Derivative of a Polynomial Function

Question

 Air is being pumped into a spherical balloon. The volume, V, in cubic centimetres, of the balloon is V=4/3πr^3, where the radius, r, is in centimetres. 

 

a) Determine the instantaneous rate of change of the volume of the balloon when its radius is 1.5cm

b) Find equations for the tangent lines

Answer 1

"Air is being pumped into a spherical balloon. The volume, V, in cubic centimeters, of the balloon is V=4/3πr^3, where the radius, r, is in centimeters"

 

Usually this type problem is given when the students are studying Related Rates.

 

If so, both volume and radius would be treated as functions of time. When you take the derivative with respect to time you get:

 

V(t) = 4/3 pi r(t)^3

 

dV/dt = 4/3 pi 3 r(t)^2 dr/dt

 

This equation has three variables: dV/dt, r(t), and dr/dt.

 

You would need to know the values of two of them to find the value of the third.

 

But we are given only that r = 1.5 cm and asked to find dV/dt. We need to know dr/dt.

Answer 2

V = 4/3 Π r³

 

(a) instantaneous rate of change of volume with respect to radius = dV/dr =4/3 Π 3r²

 

dV/dr at r= 1.5 = 4/3 Π  3 (1.5)² = 9∏

 

(b) equation of tangent at r = 1.5

 

slope = dV/dr

 

at r = 1.5, dV/dr = 9Π

 

Treat this as a graph with r on x axis and V on y axis

 

for r = 1.5 , V = 4/3 Π (1.5)³ = 4.5 Π

 

Slope intercept form of line equation

y = mx +b

 

for y = V = 4.5Π, x = r = 1.5 amd m= 9Π

 

4.5Π = 9Π(1.5) + b

 

b = -9∏

 

Equation of tanget

 

y = 9Πx  -9∏

 

:)