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4 years ago

Enter the solution (x,y) to the system of equations shown. y=5 , y=2x+16

Mathematics

1 answer:

ivolga24 [154]4 years ago

6 0

2x + 16 = 5

2x = -11

x = -5.5

y = 5

(-5.5,5)

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find an equation of the tangent plane to the given parametric surface at the specified point. x = u v, y = 2u2, z = u − v; (2, 2

Alexxandr [17]

The surface is parameterized by

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and the normal to the surface is given by the cross product of the partial derivatives of $\vec s$ :

$\vec n = \dfrac{\partial \vec s}{\partial u} \times \dfrac{\partial \vec s}{\partial v}$

It looks like you're given

$\begin{cases}x(u, v) = u + v\\y(u, v) = 2u^2\\z(u, v) = u - v\end{cases}$

Then the normal vector is

$\vec n = \left(\vec\imath + 4u \, \vec\jmath + \vec k\right) \times \left(\vec \imath - \vec k\right) = -4u\,\vec\imath + 2 \,\vec\jmath - 4u\,\vec k$

Now, the point (2, 2, 0) corresponds to u and v such that

$\begin{cases}u + v = 2\\2u^2 = 2\\u - v = 0\end{cases}$

and solving gives $u = v = 1$ , so the normal vector at the point we care about is

$\vec n = -4\,\vec\imath+2\,\vec\jmath-4\,\vec k$

Then the equation of the tangent plane is

$\left(-4\,\vec\imath + 2\,\vec\jmath - 4\,\vec k\right) \cdot \left((x-2)\,\vec\imath + (y-2)\,\vec\jmath + (z-0)\,\vec k\right) = 0$

$-4(x-2) + 2(y-2) - 4z = 0$

$\boxed{2x - y + 2z = 2}$

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Read 2 more answers

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Answer:

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cestrela7 [59]

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