Chapter 3: Applications of Differentiation
Section 3.6: Related Rates
Example 3.6.5
A right-circular conical tank, whose cross-section through its axis is shown in Figure 3.6.4, is being filled with water at the constant rate λ.
At time t^, find h.t^, the rate of change of the height of the water, where t^ is the moment when the volume is k times the volume of the tank, 0<k≤1.
The dimensions of the tank and the rate of fill are all in consistent units. The height of the tank is H, while the radius of the opening is R. The varying radius of the circle at the level of the water is rt (green dotted line in Figure 3.6.5(a)), and the varying height of the water is ht.
Hint: The volume of the tank is π3 R2H
p1:=plot([[0,0],[2,5],[-2,5],[0,0]],style=line,color=black): p2:=plot([[0,0],[0,5]],style=line,linestyle=dot,color=red): p3:=plot([[0,3.5],[7/5,3.5]],style=line,linestyle=dot,color=green): p4:=plots:-textplot({[1,5.2,typeset(R)],[-1,5.2,typeset(R)],[-.3,4.3,typeset(H)],[.7,3.3,typeset(r(t))],[.3,2.4,typeset(h(t))]},font=[Times,12]): p5:=plots:-textplot({[-.2,3.5,A],[1.6,3.5,B],[.2,0,O],[0,5.2,C],[2,5.2,E]},font=[Times,BoldRoman,14]): plots:-display(p||(1..5),scaling=constrained, axes=none);
Figure 3.6.5(a) Conical tank
Solution
Mathematical Solution
In Figure 3.6.5(a), triangles OAB and OCE are similar, from which the proportion rtht=RH and the equation rt=RHht follows.
The time-varying volume of water in the tank is then Vt=π3RHht2ht=π3RH2h3t.
The height of the water at time t^, that is, ht^=H k1/3, is found from the equation
Vt^=π3RH2h3t^=k π3 R2H
The rate of change of ht is obtained by setting V. equal to λ, and evaluating at t=t^.
Vt
=π3RH2h3t
V.t
=π RH2h2t h.t
λ
=π RH2 H k1/32 h.t^
=π R2 k2/3 h.t^
The result is the surprising h.t^=λπ R2 k2/3, a quantity independent of the height of the tank, but not the radius.
Maple Solution
The following interactive calculations are an implementation of those in the Mathematical Solution, with the one exception that the symbol t^ is not used. (It's use is problematic for the Context Panel.)
Solve the equation Vt=k Vmax for ht^
Control-drag or type the equation Vt=k Vmax. Press the Enter key.
Context Panel: Solve≻Isolate Expression for≻ht
Context Panel: All Values
Context Panel: Select Element≻1
Context Panel: Simplify≻Assuming Positive
π3RH2h3t=k π3 R2 H
13πR2ht3H2=13kπR2H
→isolate for h(t)
ht=RootOf−H3k+_Z3
→all values
ht=kH31/3,ht=kH31/3−12/3,ht=−kH31/3−11/3
→select entry 1
ht=kH31/3
→assuming positive
ht=Hk1/3
Obtain V.t
Control-drag (or type) the expression for Vt. Press the Enter key.
Context Panel: Differentiate≻With Respect To≻t
π3RH2h3t
13πR2ht3H2
→differentiate w.r.t. t
πR2ht2ⅆⅆthtH2
Solve the equation V.t=λ for h.t^
Use the equation label to write the equation V.t=λ. Press the Enter key.
Context Panel: Solve≻Isolate Expression for≻diff(h(t),t)
Context Panel: Right-hand Side
=λ
πR2ht2ⅆⅆthtH2=λ
→isolate for diff(h(t),t)
ⅆⅆtht=λH2πR2ht2
→right hand side
λH2πR2ht2
In the expression for h.t^, replace ht^
Expression palette: Evaluation template Reference quantities via equation labels
Context Panel: Evaluate and Display Inline
x=a|f(x) = λπR2k2/3
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