#title: Uso de diferencial para cálculos aproximados 10 A

#author: Smirnov

#let:

n n1 = [2,3,4,5];
n m1 = [2,3,4,5];
n k1 = [3,4];

n a = [5,6,7,8,9];
n b = [5,6,7,8,9];
n d = [2,3,4];
f c = #a^#n1+#b^#m1-#d^#k1;
n da = [-0.003,-0.001,0.001,0.003];
n db = [-0.002,0.002];
f a1 =#a+#da;
f b1 =#b+#db;

f n2 = #n1-1;
f m2 = #m1-1;
f k2 = 1-#k1;

f r = #d+ #n1* #a^#n2* #d^#k2 *#da/#k1 + #m1* #b^#m2 *#d^#k2* #db/#k1;


#question:

\noindent Usando diferenciais, calcule um valor aproximado de $S=\sqrt[{#k1}]{{#a1}^{#n1}+{#b1}^{#m1}-#c}$.

#sugestion:



#resolution: 

Sejam $f(x,y)=\sqrt[{#k1}]{x^{#n1}+y^{#m1}-#c}$, $x=#a$, $y=#b$, $\Delta x=#da$ e $\Delta y=#db$. Então temos
$$
 S=f(x+\Delta x,y+\Delta y)\approx f(x,y)+
 \frac{\partial f(x,y)}{\partial x}\Delta x
 +
 \frac{\partial f(x,y)}{\partial y}\Delta y
$$
$$
 = #d+\frac{#n1}{#k1}  #a^{#n2} #d^{#k2} ( #da ) +
 \frac{#m1}{#k1}  #b^{#m2} #d^{#k2}( #db)
 = #r.
$$






#result:
 #r


#verification:

 #r type=f error=10 ** (-3) ;

#com precisão 10^-3