# How do you find the linearization of the function ##z=xsqrt(y)## at the point (-7, 64)?

The linear function that best aproximates ##z=x sqrt(y)## at ##(-7, 64)## is ##z = -56 + 8(x+7) – 7/16(y-64) = 28 + 8x – 7/16y##.

To get this result, we must first notice that ##z## is a function of the two variables ##x## and ##y##. Let’s write ##z=f(x,y)##. So, the best linear approximation ##L_(r_0)(x,y)## of ##f## at ##r_0 = (x_0,y_0) = (-7,64)## is given by

##L_(r_0) (x,y)= f(x_0, y_0) + vec(grad)f(x_0, y_0) * ((x,y)-(x_0,y_0))##

Where ##vec(grad)f## is the gradient of ##f## and ##*## is the dot product.

Geometrically, this linear approxiamtion is the tangent plane of ##f## at ##r_0##. The deduction of this equation is very similar to the deduction of the equation for the tangent line of a real function at a point, with the gradient ##vec(grad)f## playing the role of the derivative.

Now we need to calculate the components of the equations for the linear aproximation. ##f(x_0, y_0)## is simply the value of the function at ##(x_0, y_0)##:

##f(x_0, y_0) = f(-7, 64) = -7 times sqrt(64) = -56##

The gradient ##vec(grad)f(x,y)## of ##f## is given by the expression

##vec(grad)f(x, y) = ((del f)/(del x), (del f)/(del y)) = (sqrt(y), x/(2sqrt(y)))##

So, ##vec(grad)f(x_0, y_0) = (sqrt(64), -7/(2sqrt(64))) = (8, -7/16)##

Finally, we have:

##L_(r_0) (x,y)= -56 + (8, -7/16) * ((x,y)-(-7,64)) =## ##= – 56 + (8, -7/16) * (x+7, y-64) =## ##= -56 + 8 (x – 7) – 7/16 (y – 64) =28 + 8x – 7/16y##