Laboratory 10

More Related Rates

In the last lab we investigated the chain rule and its application to related rate equations. We now look at examples of how we can use these related rate equations in solving certain problems.  The most common way these equations are used is to solve for certain rates of change when we know a related rate. 
 

Example 1 (spherical volume): 
 

The radius of a spherical cell is increasing at a constant rate of 0.4 cubic μm per day.  At what rate is the cell volume increasing when the radius is 10 μm? 
 

Before starting your calculation and writing your Matlab code you need to analyze the problem. The questions below should help you to solve related rates problems. 
 

The Questions to Ask Yourself When Doing Related Rates Problems 
 

Question 1: What is the independent variable in this problem? The independent variable is usually (but not always) time, which you traditionally notate using the symbol t. The key phrases that indicate time to be the independent variable are "how fast," "how often," "at what speed," "at what rate is this or that changing," and so on. In rare cases a related rate problem might have something other than time as its independent variable. The word, "per," or the phrase, "for each," gives you a clue that the word or phrase that follows is likely to be an independent variable.

In Example 1 the question is " At what rate…", which clear indicate that time is our independent variable. 
 

Question 2: What are the dependent variables? In the case of the problem above, the volume of the cell is clearly a dependent variable, since the problem asks you about it. Any variable that the problem mentions is probably going to be a dependent variable. Since the problem mentions the radius of the cell, it  is also a dependent variable.  
 

Question 3: What are the relationships between the variables? Again this often involves what you learned in geometry and trig. In the example above, we know that the cell has a spherical shape. And we know that the volume of the sphere is . That gives us the relationship between dependent variable, radius, and dependent variable, volume.  
 

Question 4: What are the derivatives of dependent variables? This just means you have to take some derivatives. You will almost always have to apply the chain rule, because one of your dependent variables will very likely be squared or cubed or under a square root, or be the argument of a trig function. Here for the example 1, using t as the independent variable:  
 

Question 5: What is the problem asking for? Very often it is asking for a rate. That means it is asking for the derivative of one of your dependent variables. The problem ask for the rate at which the volume increases. We see "rate" and "volume" in the same phrase, and that indicates that the problem is asking for the derivative of volume with respect to the independent variable, which in this case is time. We declared V to be our symbol for the dependent variable, volume, and t to be our symbol for the independent variable, time. So what the problem is asking you to solve for is. Look at the equations you have and identify where this term occurs and how you might solve for it in terms of the other variables.  
 

Question 6: What variables and derivatives does the problem give actual values for? Example 1 stated the rate the cell was growing, and the radius at which you were to establish the rate. We did identify the radius as a dependent variable. We observe that the problem gives a rate for that, which means that it assigns a value to the derivative of radius. It also assigns a value to the radius . Once you have identified all the values given, go back to the equation where you solved for what the problem was asking for and plug all the values in. That will enable you to get an answer.  
 
 
 

Question 7: What are the units for the answer?  The units for your answer depend on the units of the given values. In example 1 the radius is given in μm and the derivative of radius is in μm/day. We calculated the volume as a cube of the radius so the volume units are μm³ and the rate of volume change is μm³/day 
 

Now let's do the calculation in Matlab

syms r

v=4/3*pi*r^3

r10=10

dr_dt10=0.4

dv_dr=diff(v,r)

dv_dr10=subs(dv_dr,r,r10)

dv_dt10=dv_dr10*dr_dt10  
 

v =

4/3*pi*r^3

r10 =

    10

dr_dt10 =

    0.4000

dv_dr =

4*pi*r^2

dv_dr10 =

  1.2566e+003

dv_dt10 =

  502.6548

The cell is growing at a rate of 500 μm per day

Example 2 (continuous drug administration)

Sometimes it is important to deliver a drug with a continuous slow rate. The device research