Question

Suppose F(m,r)=Gm/r^2.

(a) At any point (m,r), the differential is
dF=


(b) At the point (100,9), the differential is
dF=


(c) At the point (100,9) with dm=0.2 and dr=−0.2, the differential is
dF=

Answer 1

$(a)\\\displaystyle F = \frac{Gm}{r^2}\\ \frac{\partial F}{\partial r} = -2\frac{Gm}{r^3}, \,\frac{\partial F}{\partial m} = \frac{G}{r^2} \\ \textsf{The total differential}\hspace{0.1cm}\\ \mathrm{d}F\hspace{0.1cm}\textsf{at}\hspace{0.1cm} (m, r)\hspace{0.1cm}\textsf{is given by}\\ \mathrm{d}F = \frac{\partial F}{\partial r}\delta r + \frac{\partial F}{\partial m}\delta m\\ \mathrm{d}F = \frac{G}{r^2}\cdot\delta m - 2\frac{Gm}{r^3}\delta r\\ (b)\\ \textsf{At}\hspace{0.1cm} (100, 9)\\\mathrm{d}F = \frac{G}{9^2}\delta m - 2\cdot\frac{100G}{9^3}\delta r\\ \mathrm{d}F = \frac{G}{81}\delta m - \frac{200G}{729}\delta r\\ (c)\\ \textsf{Given that}\hspace{0.1cm} \delta r = -0.2, \delta m = 0.2\\ \begin{aligned} \mathrm{d}F &= \frac{G}{81}(0.2) - \frac{200G}{729}(-0.2) \\&= 0.2G\left(\frac{200}{729} +\frac{1}{81}\right) \\&= 0.2G\cdot\frac{209}{729} = \frac{209G}{3645} \end{aligned}$