3 years ago

(2-6i)+(4+3i) standard form

Mathematics

1 answer:

Mazyrski [523]3 years ago

5 0

6-3i. You add common factors. 2+4 is 6 and -6i and 3i is negative 3i

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Natali [406]

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$\dfrac{\partial z}{\partial u}=\dfrac{\partial z}{\partial x}\dfrac{\partial x}{\partial u}+\dfrac{\partial z}{\partial y}\dfrac{\partial y}{\partial u}$

where $u\in\{r,t\}$ .

We have component partial derivatives

$\dfrac{\partial z}{\partial x}=\dfrac{2x}{x^2+y}=\dfrac{2re^t}{r^2e^{2t}+te^r}$

$\dfrac{\partial z}{\partial y}=\dfrac1{x^2+y}=\dfrac1{r^2e^{2t}+te^r}$

$\dfrac{\partial x}{\partial r}=e^t$

$\dfrac{\partial x}{\partial t}=re^t$

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$\dfrac{\partial y}{\partial t}=e^r$

Putting the appropriate pieces together and setting $(r,t)=(1,2)$ , we get

$\dfrac{\partial z}{\partial r}(1,2)=\dfrac{2e^3+2}{e^3+2}$

$\dfrac{\partial z}{\partial t}(1,2)=\dfrac{2e^3+1}{e^3+2}$

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3 years ago