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3 years ago

NEED HELP!!?? 40 pts See attached image (Pre-Cal)

Mathematics

1 answer:

Naya [18.7K]3 years ago

7 0

Answer:

8(cos(π) +i·sin(π))

Step-by-step explanation:

The number in parentheses has a magnitude of ...

√(1^2 +(√3)^2) = √4 = 2

Then the cube of that number will have a magnitude of 8, eliminating the 2nd and 4th choices

The angle of the number in parentheses is ...

arctan(-√3/1) = -π/3

Then 3 times that angle will be -π, also π.

The number of interest has a magnitude of 8 and an angle of π, so is written in the desired form as ...

8(cos(π) +i·sin(π)) . . . . . matches the 3rd choice

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The level curves F(t,z) = C for any constant C in the real numbers

where

$F(t,z)=z^3t^2+e^{tz}-4t+2z$

Step-by-step explanation:

Let's call

$M(t,z)=2tz^3+ze^{tz}-4$

$N(t,z)=3t^2z^2+te^{tz}+2$

Then our differential equation can be written in the form

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To see that is an exact differential equation, we have to show that

2) $\frac{\partial M}{\partial z}=\frac{\partial N}{\partial t}$

But

$\frac{\partial M}{\partial z}=\frac{\partial (2tz^3+ze^{tz}-4)}{\partial z}=6tz^2+e^{tz}+zte^{tz}$

In this case we are considering t as a constant.

Similarly, now considering z as a constant, we obtain

$\frac{\partial N}{\partial t}=\frac{\partial (3t^2z^2+te^{tz}+2)}{\partial t}=6tz^2+e^{tz}+zte^{tz}$

So, equation 2) holds and then, the differential equation 1) is exact.

Now, we know that there exists a function F(t,z) such that

3) $\frac{\partial F}{\partial t}=M(t,z)$

AND

4) $\frac{\partial F}{\partial z}=N(t,z)$

We have then,

$\frac{\partial F}{\partial t}=2tz^3+ze^{tz}-4$

Integrating on both sides

$F(t,z)=\int (2tz^3+ze^{tz}-4)dt=2z^3\int tdt+z\int e^{tz}dt-4\int dt+g(z)$

where g(z) is a function that does not depend on t

so,

$F(t,z)=\frac{2z^3t^2}{2}+z\frac{e^{tz}}{z}-4t+g(z)=z^3t^2+e^{tz}-4t+g(z)$

Taking the derivative of F with respect to z, we get

$\frac{\partial F}{\partial z}=3z^2t^2+te^{tz}+g'(z)$

Using equation 4)

$3z^2t^2+te^{tz}+g'(z)=3z^2t^2+te^{tz}+2$

Hence

$g'(z)=2\Rightarrow g(z)=2z$

The function F(t,z) we were looking for is then

$F(t,z)=z^3t^2+e^{tz}-4t+2z$

The level curves of this function F and not the function F itself (which is a surface in the space) represent the solutions of the equation 1) given in an implicit form.

That is to say,

The solutions of equation 1) are the curves F(t,z) = C for any constant C in the real numbers.

Attached, there are represented several solutions (for c = 1, 5 and 10)

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3 years ago