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

1 Answer

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#