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Iteru [2.4K]
3 years ago
11

Suppose Upper F Superscript prime Baseline left-parenthesis x right-parenthesis equals 3 x Superscript 2 Baseline plus 7 and Upp

er F left-parenthesis 0 right-parenthesis equals 5. Find the value of Upper F left-parenthesis b right-parenthesis for b equals 0 comma 0.1 comma 0.2 comma 0.5 comma and 2.0.
Mathematics
1 answer:
Sedaia [141]3 years ago
4 0

It looks like you're given

<em>F'(x)</em> = 3<em>x</em>² + 7

and

<em>F</em> (0) = 5

and you're asked to find <em>F(b)</em> for the values of <em>b</em> in the list {0, 0.1, 0.2, 0.5, 2.0}.

The first is done for you, <em>F</em> (0) = 5.

For the remaining <em>b</em>, you can solve for <em>F(x)</em> exactly by using the fundamental theorem of calculus:

F(x)=F(0)+\displaystyle\int_0^x F'(t)\,\mathrm dt

F(x)=5+\displaystyle\int_0^x(3t^2+7)\,\mathrm dt

F(x)=5+(t^3+7t)\bigg|_0^x

F(x)=5+x^3+7x

Then <em>F</em> (0.1) = 5.701, <em>F</em> (0.2) = 6.408, <em>F</em> (0.5) = 8.625, and <em>F</em> (2.0) = 27.

On the other hand, if you're expected to <em>approximate</em> <em>F</em> at the given <em>b</em>, you can use the linear approximation to <em>F(x)</em> around <em>x</em> = 0, which is

<em>F(x)</em> ≈ <em>L(x)</em> = <em>F</em> (0) + <em>F'</em> (0) (<em>x</em> - 0) = 5 + 7<em>x</em>

Then <em>F</em> (0) = 5, <em>F</em> (0.1) ≈ 5.7, <em>F</em> (0.2) ≈ 6.4, <em>F</em> (0.5) ≈ 8.5, and <em>F</em> (2.0) ≈ 19. Notice how the error gets larger the further away <em>b </em>gets from 0.

A <em>better</em> numerical method would be Euler's method. Given <em>F'(x)</em>, we iteratively use the linear approximation at successive points to get closer approximations to the actual values of <em>F(x)</em>.

Let <em>y(x)</em> = <em>F(x)</em>. Starting with <em>x</em>₀ = 0 and <em>y</em>₀ = <em>F(x</em>₀<em>)</em> = 5, we have

<em>x</em>₁ = <em>x</em>₀ + 0.1 = 0.1

<em>y</em>₁ = <em>y</em>₀ + <em>F'(x</em>₀<em>)</em> (<em>x</em>₁ - <em>x</em>₀) = 5 + 7 (0.1 - 0)   →   <em>F</em> (0.1) ≈ 5.7

<em>x</em>₂ = <em>x</em>₁ + 0.1 = 0.2

<em>y</em>₂ = <em>y</em>₁ + <em>F'(x</em>₁<em>)</em> (<em>x</em>₂ - <em>x</em>₁) = 5.7 + 7.03 (0.2 - 0.1)   →   <em>F</em> (0.2) ≈ 6.403

<em>x</em>₃ = <em>x</em>₂ + 0.3 = 0.5

<em>y</em>₃ = <em>y</em>₂ + <em>F'(x</em>₂<em>)</em> (<em>x</em>₃ - <em>x</em>₂) = 6.403 + 7.12 (0.5 - 0.2)   →   <em>F</em> (0.5) ≈ 8.539

<em>x</em>₄ = <em>x</em>₃ + 1.5 = 2.0

<em>y</em>₄ = <em>y</em>₃ + <em>F'(x</em>₃<em>)</em> (<em>x</em>₄ - <em>x</em>₃) = 8.539 + 7.75 (2.0 - 0.5)   →   <em>F</em> (2.0) ≈ 20.164

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Answer:

x=# of boys

x+7=# of girls

x+x+7=25

2x+7=25

2x=25-7

2x=18

x=18/2

x=9 boys

x+7=16 girls

 

If you don't understand variables use trial and error.

 

Boys     Girls

1           8

2           9

3           10

4           11

5           12

6           13

7           14

8           15

9           16   9+16=25 students

 

You could start this way...

 

Boys     Girls

12        13

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3 years ago
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Answer:

P(t)=3,000,000-3,000,000e^{0.0138t}

Step-by-step explanation:

Since P(t) increases at a rate proportional to the number of people still unaware of the product, we have

P'(t)=K(3,000,000-P(t))

Since no one was aware of the product at the beginning of the campaign and 50% of the people were aware of the product after 50 days of advertising

<em>P(0) = 0 and P(50) = 1,500,000 </em>

We have and ordinary differential equation of first order that we can write

P'(t)+KP(t)= 3,000,000K

The <em>integrating factor </em>is

e^{Kt}

Multiplying both sides of the equation by the integrating factor

e^{Kt}P'(t)+e^{Kt}KP(t)= e^{Kt}3,000,000*K

Hence

(e^{Kt}P(t))'=3,000,000Ke^{Kt}

Integrating both sides

e^{Kt}P(t)=3,000,000K \int e^{Kt}dt +C

e^{Kt}P(t)=3,000,000K(\frac{e^{Kt}}{K})+C

P(t)=3,000,000+Ce^{-Kt}

But P(0) = 0, so C = -3,000,000

and P(50) = 1,500,000

so

e^{-50K}=\frac{1}{2}\Rightarrow K=-\frac{log(0.5)}{50}=0.0138

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P(t)=3,000,000-3,000,000e^{0.0138t}

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