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 c

Question

3 months ago

13

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 c

one, 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.

Mathematics

1 answer:

AnnZ [12.3K] · Answer 1 · 2026-02-22T05:15:38+03:00

Answer:

The estimated change in volume is 17.20 $\rm{in^3}$

Step-by-step explanation:

The linear approximation of a function $f(x)$ is expressed as:

$f(x_0+dx) \approx f(x_0) + df(x)|_{x_0}$ where $df$ indicates the total differential of the function at the specified point.

<pFor the function provided, the linearization formula is:

$V(R_0+dR, h_0+dh) = V(R_0, h_0) + \frac{\partial V(R_0, h_0)}{\partial R}dR + \frac{\partial V(R_0, h_0)}{\partial h}dh$

Using $R_0=1.5$ inches and $h=3$ inches, we evaluate the partial derivatives to get:

$V(R_0+dR, h_0+dh) = V(R_0, h_0) + \frac{\partial V(R_0, h_0)}{\partial R}dR + \frac{\partial V(R_0, h_0)}{\partial h}dh\\V(R, h) = V(R_0, h_0) + (\frac{2 h \pi r}{3} + 2 \pi r^2)dR + (\frac{\pi r^2}{3} )dh$

Substituting the respective values and taking $dx=0.1$ and $dh=0.3$ inches, we arrive at:

$V(R_0+dR, h_0+dh) =V(R_0, h_0) + (\frac{2 h \pi r}{3} + 2 \pi r^2)dR + (\frac{\pi r^2}{3} )dh\\V(1.5+0.1, 3+0.3) =V(1.5, 3) + (\frac{2 \cdot 3 \pi \cdot 1.5}{3} + 2 \pi 1.5^2)\cdot 0.1 + (\frac{\pi 1.5^2}{3} )\cdot 0.3\\V(1.5+0.1, 3+0.3) = 17.2002\\\boxed{V(1.5+0.1, 3+0.3) \approx 17.20}$

Thus, the estimated change in volume is 17.20 $\rm{in^3}$