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Gradient
The gradient of a scalar is basically the scalar multiplied by a vector to form another vector
...
If we are getting a gradient, we are after converting non-vector functions to
vector functions in terms of the variables in question (x,y,z)
The scalar is multiplied by ∇ called nabla
∇=

𝜕
𝜕𝑥

𝑖+

𝜕
𝜕𝑦

𝑗+

𝜕
𝜕𝑧

𝑘
𝑑

And we know that partial derivatives are different from ordinary derivatives ( )
...


For the above case, the solution of the second part is 0 (since y4 z2 has no x variable in it, it will
be zero) and the first part is 2xy3 (since the derivative of x2is 2x and we are keeping other
variables constant)
...


𝜕𝑥

If we are finding the partial derivative with respect to y, we only derive y and leave other
variables
...

The gradient of a scalar is then ∇ø (product of nabla and ø; where ø is the scalar given in terms
of x,y and z)
Gradient of a vector, ∇ø=

𝜕
𝜕𝑥

(ø)𝑖 +

𝜕
𝜕𝑦

𝜕

(ø)𝑗 +

Example

𝜕𝑧

If we are to find the gradient of x5 y3+y2 z4+xz2
We apply
∇ø=

𝜕
𝜕𝑥

(ø)𝑖 +

𝜕
𝜕𝑦

(ø)𝑗 +

𝜕
𝜕𝑧

(ø)𝑘

Where ø = x5 y3+y2z4+xz2
𝜕
𝜕𝑥

(ø )i =

𝜕
𝜕𝑥

(x5y3+y2 z4+xz2) i

(ø)𝑘

𝜕

= ( x5 y3+
𝜕𝑥

𝜕 2 4
𝜕
y z + xz2)
𝜕𝑥
𝜕𝑥

i

= (5x4y3+0+z2) i (since y2 z4 has no x variable, it is equal to zero
...
(1)
𝜕
𝜕𝑦

(ø )j =

𝜕
𝜕𝑦

(x5y3+y2 z4+xz2) j

𝜕

𝜕

𝜕𝑦

𝜕𝑦

= ( x5 y3+

y2z4+

𝜕

xz2) j

𝜕𝑦

= (3x5y2+2yz4+0) j (since xz2 has no y variable, it is equal to zero) ………………
...
(3)
Recall
∇ø=

𝜕
𝜕𝑥

(ø)𝑖 +

𝜕
𝜕𝑥

(ø)𝑗 +

𝜕
𝜕𝑥

(ø)𝑘

So we add the three together
∇ø = (5x4y3+0+z2 ) i + (3x5 y2+2yz4+0) j + (0+4y2z3+2xz) k
∇ø = (5x4y3+z2) i + (3x5y2+2yz4 ) j + (4y2 z3+2xz) k
And that is the gradient
...
For
example if the question says find the gradient at (1,2,3) it means that you need to substitute 1 for
x, 2 for y and 3 for z
...
This is the background to any question you
might encounter about the gradient of a scalar
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Divergent
Unlike the gradient, the divergent is the dot product of two vectors to produce a scalar
...
v
∇=

𝜕
𝜕𝑥

𝑖+

𝜕
𝜕𝑦

𝑗+

𝜕
𝜕𝑧

𝑘

The knowledge of dot product is used in this situation
...
i=1

i
...
k=0

j
...
k=0

k
...
k=1

j
...
j=0

Basically, the dot product of two like ones is 1 and the dot product of two un-likes is zero
For example,
(a)i+(b)j+(c)k
...
i + (b*e)j
...
k
= (a*d)*1 + (b*e)*1 + (c*f)*1
= ad + be + cf
As you can see, it was initially the product of two vectors making a scalar by dot product
...
v = (

𝜕
𝜕𝑥

𝑖+

𝜕
𝜕𝑦

𝑗+

𝜕
𝜕𝑧

𝑘)
...
v =

𝜕𝑉1
𝜕𝑉2
𝜕𝑉3
𝑖+
𝑗+
𝑘
𝜕𝑥
𝜕𝑦
𝜕𝑧

Example
Let’s find the divergent of x5 y3i +y2 z4j- xz2k at (1,2,3)
∇
...
(x5 y3i+ y2 z4j- xz2k)
∇
...
𝑖 +
𝑗
...
𝑘
𝜕𝑥
𝜕𝑦
𝜕𝑧

∇
...
v =
+
−
𝜕𝑥
𝜕𝑦
𝜕𝑧
∇
...
v (1,2,3) =5(1)4(2)3 + 2(2)(3)4 - 2(1)(3)
∇
...
v (1,2,3) = 358
The Curl
The curl is the product of two vectors delivering the result in vector form
...
This is accomplished by applying cross product
...

Example
Find the curl of x5y3 i +y2 z4 j+ xz2k at (1,2,3)
i

=i(

𝜕𝑥𝑧 2
𝜕𝑦

−

j

k

𝜕

𝜕

𝜕

𝜕𝑥

𝜕𝑦

𝜕𝑧

x5y3

y2 z4

xz2

𝜕𝑦 2 𝑧 4
𝜕𝑧

)–j(

𝜕𝑥𝑧 2
𝜕𝑥

−

𝜕𝑥 5 𝑦 3
𝜕𝑧

)+k(

𝜕𝑦 2𝑧 4
𝜕𝑥

−

𝜕𝑥 5 𝑦 3
𝜕𝑦

)

= i (0- 4y2 z3) – j (z2 -0) + k(0-3x5y2)
= i (-4y2 z3) – j (z2) + k (-3x5 y2)
∇ X v = -4y2 z3 i – z2 j - 3x5y2 k
∇ X v (1,2,3) = -4(2)2(3)3 i – (3)2 j – 3(1)5(2)2 k
∇ X v (1,2,3) = −432𝑖 − 9𝑗 − 12𝑘



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