Math 121 - Calculus for Biology I Spring Semester, 2011 Derivative of ex and ln(x) - Examples
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1. Differentiation of Exponentials and Logarithms
2. Graphing Exponentials and Logarithms
3. Polymer Drug Delivery System
4. Radioactive Decay
5. von Bertalanffy Equation 

Example 1: Find the derivatives of the following functions:

Solution: The derivative of f(x)

• The rules of differentiation give

• For g(x), we first use the properties of logarithms to change the expression so that it only includes ln(x)

 

• The domain of this function is all values of x
• The y-intercept satisfies

• The x-intercept satisfies

• For large values of x, the exponential function decays to zero
• There is a horizontal asymptote to the right with

y = -1.

• Often when there is an exponential function, there is a horizontal asymptote going either to the right (negative constant times x) or to the left (positive constant times x).

• The graph of the function is given by

• The derivative of this function satisfies

y' = 2(-0.2) e^-0.2x = -0.4 e^-0.2x.

• Since the exponential function is always positive, the derivative is always negative
• The derivative does approach zero as x becomes large (approaching the horizontal asymptote)
• This function is always decreasing

• The derivative for this function is given by

• The second derivative is given by

• This function is only defined for x > 0
• There is no y-intercept
• Tthere is a vertical asymptote at x = 0

• Extrema are found by setting the derivative equal to zero
• The derivative is at x = 1
• This gives an extremum at (1, 1)

• The second derivative is always positive
• This function is concave upwards
• The extremum is a minimum

The graph of the function is given by

Example 4: Polymer Drug Delivery System

• Drugs are often administered by a pill or an injection
  • The body receives a high dose rapidly
  • The drug remaining in the blood disappears exponentially
• Scientists invented polymers that are implanted to deliver a drug or hormone
  • These deliver the drug (or hormone) for a much longer period of time
  • Drug doses can be lower
• There are several long term birth control devices that are injected just below the skin
  • These devices deliver the hormones estrogen and progesterone
  • The delivery gives a more uniform level of the hormones over extended periods of time to prevent pregnancy
• These new drug delivery devices are a hot area of research for a variety of medical conditions
  • Diabetes sufferers could receive a more uniform level of insulin
  • Chemotherapeutic drugs to cancer patients could extend over a much longer period of time at lower doses to maximize their efficacy

Model for Injection of a Drug

• The body clears a drug relatively quickly by either filtration of the kidneys or metabolism of the drug
• Mathematically, this is described by

• k(t) is the concentration of the drug
• A₀ relates to the dose in the polymer delivery device
• q is a kinetic constant depending on how the patient clears the drug

Model for a Polymer Drug Delivery Device

• The polymer drug delivery devices have a long time period of decaying release of a drug or hormone
• The body clears the drug or hormone out relatively quickly
• Mathematically, this is described by two decaying exponentials

• c(t) is the concentration of the drug
• C₀ relates to the dose in the polymer delivery device
• r relates to the decay of the polymer, releasing the drug (q > r)
• q is a kinetic constant depending on how the patient clears the drug

• The amounts of drug are the same when

A₀ = C₀/r.

• Consider the specific example where the injected drug satisfies

k(t) = 1000e^-0.2t,

• k(t) is a concentration in mg/dl and the time t is in days

• The same amount of drug is delivered by a polymer drug delivery device satisfies

• c(t) is a concentration in mg/dl

• Find the rate of change in concentration for both k(t) and c(t) at t = 5 and 20
• Determine the maximum concentration of c(t) and when it occurs
• Graph each of these functions

Solution:

• The derivatives of k(t) and c(t) are

k '(t) = (-0.2)1000e^-0.2t = -200e^-0.2t

c'(t) = 10 (-0.01e^-0.01t - (-0.2)e^-0.2t) = 2 e^-0.2t - 0.1 e^-0.01t.

• The rate of change of the drug concentrations at times t = 5 and 20 for the injected drug is

k '(5) = -200 e^-0.2(5) = -73.58 mg/dl/day

k '(20) = -200 e^-0.2(20) = -3.66 mg/dl/day

• The rate of change of the polymer delivered drug is

c '(5) = 2 e^-0.2(5) - 0.1 e^-0.01(5) = 0.64 mg/dl/day

c '(20) = 2 e^-0.2(20) - 0.1 e^-0.01(20) = -0.045 mg/dl/day

Maximum for c(t)

• The maximum is found by taking the derivative of c(t) and setting it equal to zero

• Thus, t = ln(20)/0.19 = 15.767 days
• The maximum occurs at c(15.767) = 8.11 mg/dl.

• Below is the graph of the two functions, k(t) and c(t)

• This graph shows the large amount from the injection, which decays very rapidly
• The functions are equal when t = 24.29 days with concentrations of 7.766 mg/dl
• The injected drug falls off very rapidly, continuing its decline, while the polymer delivered drug maintains a relatively constant level over a much longer time

• Below is the graph of the polymer delivered drug over a longer period of time
• Its maximum concentration is easily seen in the graph
• These graphs show the obvious advantages of the time released drug if it has serious side effects or toxicity

Example 5:

• Naturally-occurring potassium is not radioactive
• Radioactive isotopes are manufactured from natural potassium
• Potassium isotope 43 decays at a rate proportional to the amount of the radioactive potassium
• After it emits a beta ray, it reverts to calcium (⁴³K goes through a b- decay with a neutron changing to a proton to produce ⁴³Ca.)
• This can be used as a tracer for studying the effectiveness of potassium absorption in the body

• Suppose we begin with a 10 mCi (millicurie) sample of ⁴³K
• This substance has a half-life of 22 hours
• Find the amount of ⁴³K after 5, 20, and 50 hours
• Also, find the rate of loss of ⁴³K after 5, 20, and 50 hours.

Solution:

• If a radioactive substance has a half-life of x hours, then it decays exponentially to half its original amount in x hours
• Let K(t) be the amount of ⁴³K, then the amount of ⁴³K satisfies

• where the constant k must be determined based on the half-life
• The amount of K(t) drops to 5 mg in 22 hours, so

• Taking the natural log of both sides,

• The expression for K(t) is

• With this expression for K(t), we can find the amount of ⁴³K after 5, 20, and 50 hours

K(5) = 10e^-5k ~ 10 e^-0.1575 ~ 8.54 mCi,

K(20) = 10e^-20k ~ 10 e^-0.6301 ~ 5.325 mCi,

K(50) = 10e^-50k ~ 10 e^-1.575 ~ 2.07 mCi.

• The rate of change in the amount of K(t) is given by the derivative of K(t)

• This expression for K '(t) gives the rate of change in the amount of ⁴³K after 5, 20, and 50 hours

• This gives an estimate at each time of how much beta radiation is coming from the sample of ⁴³K.

Example 6 (von Bertalanffy Equation):

• As a fish ages, it reaches a maximum size

a. Data for an average a lake trout shows

• 5.5 years to reach 2 kg
• 15 years to reach 5 kg
• The von Bertalanffy equation is

W(a) = 20.2(1 - e^-0.019a).

• Find the rate of change of weight, W, with respect to the age, a
• Graph the solution of the von Bertalanffy equation showing the intercepts and any asymptotes.

b. Solve the above equation for age, a, as a function of the weight, W

• Differentiate this function, finding the rate of change of age with respect to weight
• Graph this function showing any intercepts and asymptotes.

Solution: a. Write the von Bertalanffy equation as follows,

W(a) = 20.2 - 20.2 e^-0.019a

• Differentiating with respect to the age, a, gives

• This function is monotonically increasing (as we would expect for growth of a fish).

Graphing

• This equation goes through the origin
• For large values of a, the exponential decays to zero
• Asymptotically the fish grows to a weight of 20.2 kg (horizontal asymptote)

Below is a graph of the von Bertalanffy equation

b. Solve the equation

W = 20.2 - 20.2 e^-0.019a

It follows that

• The age, a, as a function of the weight, W is

a(W) = 158.2 - 52.63 ln(20.2 - W).

• This expression cannot be directly differentiated without the chain rule
• Try a substitution of Z = 20.2 - W (Note that dZ/dW = -1.)

a(Z) = 158.2 - 52.63 ln(Z),

and

• We will show

• Since Z = 20.2 - W and dZ/dW = -1, the formula gives

Graph (Inverse function for the von Bertalanffy equation)

• a(W) has a domain of W < 20.2
• There is a vertical asymptote at W = 20.2
• The derivative shows that this function is strictly increasing
• Since the function W(a) passes through the origin, its inverse function also passes through the origin

a(0) = 158.2 - 52.63 ln(20.2) = 0.

Below is the graph of this inverse function.