1s and y(s) = y_0 + u_2s
And according to the definition,

15. EXAMPLE
➼You can use a partial derivative to
measure a rate of change in a coordinate
direction in three dimensions. To do this,
you visualize a function of two variables
z = f(x, y) as a surface floating over the
Xy-plane of a 3-D Cartesian graph.
The following figure contains a sample
function.

16. So intuitively, you expect that the partial
derivative
dz/dy(the rate of change in the direction
of the y-axis)
is 1.
You also expect that the partial
derivative
dz/dx (the rate of change in the
direction of the x-axis)
is 0.

17. GRADIENT

18. The gradient stores all the partial derivative information of a multivariable function. But it's
more than a mere storage device, it has several wonderful interpretations and many, many
uses.
The gradient of a function w=f(x,y,z) is the vector function
For a function of two variables z=f(x,y), the gradient is the two-dimensional vector
<fx(x,y),fy(x,y)>. This definition generalizes in a natural way to functions of more than three
variables.
A differential operator applied to a three-dimensional vector-valued function to yield a
vector whose three components are the partial derivatives of the function with respect to its
three variables.
GRADIENT

19. 1) The skier is located at the point with xy-coordinates (1, 1), and wants to ski downhill along
the steepest possible path. In which direction (indicated by a vector (a, b) in the xy-plane)
should the skier begin skiing?
Direction of greatest rate of decrease is opposite of direction of gradient.
∇g(x, y) = (−0.8x, −0.6y)
∇g(1, 1) = (−0.8, −0.6) ||∇g(1, 1)|| = 1
Gradient vector is already a unit vector, so unit vector in opposite direction is
u = −∇g(1, 1) = (0.8, 0.6)
A SKIER IS ON A MOUNTAIN WITH EQUATION:
Z= 100-0.4x2-0.3y2

20. 2) The skier begins skiing in the direction given by the xy-vector (a, b) you found in part (a), so the
skier heads in a direction in space given by the vector (a, b, c). Find the value of c.
The directional derivative in the direction u (or (a, b)),
gives the slope. which is the ratio of vertical change to horizontal change. In the direction of the
vector (a, b, c), this ratio is . So

21. We could find approximate values of directional derivatives from level curves by using
the techniques of the last section to estimate the x- and y-derivatives and then
applying Theorem 1. It is easier, however, to estimate a directional derivative directly
from the level curves by estimating an average rate of change in the specified
direction
ESTIMATING DIRECTIONAL DERIVATIVES FROM LEVEL
CURVES

22. EXAMPLE OF LEVEL CURVES
One common example of level curves occurs in topographic maps of
mountainous regions, such as the map shown below .
Gran Canyon topographic map
The level curves are curves of
constant elevation of the Gran
canyon

23. FUNCTION OF 3 VARIABLES:
GRADIENT IN 3 DIMENSIONS:
SOLUTION:
EXAMPLE:
Find the gradient of f(x,y,z)= sin2x+ sin2y+ sin2z

24. 1.) ∇(f+g) = ∇(f) + ∇(g)
2.) ∇(cf) = c∇(f) [c = constant]
3.) ∇(fg) = f(∇(g)) + g(∇(f))
4.) The gradient of a horizontal line is zero and hence the gradient of the x-axis
is zero.
5.) The gradient of a vertical line is undefined and hence the gradient of the y-
axis is undefined.
PROPERTIES OF GRADIENT

26. If the gradient of a function is non-zero at a point p, the
direction of the gradient is the direction in which the function
increases most quickly from p, and the magnitude of the
gradient is the rate of increase in that direction, the greatest
absolute directional derivative.

27. If you’ve ever had to walk up a steep hill, or been faced with descent down a slippery ski slope,
then you know that gradients are important. Suppose we are faced with a choice of walking up
one of these two hills:
APPLICATION OF GRADIENT
It’s pretty clear which one most of us would choose. We can tell just by looking that the hill on the right
is much steeper than the hill on the left, and we all know from bitter experience that each step up its
slope will be much harder work.

28. APPLICATION OF GRADIENT-2
GRADIENT IN STATION YARD:-
The gradient in station yards are quite flat due to following reason:
● To prevent standing vehicles from rolling and moving away
from yard due to the combined effect of gravity and strong
winds.
● To reduce the additional resistive forces required to start a
locomotive to the extent possible.

29. GRADIENT IN REAL LIFE

30. EXAMPLE:-
y=2/3 x^3-1 -
x^2
dy/dx= 2x^2 - 2x
2x^2 - 2x = 4
2x^2 - 2x -4 = 0
x^2 - x - 2 = 0
p = -2 , s = -1
f = -2x , 1x
x^2 - 2x + 1x -2 = 0
x(x-2) + 1(x-2) = 0
(x+1)(x-2)=0
x+1=0 or x-2=0
x = -1
or
x = 2
Find the values of x for which the gradient of the curve y=2/3 x^3 -
x^2 is 4?
Solution

31. Summary of Ideas: Directional Derivatives and the Gradient Vector
•The gradient vector of a function f, denoted rf or grad(f), is a vectors whose entries are the
partial derivatives of f.
• The gradient points in the direction of steepest ascent.
• The directional derivative, denoted Dvf(x, y), is a derivative of a f(x, y) in the direction of a
vector ~v. It is the scalar projection of the gradient onto ~v
This produces a vector whose magnitude represents the rate a function ascends (how steep it
is) at point (x, y) in the direction of ~v
• Both the gradient and the directional derivative work the same in higher variables.
• The maximum directional derivative is always |f|.
• This happens in the direction of the unit vector.