Calculus: The Language of Change

Exercise Set 7.5(CD7.1)
Implicitly Linked Variables

by John Robeson

Exercise 1) Blowing Up a Balloon

A child is blowing up a balloon by adding air at the rate of 2 cubic inches per second.  Well before it bursts, its radius is 6 inches. Assume that the balloon is a perfect sphere and find the surface area S as an explicit function of volume V. Use your explicit function to say how fast the surface is stretching at this point. 

HINTS: Give variables for surface area, volume, time, and whatever else you need. Solve for S in terms of V. The "chain" in this exercise is that volume depends on time (through the child's  efforts), V=V[t] and surface area depends on volume, S=S[V]. We want to know how area changes with time,(dS/dt) , where S=S[V[t]]. (See review Exercise CD-28.3.2 to express surface area as an explicit function of volume.)

Solution to Exercise 1)

We assume the following and will not consider the effect of pressure.

		V =((4 π (r^3))/3)
		
		S =4 π (r^2)
		
We know that (dV/dt) is 2(( (in^3))/sec)and that we need a relationship between V and S (which are both dependent on the radius r).  So we first solve both of the above equations for r.

		r = (((((3 V)/(4 π))))^(1/3))
		
		r = (Sqrt[(S/(4 π))])

\(We\ can\ now\ set\ these\ equal\ to\ each\ other\ and\ then\ solve
              \ for\ S . 

		\((\(3\ V\)\/\(4\ π\))\)\^\(1/3\)\  = \ 
            \@\(S\/\(4\ π\)\)\)		\(Raising\ both\ sides\ to\ the\ 6  th\ power\ gives\ 
              \(us : 		
		
		\((\(3\ V\)\/\(4\ π\))\)\^2\) = 
            \((S\/\(4\ π\))\)\^3\)		\(		\(9\ V\^2\)\/\(16\ π\^2\) = S\^3\/\(64\ π\^3\)\)

We now solve this for S
		\(		S\^3\  = \ 36\ π\ V\^2\)
		\(so		S\  = \((36\ π\ V\^2)\)\^\(1/3\)\)
\(Since\ V\ is\ a\ function\ of\ time, \ 
        we\ can\ now\ find\ \(dS\/dt\) . 


		\(dS\/dt\) = \ 
          \(\(dS\/dV\) dV\/dt\  = \ \ 
            \(\(\((36\ π\ V\^2)\)\^\(\(-2\)/3\)\ \((72\ π\ V)\)\)\/3\ 
                dV\/dt
		
		
		 = \ 
              \(24\ π\ V\)\/\((36\ π\ V\^2)\)\^\(2/3\)\ dV\/dt\)\)\)

\(TextForm
    \`We\ know\ that\ dV\/dt\ is\ 2 \(\(\ in\^3\)\/sec\) and\ that\ when\ r\ 
      is\ 6\ inches, \ 
    V\  = \ \(\(4\ π\ 6\^3\)\/3\  = \ 
        \(288\ \ π\ \ cubic\ inches . 

So\ \ \ \ 	dS\/dt\  = \ 
          \(24\ π\ 
                \((\ 288\ \ π)\)\)\/\((36\ π\ \((288\ π)\)\^2)
                  \)\^\(2/3\)\ \((2)\)\)\)\)

\(\(

\ 
    \(24\ π\ 
          \((\ 288\ \ π)\)\)\/\((36\ π\ \((288\ π)\)\^2)\)\^\(2/3
            \)\ \((2)\)\)\)

So (dS/dt) = (2/3)((in^2)/sec) which is the same answer we obtained implicitly in Example CD-7.4.

Exercise 2) The Explicit Fast Ladder

Solve the equation (x^2) + (y^2) = (L^2) for y = (Sqrt[(L^2) - (x^2)]).  Use the fact that x = x[t] is a function of time together with the Chain Rule to show that

	y'[t] = (dy/dt) = (dy/dx) . (dx/dt) = -x[t]/(Sqrt[(L^2) - (((x[t]))^2)]) . x'[t]
	
Verify that this model predicts that as x approaches L, the speed of the tip resting on the wall tends to infinity. 

The result of the previous exercise is wrong for a real ladder. The tip of such a ladder cannot go faster than the speed of light. There is a physical condition that this simple mathematical model does not take into account. The real falling ladder is explored in the Scientific Projects Chapter on Mechanics. It uses Galileo's Law of Gravity from Chapter 9.

Solution to Exercise 2)

See Example CD-7.5, The Fast Ladder, for the implicit solution.

Here we will solve for y explicitly.

		(x^2) + (y^2) = (L^2)
		
		(y^2) = (L^2) - (x^2)
		
		y = (Sqrt[(L^2) - (x^2)])
		
The length of the ladder L is constant, but x and y vary with time.  So

		(dy/dt) =  (dy/dx). (dx/dt) = (1/2) (1/(Sqrt[(L^2) - (x^2)])) (-2 x) \(\(dx\/dt
\)\) 
		= ((-x)/(Sqrt[(L^2) - (x^2)])) (dx/dt) = ((-x)/y) (dx/dt) 
		
In this problem we consider  (dx/dt) as a given constant.  So we 
now find the limit of (dy/dt) (which is the speed of the tip resting on the wall) as x goes to L (which is as y goes to zero.)

\(\(Limit[\(\(-x\)\ c\)\/\@\(L\^2 - x\^2\), x -> L]

\)\)

Exercise 3) Related Rates Drill

1. Each edge of a cube is expanding at the rate of 3 inches per second. How fast is the volume expanding at the point where the volume equals 64 cubic inches? 
2. Each edge of a cube is expanding at the rate of 3 inches per second. How fast is the surface area expanding at the point where the volume equals 64 cubic inches? 
3. A 6-foot man walks away from an 8-foot lamp at the rate of 5 feet per second.  How fast is his shadow growing at the point where he is 7 feet from the lamp? 
4. A snowball melts at a rate proportional to its surface area, that is, it loses volume per unit time in this proportion. Say the constant of proportionality is k.  What is (dr/dt) , the rate of change of radius with respect to time? 
5. Your snow-cone has melted filling the conical holder with sticky liquid. The liquid is dripping from the point at the bottom at the rate of 1 liter per hour. If the cone is 7 centimeters high and 5 centimeters in diameter at the top, what is the rate of change of the height of the liquid when half of it has leaked out? (Note: The volume of a cone of height h and base radius r is V = (1/3) π (r^2) h.)

Solution to Exercise 3)

Solution to 3.1)

The volume of the cube is
		V = (x^3)
where x is the length of a side.  We know x is a function of t and that
		(dV/dt) = (dV/dx). (dx/dt)

		
		(dV/dx) = 3 (x^2) and (dx/dt) = 3 in/sec
		
so		(dV/dt) =  (3 (x^2) ) 3 = 9 (x^2)

When V = 64, x = 4 so at this point,

		(dV/dt) = (9) (4^2)= 144 ((in^3)/sec)

Solution to 3.2)

The surface area of the cube is
		S = 6 (x^2)
where x is the length of a side.  We know x is a function of t and that
		(dS/dt) = (dS/dx). (dx/dt)

		
		(dS/dx) = 12 x and (dx/dt) = 3 in/sec
		
so		(dS/dt) =  (12 x ) 3 = 36 x

When V = 64, x = 4 so at this point,

		(dS/dt) = (36) 4= 144 ((in^2)/sec)

Solution to 3.3)

To solve this problem we use two similar right triangles.  The larger one has a vertical side of 8 feet and a horizontal side of x + s, where x is the man's horizontal distance from the lamp and s is his shadow's length.  The smaller right triangle has a vertical side of 6 feet and a horizontal side of s.  For these two similar triangles we know that the ratio of their sides is equal.

		(8/(x + s)) = (6/s)
		
Solving for s we get

		8s = 6x + 6s
		2s = 6x
		s = 3x
We also know that

		(ds/dt) = (ds/dx) . (dx/dt)
		
		(ds/dx) = 3 and (dx/dt) = 5
		
		
so		(ds/dt) = (3) (5) = 15 (ft/sec)


Note that both (ds/dx) and  (dx/dt) are constant, so (ds/dt) is 15 (ft/sec) no matter what the man's distance is from the lamp.

Solution to 3.4)

We assume our snowball is a perfect sphere so

		V =((4 π (r^3))/3)
		
and		S =4 π (r^2)

The information given tells us that

		(dV/dt) = -k S
		
		
We know	(dr/dt) = (dr/dV) . (dV/dt)

Since		V =((4 π (r^3))/3)

Implicitly
		dV = ((4 π)/3)(3 (r^2)) dr
		
		
So		(dr/dV) = (1/(4 π (r^2)))

Therefore	(dr/dt) = (dr/dV) . (dV/dt)

		= ((1/(4 π (r^2)))) ( -k S)
		
But		S = 4 π (r^2)

So		(dr/dt) = -k

Solution to 3.5)

We again use similar triangles in solving this problem.  Our cone is 7 cm high and has a radius of 2.5 cm (half of the 5 cm diameter).  As the liquid leaks out, the ratio of liquid height to liquid radius remains(7/2.5)=  2.8

So		h = 2.8 r
or		r =(1/2.8) h

Implicitly	dr = (1/2.8) dh

We also know that the volume of the cone is

		V = (1/3) π (r^2) h
		
Implicitly	dV = (1/3) π  ( (r^2) dh + h 2 r dr)

Since		dr = (1/2.8) dh


we have	dV = (1/3) π  ( (r^2) dh + h 2 r  (1/2.8) dh)


or		(dV/dt) =  (1/3) π  ( (r^2)  + h 2 r  (1/2.8) ) (dh/dt)


		= (1/3) π ( ((((h/2.8)))^2) + ((2 (h^2))/(2.8^2)) )(dh/dt)


		= π ((((h/2.8)))^2) (dh/dt)


so		 (dh/dt) = ((2.8^2)/(π (h^2))) (dV/dt)

We are given that dV/dt = 1 (liter/hour) = 1000 ((cm^3)/hour)


Therefore	(dh/dt) = ((2.8^2)/(π (h^2))) 1000


		(dh/dt) = (7840/(π (h^2)))

Now we need to determine what h is when half of the liquid has leaked out.  The total volume of the cone is

		(Subscript[V, c])  = (1/3) π (r^2) h = (1/3) π (2.5^2) 7 =  ((43.75 π)/3)

Half of that is ((43.75 π)/6)(cm^3)

For this volume of liquid in our cone

		V =  ((43.75 π)/6) = (1/3) π (r^2) h
		
		
Recall that 	r =(1/2.8) h


So		V =  (1/3) π ((((1/2.8) h))^2) h


		V = (π/((3) (2.8^2)))(h^3) =((π (h^3))/23.52)
		
and		h = (((((23.52 V)/π)))^(1/3))

So when V is ((43.75 π)/6) ,


		h =(((((23.52 (43.75 π))/(6 π))))^(1/3))

\(\(\((\(23.52\ \((43.75\ π)\)\)\/\(6\ π\))\)\^\(1/3\)
\)\)

Finally we have

		(dh/dt) = (7840/(π (h^2))) =(7840/(π (5.5559^2)))

(7840/(π (5.5559^2)))

Therefore the rate of change of the height of the liquid when half of the liquid has leaked out 80.8458 (cm/hr).