How do you find the linearization at #(5,16)# of #f(x,y) = x sqrt(y)#?
Consider the surface
so
The linearization in the desired point is
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To find the linearization of ( f(x, y) = x\sqrt{y} ) at the point ( (5, 16) ), we first compute the partial derivatives of ( f(x, y) ) with respect to ( x ) and ( y ). Then, we evaluate these derivatives at the given point ( (5, 16) ). The linearization is then given by the equation of the tangent plane to the surface at that point, which is:
[ L(x, y) = f(a, b) + \frac{\partial f}{\partial x}(a, b)(x  a) + \frac{\partial f}{\partial y}(a, b)(y  b) ]
where ( (a, b) ) is the point of tangency. Substituting the given point ( (5, 16) ) into the equation, and plugging in the computed partial derivatives, we obtain the linearization.
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To find the linearization of ( f(x,y) = x\sqrt{y} ) at the point (5,16), follow these steps:

Find the partial derivatives of ( f(x,y) ) with respect to ( x ) and ( y ). [ f_x = \sqrt{y} ] [ f_y = \frac{x}{2\sqrt{y}} ]

Evaluate the partial derivatives at the given point (5,16). [ f_x(5,16) = \sqrt{16} = 4 ] [ f_y(5,16) = \frac{5}{2\sqrt{16}} = \frac{5}{8} ]

Use the formula for the linearization: [ L(x,y) = f(a,b) + f_x(a,b)(xa) + f_y(a,b)(yb) ]

Substitute the values into the formula: [ L(x,y) = f(5,16) + f_x(5,16)(x5) + f_y(5,16)(y16) ] [ L(x,y) = 5\sqrt{16} + 4(x5) + \frac{5}{8}(y16) ] [ L(x,y) = 20 + 4(x5) + \frac{5}{8}(y16) ]
Therefore, the linearization of ( f(x,y) = x\sqrt{y} ) at the point (5,16) is: [ L(x,y) = 20 + 4(x5) + \frac{5}{8}(y16) ]
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