The volume V of an ice cream cone is given by V = 2 3 πR3 + 1 3 πR2h where R is the common radius of the spherical cap and the cone, and h is the height of the cone. Use linearization to estimate the change in the volume when R changes from R = 1.5 inches to R = 1.6 inches, and h changes from h = 3 inches to h = 3.3 inches. Give your answer to two decimal places.

Answer:The change in volume is estimated to be 17.20 [tex]\rm{in^3}[/tex]Step-by-step explanation:The linearization or linear approximation of a function [tex]f(x)[/tex] is given by:[tex]f(x_0+dx) \approx f(x_0) + df(x)|_{x_0}[/tex] where [tex]df[/tex] is the total differential of the function evaluated in the given point.For the given function, the linearization is:[tex]V(R_0+dR, h_0+dh) = V(R_0, h_0) + \frac{\partial V(R_0, h_0)}{\partial R}dR + \frac{\partial V(R_0, h_0)}{\partial h}dh[/tex]Taking [tex]R_0=1.5[/tex] inches and [tex]h=3[/tex] inches and evaluating the partial derivatives we obtain:[tex]V(R_0+dR, h_0+dh) = V(R_0, h_0) + \frac{\partial V(R_0, h_0)}{\partial R}dR + \frac{\partial V(R_0, h_0)}{\partial h}dh\\V(R, h) = V(R_0, h_0) + (\frac{2 h \pi r}{3}  + 2 \pi r^2)dR + (\frac{\pi r^2}{3} )dh[/tex]substituting the values and taking [tex]dx=0.1[/tex] and [tex]dh=0.3[/tex] inches we have:[tex]V(R_0+dR, h_0+dh) =V(R_0, h_0) + (\frac{2 h \pi r}{3}  + 2 \pi r^2)dR + (\frac{\pi r^2}{3} )dh\\V(1.5+0.1, 3+0.3) =V(1.5, 3) + (\frac{2 \cdot 3 \pi \cdot 1.5}{3}  + 2 \pi 1.5^2)\cdot 0.1 + (\frac{\pi 1.5^2}{3} )\cdot 0.3\\V(1.5+0.1, 3+0.3) = 17.2002\\\boxed{V(1.5+0.1, 3+0.3) \approx 17.20}[/tex]Therefore the change in volume is estimated to be 17.20 [tex]\rm{in^3}[/tex]