## University Calculus: Early Transcendentals (3rd Edition)

We are given that the % change in internal volume is 1% % change in external surface area is 1% let the diameter of the cylinder be x(internal) and y(external) then radius $r=\frac{x}{2}$ Volume of cylinder $V=\pi r^2 h=\pi (\frac{x}{2})^2h=\pi (\frac{x^2}{4})h$ on differentiating the above equation: $dV=\frac{\pi}{4}(2xhdx+x^2(dh))$ height is not changing so $dh=0$ $dV=\frac{\pi}{4}(2xh(dx))=\frac{\pi}{2}(xh(dx))$ so %change in volume =change in volume/original volume $1=\frac{dV}{V}$ $\frac{\frac{\pi}{2}(xh(dx))}{\pi (\frac{x^2}{4})h}=1$ ${\frac{\pi}{2}(xh(dx))}={\pi (\frac{x^2}{4})h}$ $\frac{dx}{x}=\frac{1}{2}$ so the internal diameter of the cylinder is changing by 0.5% for exterior diameter change external surface area of cylinder $S=2\pi r h=2\pi (\frac{y}{2})h=\pi ({y})h$ on differentiating the above euation $dS={\pi}(hdy+y(dh))$ height is not changing so $dh=0$ $dS={\pi}(h(dy))$ so %change in surface area =change in surface area/original surface area $5=\frac{dS}{S}$ $\frac{{\pi}(h(dy))}{\pi ({y})h}=5$ $\frac{dy}{y}=5$ so % error in external diameter is 5% and % change in internal diameter is of 0.5%