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kodGreya [7K]
4 years ago
14

Use implicit differentiation to find ∂z/∂x and ∂z/∂y. x2 + 4y2 + 7z2 = 1 ∂z/∂x = ________ ∂z/∂y = __________

Mathematics
1 answer:
Paha777 [63]4 years ago
8 0

Assuming z=z(x,y), differentiating both sides wrt x gives

2x+14z\dfrac{\partial z}{\partial x}=0\implies\dfrac{\partial z}{\partial x}=-\dfrac x{17z}

and wrt y gives

8y+14z\dfrac{\partial z}{\partial y}=0\implies\dfrac{\partial z}{\partial y}=-\dfrac4{7z}

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rusak2 [61]
Given

w-x+2y+4z=0 \\ 5w+2x-y+3z=0

We can rewrite it in matrix form as:

\left[\begin{array}{cccc}1&-1&2&4\\5&2&-1&3\\0&0&0&0\\0&0&0&0\end{array}\right]   \left[\begin{array}{c}w\\x\\y\\z\end{array}\right] =\left[\begin{array}{c}0\\0\\0\\0\end{array}\right] \\  \\ \left[\begin{array}{cccc}1&-1&2&4\\5&2&-1&3\\0&0&0&0\\0&0&0&0\end{array}\right|\left.\begin{array}{c}0\\0\\0\\0\end{array}\right]\ \ \ \ \ -5R_1+R_2\rightarrow R_2
\left[\begin{array}{cccc}1&-1&2&4\\0&7&-11&-17\\0&0&0&0\\0&0&0&0\end{array}\right|\left.\begin{array}{c}0\\0\\0\\0\end{array}\right]\ \ \ \ \  \frac{1}{7} R_2\rightarrow R_2 \\  \\ \left[\begin{array}{cccc}1&-1&2&4\\0&1&-\frac{11}{7}&-\frac{17}{7}\\0&0&0&0\\0&0&0&0\end{array}\right|\left.\begin{array}{c}0\\0\\0\\0\end{array}\right]\ \ \ \ \ R_1+R_2\rightarrow R_1
\\  \\ \left[\begin{array}{cccc}1&0&\frac{3}{7}&\frac{11}{7}\\0&1&-\frac{11}{7}&-\frac{17}{7}\\0&0&0&0\\0&0&0&0\end{array}\right|\left.\begin{array}{c}0\\0\\0\\0\end{array}\right] \\  \\ \Rightarrow w= -\frac{3}{7} y- \frac{11}{7} z \\ x=\frac{11}{7} y+ \frac{17}{7} z \\ y=free \\ z=free \\  \\ =y\left\ \textless \ -\frac{3}{7},\frac{11}{7}1,0\right\ \textgreater \ +z\left\ \textless \ - \frac{11}{7},\frac{17}{7},0,1\right\ \textgreater \

Thus, the solution set is a span of \{\left\ \textless \ -\frac{3}{7},\frac{11}{7}1,0\right\ \textgreater \ ,\left\ \textless \ - \frac{11}{7},\frac{17}{7},0,1\right\ \textgreater \ \}
7 0
3 years ago
HELP PLEASE ill give brainliest to whoever answers good
julsineya [31]

Answer:

ITS D

Step-by-step explanation:

I JUST DID THIS LOL

7 0
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