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# Calculate the slope of the tangent line to the intersection curve of the surface Z= ½√24-X² -2Y² with the plane Y=2 at the point $2,2, √3$

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## Answer to a math question Calculate the slope of the tangent line to the intersection curve of the surface Z= ½√24-X² -2Y² with the plane Y=2 at the point $2,2, √3$

4.7
First, we need to find the partial derivatives of the surface:
Z = \frac{1}{2} \sqrt{24 - X^2 - 2Y^2}

The partial derivatives with respect to $X$ and $Y$ are:
\frac{\partial Z}{\partial X} = \frac{\partial}{\partial X} \left$\frac{1}{2} \sqrt{24 - X^2 - 2Y^2} \right$
\frac{\partial Z}{\partial Y} = \frac{\partial}{\partial Y} \left$\frac{1}{2} \sqrt{24 - X^2 - 2Y^2} \right$

Calculate these partial derivatives:
\frac{\partial Z}{\partial X} = \frac{\partial}{\partial X} \left$\frac{1}{2} \sqrt{24 - X^2 - 2Y^2} \right$ = \frac{1}{2} \cdot \frac{1}{2 \sqrt{24 - X^2 - 2Y^2}} \cdot $-2X$ = -\frac{X}{2\sqrt{24 - X^2 - 2Y^2}}
\frac{\partial Z}{\partial Y} = \frac{\partial}{\partial Y} \left$\frac{1}{2} \sqrt{24 - X^2 - 2Y^2} \right$ = \frac{1}{2} \cdot \frac{1}{2 \sqrt{24 - X^2 - 2Y^2}} \cdot $-4Y$ = -\frac{2Y}{2\sqrt{24 - X^2 - 2Y^2}} = -\frac{Y}{\sqrt{24 - X^2 - 2Y^2}}

Evaluate the partial derivatives at the given point $$2, 2, \sqrt{3}$$:

Partial derivative with respect to $X$:
\frac{\partial Z}{\partial X}$2, 2, \sqrt{3}$ = -\frac{2}{2\sqrt{24 - 2^2 - 2$2$^2}} = -\frac{2}{2\sqrt{24 - 4 - 8}} = -\frac{2}{2\sqrt{12}} = -\frac{2}{2$2\sqrt{3}$} = -\frac{1}{2\sqrt{3}} = -\frac{\sqrt{3}}{6}

Partial derivative with respect to $y$:
\frac{\partial Z}{\partial Y}$2, 2, \sqrt{3}$ = -\frac{2}{\sqrt{24 - 2^2 - 2$2$^2}} = -\frac{2}{\sqrt{24 - 4 - 8}} = -\frac{2}{\sqrt{12}} = -\frac{2}{2\sqrt{3}} = -\frac{1}{\sqrt{3}} = \frac{-1}{\sqrt{3}} = -\sqrt{3}

The slope of the tangent plane at $$2,2,\sqrt{3}$$ in the direction of $Y$ is given by:
m = -\frac{\partial Z}{\partial Y} = -\sqrt{3}

Therefore, the slope of the tangent line in the direction of $Y$ when $Y = 2$:
m = -2

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