UPSC 2019 Maths Optional Paper 2 Q2c — Step-by-Step Solution

15 marks · Section A

Partial derivatives · Calculus · asked 8× in 13 yrs · Read the full method →

Question

Using differentials, find an approximate value of $f(4.1,4.9)$ where $f(x,y)=(x^3+x^2y)^{1/2}$ .

Total differential $df=f_x\,dx+f_y\,dy$ about a convenient base point $(4,5)$ where $f$ is an integer.

Take $(x_0,y_0)=(4,5)$ , near $(4.1,4.9)$ , where the value is clean:

f(4,5)=\sqrt{4^3+4^2\cdot5}=\sqrt{64+80}=\sqrt{144}=12.

Increments: $\Delta x=0.1,\ \Delta y=-0.1$ .

With $f=(x^3+x^2y)^{1/2}$ ,

f_x=\frac{3x^2+2xy}{2\sqrt{x^3+x^2y}},\qquad f_y=\frac{x^2}{2\sqrt{x^3+x^2y}}.

At $(4,5)$ , the radical is $12$ :

f_x(4,5)=\frac{3\cdot16+2\cdot4\cdot5}{2\cdot12}=\frac{48+40}{24}=\frac{88}{24}=\frac{11}{3},

f_y(4,5)=\frac{16}{24}=\frac{2}{3}.

df=f_x\,\Delta x+f_y\,\Delta y=\frac{11}{3}(0.1)+\frac{2}{3}(-0.1)=\frac{0.1}{3}(11-2)=\frac{0.9}{3}=0.3.

Hence

f(4.1,4.9)\approx f(4,5)+df=12+0.3.

\boxed{\;f(4.1,4.9)\approx 12.3.\;}