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otez555 [7]
2 years ago
14

Let f be a scalar field and F be a vector field. The following expressions either represent scalar fields, vector fields, or are

completely meaningless. Determine which of the three applies to each expression and briefly explain why.
(a) curl f
(b) Δf
(c) divF
(d) curl(Δf)
(e) ΔF
(f) Δ(divF)
(g) div(Δf)
(h) Δ(div f)
(i) curl(curlF)
(j) div(divF)
(k) (Δf) x (divF)
(l) div(curl(Δf))

Δ should be upside down as for a gradient sign. Thanks for any help you give! No need for in depth explanation, but it would help if you put a couple words down for how you got an answer.

Videos
Mathematics
1 answer:
Mama L [17]2 years ago
5 0

Answer:

(a) curl f - meaningless

(b) grad f -Δf- vector field

(c) div F - scalar field

(d) curl( grad f )curl (Δf)- vector field (e) grad F -ΔF- meaningless

(f) grad( div F )-Δ(divF) - vector field

(g) div( grad f ) - div(Δf)- scalar field

(h) grad ( div f ) -Δ(div f)-meaningless

(i) curl ( curl F ) - vector field

(j) div( div F ) - meaningless

(k) ( grad f ) x ( div F ) -(Δf) x (divF)-meaningless

(l) div( curl( grad f )) -div(curl(Δf))-scalar field

Step-by-step explanation:

(a) curl f - meaningless; a curl can only be taken of a vector field

(b) grad f - vector field; a gradient results in a vector field

(c) div F - scalar field; a divergence results in a scalar field

(d) curl( grad f ) - vector field; the curl of a vector field results in a vector field

(e) grad F - meaningless; a gradient can only be taken of a scalar field

(f) grad( div F ) -vector field ; the gradient of a scalar field is a vector field

(g) div( grad f ) - scalar field; the divergence of a vector field is a scalar field

(h) grad ( div f ) - meaningless; the divergence of a scalar field can not be taken

(i) curl ( curl F ) - vector field; the curl of a vector field is a vector field

(j) div( div F ) - meaningless; the divergence of a scalar field can not be taken

(k) ( grad f ) x ( div F ) - meaningless; a vector and scalar field cannon be crossed

(l) div( curl( grad f )) - scalar field; the divergence of a vector field is a scalar field

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Find T5(x) : Taylor polynomial of degree 5 of the function f(x)=cos(x) at a=0 . (You need to enter function.) T5(x)= Find all va
Burka [1]

Answer:

\bf cos(x)\approx1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{4!}=\\\\=1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}

The polynomial is an approximation with an error less than or equals to <em>0.002652</em> for x in the interval

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Step-by-step explanation:

According to Taylor's theorem

\bf f(x)=f(0)+f'(0)x+f''(0)\displaystyle\frac{x^2}{2}+f^{(3)}(0)\displaystyle\frac{x^3}{3!}+f^{(4)}(0)\displaystyle\frac{x^4}{4!}+f^{(5)}(0)\displaystyle\frac{x^5}{5!}+R_6(x)

with

\bf R_6(x)=f^{(6)}(c)\displaystyle\frac{x^6}{6!}

for some c in the interval (-x, x)

In the particular case f

<em>f(x)=cos(x) </em>

<em> </em>

we have

\bf f'(x)=-sin(x)\\f''(x)=-cos(x)\\f^{(3)}(x)=sin(x)\\f^{(4)}(x)=cos(x)\\f^{(5)}(x)=-sin(x)\\f^{(6)}(x)=-cos(x)

therefore

\bf f'(x)=-sin(0)=0\\f''(0)=-cos(0)=-1\\f^{(3)}(0)=sin(0)=0\\f^{(4)}(0)=cos(0)=1\\f^{(5)}(0)=-sin(0)=0

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\bf cos(x)\approx1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{4!}=\\\\=1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}

In order to find all the values of x for which this approximation is within 0.002652 of the right answer, we notice that

\bf R_6(x)=-cos(c)\displaystyle\frac{x^6}{6!}

for some c in (-x,x). So

\bf |R_6(x)|\leq|\displaystyle\frac{x^6}{6!}|=\displaystyle\frac{|x|^6}{6!}

and we must find the values of x for which

\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652

Working this inequality out, we find

\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652\Rightarrow |x|^6\leq1.90944\Rightarrow\\\\\Rightarrow |x|\leq\sqrt[6]{1.90944}\Rightarrow |x|\leq1.113826815

Therefore the polynomial is an approximation with an error less than or equals to 0.002652 for x in the interval

[-1.113826815, 1.113826815]

8 0
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Answer:

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y = amount invested at 9%

---

x + y = 20287

0.12x = 0.09y + 492.99

---

put the system of linear equations into standard form

---

x + y = 20287

0.12x - 0.09y = 492.99

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