Consider a function f(x,y) = sqrt(20-x2-7y2). Using linear approximation of f near (2,1), find how large the approximated value of f(1.95, 1.08) is from the exact value
a) 0.44%
b) 0.28%
c) 0.33%
d) 0.55%
e) None of the above

Question

Consider a function f(x,y) = sqrt(20-x2-7y2). Using linear approximation of f near (2,1), find how large the approximated value of f(1.95, 1.08) is from the exact value
a)	0.44%
b)	0.28%
c)	0.33%
d)	0.55%
e)	None of the above

anonymous · Answer

linear = f'(x_0)(x-x_0)
$$f'(x)=\frac{x+7y}{\sqrt{20-x^2-7y^2}}$$
can you do the rest?

anonymous · Answer

sry, forgot - sign infront of fraction

anonymous · Answer

No please

stamp · Answer

$$f(x,y)=\sqrt{20-x^2-7y^2}$$

stamp · Answer

@myko this is no single variable, this is multivariable

stamp · Answer

http://tutorial.math.lamar.edu/Classes/CalcIII/TangentPlanes.aspx

stamp · Answer

$$f(2,1)=3$$$$f_x(x,y)=-x/\sqrt{20-x^2-7y^2}$$$$f_y(x,y)=-7y/\sqrt{20-x^2-7y^2}$$

stamp · Answer

$$f_x(2,1)=-2/3$$$$f_y(2,1)=-7/3$$$$L(x,y)=3-\frac{2}{3}(x-2)-\frac{7}{3}(y-1)$$

stamp · Answer

$$L(1.95,1.08)=2.84667$$$$f(1.95,1.08)=2.83420$$$$2.83420x=2.84667$$$$x=2.84667/2.83420$$$$x=1.00439$$

So the linear approximation is ~ 0.44% greater than the actual value