How do you use the linear approximation to #f(x, y)=(5x^2)/(y^2+12)# at (4 ,10) to estimate f(4.1, 9.8)?

Question

Samantha Adkison · Accepted Answer

Tangent plane approximation for #f(4.1,9.8) = 0.77551#

The tangent plane to #f(x,y)-z=0# in #p_0 = {x_0,y_0,f(x_0,y_0}# can be obtained by doing

#Pi_0 = << p-p_0,vec n_0>> = 0#

where #p = {x,y,z}in Pi_0# is a generic point and #vec n_0# is the normal vector to the surface #f(x,y)-z=0# at point #p_0#

but #vec n_0 = grad (f(x,y)-z)_0 = {f_x,f_y,-1}_0#

so

#vec n_0 = {(10 x_0)/(12 + y_0^2), -(10 x_0^2 y_0)/(12 + y_0^2)^2,-1}#

or

#vec n_0 = {5/14, -25/196,-1}#

given that #p_0 = {4,10,5/7}# the tangent plane is

#55/98 + (5 x)/14 - (25 y)/196 - z=0#

Calculating the approximation for #f(4.1,9.8)# gives

#z=55/98 + (5 xx4.1)/14 - (25xx 9.8)/196 =0.77551#

and

#f(4.1,9.8) = 0.777953#

Ezra Adams · Answer

undefined To use linear approximation to estimate $ f(4.1, 9.8) $ using $ f(x, y) = \frac{5x^2}{y^2 + 12} $ at $ (4, 10) $:

1. Find the partial derivatives of $ f(x, y) $ with respect to $ x $ and $ y $ at $ (4, 10) $.
2. Compute the values of the partial derivatives at $ (4, 10) $.
3. Use the formula for linear approximation:
   $$ L(x, y) = f(a, b) + \frac{\partial f}{\partial x}(a, b)(x - a) + \frac{\partial f}{\partial y}(a, b)(y - b) $$
   where $ (a, b) $ is the point of approximation, and $ (x, y) $ is the point where you want to estimate $ f $.
4. Plug in the values into the formula:
   $$ L(4.1, 9.8) = f(4, 10) + \frac{\partial f}{\partial x}(4, 10)(4.1 - 4) + \frac{\partial f}{\partial y}(4, 10)(9.8 - 10) $$
5. Evaluate $ f(4, 10) $, $ \frac{\partial f}{\partial x}(4, 10) $, and $ \frac{\partial f}{\partial y}(4, 10) $.
6. Plug in the values into the formula and compute $ L(4.1, 9.8) $.