How do you find the linearization at (3,5,1) of # f(x,y,z) = sqrt(x^2 y^2 z^2)#?
and the parameterised tangent line :
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To find the linearization of ( f(x, y, z) = \sqrt{x^2 + y^2 + z^2} ) at the point (3, 5, 1), follow these steps:
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Compute the gradient of ( f(x, y, z) ): [ \nabla f(x, y, z) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) ] [ = \left( \frac{x}{\sqrt{x^2 + y^2 + z^2}}, \frac{y}{\sqrt{x^2 + y^2 + z^2}}, \frac{z}{\sqrt{x^2 + y^2 + z^2}} \right) ]
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Evaluate the gradient at the point (3, 5, 1): [ \nabla f(3, 5, 1) = \left( \frac{3}{\sqrt{3^2 + 5^2 + 1^2}}, \frac{5}{\sqrt{3^2 + 5^2 + 1^2}}, \frac{1}{\sqrt{3^2 + 5^2 + 1^2}} \right) ] [ = \left( \frac{3}{\sqrt{35}}, \frac{5}{\sqrt{35}}, \frac{1}{\sqrt{35}} \right) ]
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Define the linearization using the gradient and the given point: [ L(x, y, z) = f(3, 5, 1) + \nabla f(3, 5, 1) \cdot \left( \begin{array}{c} x - 3 \ y - 5 \ z - 1 \end{array} \right) ]
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Substitute the values and simplify: [ L(x, y, z) = \sqrt{35} + \left( \frac{3}{\sqrt{35}} \right)(x - 3) + \left( \frac{5}{\sqrt{35}} \right)(y - 5) + \left( \frac{1}{\sqrt{35}} \right)(z - 1) ]
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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