Question

Growth of a Cancerous Tumor The volume of a spherical cancerous tumor is given by the function $V(r) = \frac{4}{3}\pi r^3$ where $r$ is the radius of the tumor in centimeters. Find the rate of change in the volume of the tumor with respect to its radius under the following conditions. (a) $r = \frac{8}{9}$ cm $cm^3/cm$ (b) $r = \frac{3}{2}$ cm $cm^3/cm$

          Growth of a Cancerous Tumor The volume of a spherical cancerous tumor is given by the function
$V(r) = \frac{4}{3}\pi r^3$
where $r$ is the radius of the tumor in centimeters. Find the rate of change in the volume of the tumor with respect to its radius under the following conditions.
(a) $r = \frac{8}{9}$ cm
____ $cm^3/cm$
(b) $r = \frac{3}{2}$ cm
____ $cm^3/cm$

Growth of a Cancerous Tumor The volume of a spherical cancerous tumor is given by the function
V(r) = (4)/(3)π r^3
where r is the radius of the tumor in centimeters. Find the rate of change in the volume of the tumor with respect to its radius under the following conditions.
(a) r = (8)/(9) cm
cm^3/cm
(b) r = (3)/(2) cm
cm^3/cm

Added by Miguel -Ngel O.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Shaiju T

03/22/2022

Step 1

Step 1: Given the volume function V(r) = (4/3)πr^3, we need to find the rate of change in the volume of the tumor with respect to its radius. Show more…

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Growth of a Cancerous Tumor The volume of a spherical cancerous tumor is given by the function V(r) = (4/3)πr^3, where r is the radius of the tumor in centimeters. Find the rate of change in the volume of the tumor with respect to its radius under the following conditions: Given: r = 8 cm (dV/dt) = q cm^2/cm^3 To find the rate of change in the volume of the tumor with respect to its radius, we need to differentiate the volume function V(r) with respect to r. dV/dr = d/dt[(4/3)πr^3] = (4/3)π * d/dr(r^3) = (4/3)π * 3r^2 = 4πr^2 Therefore, the rate of change in the volume of the tumor with respect to its radius is 4πr^2 cm^2/cm^3.