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3 years ago

Can someone please help me .

Mathematics

1 answer:

aivan3 [116]3 years ago

8 0

The answer I got is r= -3.5

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Assume that all the given functions have continuous second-order partial derivatives. If z = f(x, y), where x = 9r cos(θ) and y

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Answer with Step-by-step explanation:

We are given that all the given functions have continuous second-order partial derivatives.

$z=f(x,y)$

Where $x=9rcos\theta,y=9r sin\theta$

We have to find

A. $\frac{\delta z}{\delta r}$

We know that

$\frac{\delta z}{\delta r}=\frac{\delta z}{\delta x}\frac{\delta x}{\delta r}+\frac{\delta z}{\delta y}\frac{\delta y}{\delta r}$

Using this formula

$\frac{\delta z}{\delta r}=9cos\theta \frac{\delta z}{\delta r}+9sin\theta\frac{\delta z}{\delta r}$

$\frac{\delta z}{\delta r}=\frac{x}{r}\frac{\delta z}{\delta r}+\frac{y}{r}\frac{\delta z}\delta y}$

B. $\frac{\delta z}{\delta \theta}$

$\frac{\delta z}{\delta \theta}=\frac{\delta z}{\delta x}\frac{\delta x}{\delta\theta }+\frac{\delta z}{\delta y}\frac{\delta y}{\delta\theta}$

$\frac{\delta z}{\delta \theta}=-9rsin\theta\frac{\delta z}{\delta x}+9rcost\theta\frac{\delta z}{\delta y}$

$\frac{\delta z}{\delta \theta}=-y\frac{\delta z}{\delta x}+x\frac{\delta z}{\delta y}$

C. $\frac{\delta^2 z}{\delta r\delta\theta}$

$\frac{\delta^2 z}{\delta r\delta\theta}=-9sin\theta\frac{\delta z}{\delta x}-y\frac{\delta^2z}{\delta x^2}(9cos\theta)+9 cos\theta\frac{\delta z}{\delta y}+x\frac{\delta^2z}{\delta y^2}(9sin\theta)$

$\frac{\delta^2 z}{\delta r\delta\theta}=-9sin\theta\frac{\delta z}{\delta x}-(81rsin\theta cos\theta)\frac{\delta^2z}{dx^2}+9cos\theta\frac{\delta z}{\delta y}+(81r cos\theta sin\theta)\frac{\delta^2z}{\delta y^2}$

$\frac{\delta^2 z}{\delta r\delta\theta}=-\frac{y}{r}\frac{\delta z}{\delta x}-\frac{xy}{r}\frac{\delta^2}{\delta x^2}+\frac{x}{r}\frac{\delta z}{\delta y}+\frac{xy}{r}\frac{\delta^2z}{\delta y^2}$

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3 years ago