Gradient and Directional Derivative - Multivariable Calculus Video Lecture - Engineering Mathematics

hi welcome back to recitation in lecture
you've learned about gradients and about
directional derivatives so I have a
problem here to test your understanding
of those of those objects and and and
your give you some practice computing
them so I have three functions here and
for each function I've given you a point
and a direction so what I'd like you to
do is compute the derivative the
functions then evaluate their gradients
at the point given and then compute the
directional derivative of the function
at the point in the direction given in
the direction of the vector given so
here for example in Part A I've given
you the function f of XY equal to x
squared y plus XY squared at the point X
equal minus 1 y equal 2 and the
direction in the direction of the vector
3 4 and then we've got two more examples
here G of X Y Z is equal to the square
root of x squared plus y squared plus Z
squared P is the point 2 6 minus 3 and V
is the direction 1 1 1 and our third
example H is a function of four
variables W X Y and C it's given by W X
plus w y plus W Z plus XY plus XZ plus y
Z at the point 2 0 minus 1 minus 1 in
the Direction 1 minus 1 1 minus 1 so why
don't you pause the video take some time
to work out these three problems come
back and we can work them out together
I hope you had some luck working out
these problems let's let's get started
on the on the first one so on the first
one our function f of XY is equal to x
squared y plus x y squared so to compute
the gradient I just have to compute the
first partials of our function and then
put them into a single vector so that's
what the gradient is so we have that the
gradient of F at X Y is equal to it's
the vector and its first component is
the is the first partial of F with
respect to X so that's going to be 2xy
plus y squared and it's second component
is the partial of F with respect to Y so
that's going to be x squared plus 2xy so
this is the gradient of our function f
now the question asks you to compute
this gradient at the particular point
minus 1/2 so we have gradient of F at
the point minus 1/2 is equal to well we
just put in X equal minus 1 Y well 2
into this formula so at X equal minus 1
y equals 2 this is 2 x minus 1 times 2
plus 2 squared so that's minus 4 plus 4
so that's just 0 and the second term is
minus 1 squared plus 2 times minus 1
times 2 so that's 1 plus 4 nope 1 plus
negative 4 so that's negative 3 all
right so this is the this is the
gradient of our function f this is the
gradient at the point minus 1/2 now we
are asked for the directional derivative
so we are asked for the directional
derivative of F in a direction so we're
given I didn't give you I gave you a
vector V but when we take directional
derivatives what we want is a unit
vector right so the vector V that I gave
you was right over here it was this
vector 3 4 and so we need to find the
unit vector U in the direction of V
in order to apply our usual formula so U
is just V divided by the length of V so
in this case the vector 3/4 by the
Pythagorean theorem it has length 5 so
this is equal to the vector 3/5 comma
4/5 and so our our directional
derivative in the direction of V which
is this this vector U is equal to it's
just equal to the gradient at that point
dotted with the with the direction
vector so this is equal to gradient of F
at that point dot u which in our case is
equal to where since we're interested at
the point P it's equal to zero minus
three dot our vector U which is 3/5 4/5
and that dot product is negative 12
fifths so in this direction the function
is decreasing at about this rate at the
given point all right that's part a
let's go on to Part B so you know you'll
probably notice that these questions are
all fairly similar I'm the only real
complication is is that I've been
increasing the number of variables but
of course you know how to take gradient
for a function of three variables as
well so in this case our function G of X
Y Z is given by the square root of x
squared plus y squared plus C squared so
if you like this is the function that
measures the distance of a point from
the origin and so we're asked first for
the gradient so the gradient of G
what do I do well again I just take
these partial derivatives so I take the
partial derivative here this is going to
be a tiny bit messy I'm afraid as I take
a partial derivative of this expression
with respect to X I've got to apply the
chain rule right because this is a
composition of functions here so okay so
I get what do I get I get the derivative
of the inside times the derivative of e
I so the outside function is a square
root so that's to the one half power so
I get 1 half times the thing inside to
the minus 1 half power so it's over 2
times the square root of x squared plus
y squared LC squared and then I need to
multiply by the derivative of the inside
which is x - X okay so and then the twos
cancel and that's x over square root of
x squared plus y squared plus Z squared
and similarly this is a nice symmetric
function and you know if I changed x and
y its it's the same so so the other
partial derivatives look very similar so
they're going to be Y over the square
root of x squared plus y squared plus Z
squared and Z over the square root of x
squared plus y squared plus Z squared
sorry that's a little bit of a long
formula there but there we have it
that's that's the derivative sorry not
the derivative that's the gradient the
vector of partial derivatives so now you
are asked also to compute this gradient
at a particular point so the point in
question I have to look back over here
and the point in question was this point
to 6 minus 3 so at the point 2 6 minus 3
we want to compute the gradient so we
just take these numbers back over to our
formula over there and we're going to
put them in so we take the gradient of G
at 2 6 minus 3 well ok so so this square
root of x squared plus y squared plus Z
squared appears in all terms so let's
compute that first so x squared is 4 y
squared is 36 and Z squared is 9 so I
have
those numbers together I get 49 and then
I take a square root of that and I get
seven so so these denominators are all
going to be seven and then up top I just
have XY and Z so okay so this is going
to be 2 over 7 comma 6 over 7 comma
minus 3 over 7 right so just what I get
by plugging the values at our point into
into this formula so this is the
gradient at that point and now once
again I want to compute the a particular
directional derivative of this function
so in order to do that I just need the
right unit vector and I need to dot it
so the direction that I asked you was
the direction V with coordinates 1 1 1
so of course again this isn't a unit
vector so we need to divide it by its
length to find the unit vector that
we're going to use so the length of this
vector is the square root of 1 squared
plus 1 squared plus 1 squared so that's
the square root of 3 so U is equal to
I'm just going to write it as 1 over the
square root of 3 times the vector 1 1 1
and the so the directional derivative D
gds in the direction u hat is again it's
what I get i dot the gradient with with
this Q so it's its gradient dot u which
in our case so that 1 over the square
root of 3 lives out front and now I dot
2 767s minus 3/7 with 1 1 1 and that's
just going to give me 2 7 s plus 6 7
minus 3/7 and ok so we can finish off
the arithmetic there and I think that's
five over seven square root of three if
I didn't make any mistake now the last
one very very similar now we have a
function of four variables but there's
really nothing new
there at all so let me rewrite so for
Part C we have H of W X Y Z is equal to
WX + w y plus W Z plus XY plus X Z plus
y Z so the gradient of H I just take the
four partial derivatives now here W is
first so I take the partial derivative
with respect to W is the first
coordinate so that's going to be X plus
y plus C and then I take the partial
derivative with respect to X as the
second coordinate so that's going to be
W plus y plus Z and similarly the last
two are W plus X plus C and W plus
that's a plus X plus y whoo okay so
those are my that's that's my gradient
those are my partial derivatives now I
asked you again for the gradient at a
particular point
that's grad H at the point 2 0 -1 -1 so
we can just plug that in here it is 0
plus minus 1 plus minus 1 is minus 2
here it's 2 plus minus 1 plus minus 1
that's a 0 here it's 2 plus 0 plus minus
1 so that's 1 and lastly it's 2 plus 0
plus minus 1 1 so that's the gradient of
this function H now the point at this
point P and now I gave you again I gave
you a vector V so this was the vector 1
minus 1 1 minus 1 and I asked you to
find the directional derivative at this
point in that direction again we need a
unit vector this isn't a unit vector so
you
is equal to V over the length of V well
what is the length of V in this case
it's over the square root of well we
square the coordinates and add them so
that's 1 plus 1 plus 1 plus 1 that's 4
square root is 2 so that's V over 2 and
finally I get that the that the
directional derivative dhts in this
direction
U is what I get when I when I dot this
gradient together with the with the
direction in which I'm in which I'm
heading so this is minus 2 0 1 1 the
gradient dotted with the unit vector of
the direction which we just computed and
so finally you could expand this out
yourself I won't write down the final
answer but it's just the same just a dot
product right all right so hopefully
this is a gotten you totally comfortable
with computing gradients and with
computing directional derivatives from
gradients so I'll end there