Using linear approximation for a two variable function to estimate $0.999^{10}(1 + \sin(0.01))$

Question

I am trying to evaluate $$0.999^{10}(1 + \sin(0.01))$$ using linear approximation for a function with two variables, but I am a little confused as to how to do that, as I don't have any x or y terms.

If someone wouldnt mind explaining how I would go about this, I would be very grateful.

Thanks Corey

Answer 1

Consider the function: $$ f(\varepsilon,\delta)=(1-\varepsilon)^{10}(1+\sin(\delta)) $$ Now expand around $(0,0)$ as far as the linear terms in $\varepsilon$ and $\delta$ and put $\varepsilon=0.001$ and $\delta=0.01$.

Answer 2

$$0.999^{10}(1 + \sin(0.01)) \approx (1-0.001)^{10}(1 + 0.01) \\\approx (1-0.01)(1+0.01) =1-0.0001 = 0.9999 $$

As a check, the OS X calculator gives 0.99994516400519 from $0.999^{10} =0.990044880209748 $ and $\sin(0.01) =0.009999833334167 $.