Chapter 3: Applications of Differentiation
Section 3.6: Related Rates
Example 3.6.4
Helium is pumped into a spherical balloon at the constant rate of 25 cu ft per min.
At what rate is the surface area of the balloon increasing at the moment when its radius is 8 ft?
Solution
The time at which dSdt is to be measured is the unknown, t^, but at this very moment the radius is 8. Hence, find S. as a function of r, without the need to obtain it as an explicit function of t.
Analysis
Given
Find
V=43π r3
S=4 π r2
V.t=25
rt^=8
S.t^
Define the functions Vt and St
Write Vt=… Context Panel: Assign Function
Vt=43 π r3t→assign as functionV
Write St=… Context Panel: Assign Function
St=4 π r2t→assign as functionS
Solve the equation V.t=25 for r.
Write the equation V.t=25 Press the Enter key.
Context Panel: Solve≻Isolate Expression for≻diff(r(t),t)
4⁢π⁢r⁡t2⁢ⅆⅆt⁢r⁡t=25
→isolate for diff(r(t),t)
ⅆⅆt⁢r⁡t=254⁢π⁢r⁡t2
In S.t, replace r. with its value from the equation V.t=25
Expression palette: Evaluation template Evaluate S.t at r. from the equation V.t=25
Context Panel: Evaluate and Display Inline
S.tx=a|f(x) = 50r⁡t
The rate at which the surface area is increasing at the moment t^, the moment when rt^=8, is 50/8=25/4 square feet per second.
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