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[SADT9] The Laplacian of a function f of n variables x 1 ,x 2 ,*** x n denoted nabla^ 2 f is defined by



nabla^ 2 f(x 1 ,x 2 ,***,x n ):= partial^ 2 f partial x 1 ^ 2 + partial^ 2 f partial x 2 ^ 2 +***+ partial^ 2 f partial x n ^ 2



Now assume that f depends only on r where r=(x 1 ^ 2 +x 2 ^ 2 +***+x n ^ 2 )^ 1 2 i.e. f(x 1 ,x 2 ,***,x n )=g(r) for some function g. Show that, for x 1 ,x 2 ,***,x n ne0 ,



nabla^ 2 f= n-1 r g^ prime (r)+g^ prime prime (r)

[SADT3] For scalar functions u and v, show that



B=( nabla u)*( nabla v)



is solenoidal and that



A= 1 2 (u nabla v-v nabla u)



is a vector potential for B, i.e. B= nabla* A

With respect to the bar of chocolate, where is their center of mass?

The initial x-coordinates of James and Ramon are -10.0 m and +10.0 m respectively,

so the x-coordinate of the center of mass is:

Formula:

Solution:

Final Answer


The Laplacian of a function f

 of n

 variables x

1

,x

2

,⋯x

n

, denoted ∇

2

f

 is defined by


2

f(x

1

,x

2

,⋯,x

n

):=∂

2

f

∂x

2

1


+∂

2

f

∂x

2

2


+⋯+∂

2

f

∂x

2

n


Now assume that f

 depends only on r

 where r=(x

2

1

+x

2

2

+⋯+x

2

n

)

1

2


, i.e. f(x

1

,x

2

,⋯,x

n

)=g(r)

, for some function g

. Show that, for x

1

,x

2

,⋯,x

n

≠0

,


2

f=n−1

r


g

(r)+g

′′

(r)



 If A

 and B

 are vector fields, prove the following:


∇(AB)=(B⋅∇)A+(A⋅∇)B+B×(∇×A)+A×(∇×B).

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Find the solution of the recurrence relation an = 4an−1 − 3an−2 + 2^n + n + 3 with


a^0 = 1 and a^1 = 4.

  1. D2((x+1)2/x-1)            ans, 8/(x-1)3
  2. D(ln(x2-2x+1))            ans, 2/x-1
  1. D(x2+2x-3)                  ans, 2x+2
  2. D2(xe3x-e4x)                 ans, 9xe3x+633x-16e4x

Solve 7x+9y=3


5x+7y=1 using matrix method

Jason estimates that his car loses 12% of its value every year. The initial value is $12,000. Which best describes the graph of the function that represents the value of the car after x years?

f(x) = 12,000(0.12)x, with a horizontal asymptote of y = 0

f(x) = 12,000(1.12)x, with a vertical asymptote of x = 0

f(x) = 12,000(0.88)x, with a horizontal asymptote of y = 0

f(x) = (12,000  0.88)x, with a vertical asymptote of x = 0


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