Lab 3: solutions

© 2005 Ben Bolker

Exercise 1*:

Quadratic: easiest to construct in the form ( $y = - (x - a)^{2} + b$ ), where $a$ is the location of the maximum and $b$ is the height. (Negative sign in front of the quadratic term to make it curve downward.) Thus $a = 5$ , $b = 1$ .
Ricker: if $y = {axe}^{- bx}$ , then (as discussed in the chapter) the location of the maximum is at $x = 1 / b$ and the height is at $a / (be)$ . Thus $b = 0.2$ , $a = 0.2 * e$ .
Triangle: let's say for example that the first segment is a line with intercept zero and slope 1/5, and the second segment has equation $- 1 * (x - 5) + 1$ .

> curve(-(x - 5)^2 + 1, from = 0, to = 10, ylim = c(0, 1.1), ylab = "")
> curve(0.2 * exp(1) * x * exp(-0.2 * x), add = TRUE, lty = 2)
> curve(ifelse(x < 5, x/5, -(x - 5) + 1), add = TRUE, lty = 3)

What else did you try? (Sinusoid, Gaussian (

\exp (- x^{2})

), ?)

Exercise 2*:

n (t) = \frac{K}{1 + (\frac{K}{n (0)} - 1) \exp (- r t)}

Since

n (0) << 1

(close to zero, or much less than 1),

K / n (0) - 1 \approx K / n (0)

. So:

n (t) \approx \frac{K}{1 + \frac{K}{n (0)} \exp (- r t)}

Provided

t

isn't too big,

K / n (0) \exp (- rt)

is also a lot larger than 1, so

n (t) \approx \frac{K}{\frac{K}{n (0)} \exp (- r t)}

Now multiply top and bottom by

n (0) / K \exp (rt)

to get the answer.

Exercise 3*: When

b = 1

, the Shepherd function reduces to

RN / (1 + aN)

, which is a form of the M-M. You should try not to be confused by the fact that earlier in class we used the form

ax / (b + x)

(asymptote=

a

, half-maximum=

b

); this is just a different parameterization of the function. To be formal about it, we could multiply the numerator and denominator of

RN / (1 + aN)

1 / a

to get our equation in the form

(R / a) N / ((1 / a) + N)

, which matches what we had before with

a = R / a

b = 1 / a

Near 0: we can do this either by evaluating the derivative

S' (N)

N = 0

(which gives

R

- see below) or by taking the limit of the whole function

S (N)

N \to 0

, which gives

RN

(because the

aN

term in the denominator becomes small relative to 1), which is a line through the origin with slope

R

For large $N$ : if

b = 1

, we know already that this is Michaelis-Menten, and in this parameterization the asymptote is

R / a

(in the limit, the 1 in the denominator becomes irrelevant and the function becomes approximately

\frac{RN}{aN} = \frac{R}{a}

). If

b

is not 1 (we'll assume it's greater than 0) we can start the same way (

1 + aN \approx aN

), but now we have

RN / (aN)^{b}

. Write this as

\frac{R}{a^{b}} N^{(} 1 - b)

. If

b > 1

N

is raised to a negative power and the function goes to zero as

N \to \infty

. If

b < 1

N

is raised to a positive power and

R (N)

approaches infinity as

N \to \infty

(it never levels off).

b = 0

then the function is just a straight line (no asymptote), with slope

R / 2

We don't really need to calculate the slope (we can figure out logically that it must be negative but decreasing in magnitude for large

N

and

b > 1

; positive and decreasing to 0 when

b = 1

; and positive and decreasing, but never reaching 0, when

b > 1

. Nevertheless, for thoroughness (writing this as a product and using the product, power, and chain rules):

\begin{matrix} (RN (1 + aN)^{- b})' & = & R (1 + aN)^{- b} + RN \cdot - b (1 + aN)^{(- b - 1)} a & (1) \\ = & R (1 + aN)^{- b} - abRN (1 + aN)^{(- b - 1)} & (2) \\ = & R (1 + aN)^{- b - 1} ((1 + aN) - abN) & (3) \\ = & R (1 + aN)^{- b - 1} (1 + aN (1 - b)) & (4) \end{matrix}

You could also do this by the quotient rule. The derivative of the numerator is

R

(easy); the derivative of the denominator is

b \cdot (1 + aN)^{b - 1} \cdot a = ab (1 + aN)^{b - 1}

(power rule/chain rule).

\begin{matrix} S (N)' & = & \frac{g (N) f' (N) - f (N) g' (N)}{{(g (N))}^{2}} & (5) \\ = & \frac{R (1 + aN)^{b} - RN (ab (1 + aN)^{b - 1})}{{(1 + aN)}^{2 b}} & (6) \\ = & \frac{R (1 + aN)^{b - 1} (1 + aN - abN)}{{(1 + aN)}^{2 b}} & (7) \end{matrix}

You can also do this with R (using D()), but it won't simplify the expression for you:

> dS = D(expression(R * N/(1 + a * N)^b), "N")
> dS

R/(1 + a * N)^b - R * N * ((1 + a * N)^(b - 1) * (b * a))/((1 + 
    a * N)^b)^2

If you want to know the value for a particular

N

, and parameter values, use eval() to evaluate the expression:

> eval(dS, list(a = 1, b = 2, R = 2, N = 2.5))

[1] -0.06997085

A function to evaluate the Shepherd (with default values

R = 1

a = 1

b = 1

> shep = function(x, R = 1, a = 1, b = 1) {
+     R * x/(1 + a * x)^b
+ }

Plotting:

> curve(shep(x, b = 0), xlim = c(0, 10), bty = "l")
> curve(shep(x, b = 0.5), add = TRUE, col = 2)
> curve(shep(x, b = 1), add = TRUE, col = 3)
> curve(shep(x, b = 1.5), add = TRUE, col = 4)
> abline(a = 0, b = 1, lty = 3, col = 5)
> abline(h = 1, col = 6, lty = 3)
> legend(0, 10, c("b=0", "b=0.5", "b=1", "b=1.5", "initial slope", 
+     "asymptote"), lty = rep(c(1, 3), c(4, 2)), col = 1:6)

extra credit: use the expression above for the derivative, and look just at the numerator. When does

(1 + aN - abN) = (1 + a (1 - b) N) = 0

? If

b \leq 1

the whole expression must always be positive (

a \geq 0

N \geq 0

). If

b > 1

then we can solve for

N

\begin{matrix} 1 + a (1 - b) N & = & 0 & (8) \\ a (b - 1) N & = & 1 & (9) \\ N = 1 / (a (b - 1)) & (10) \end{matrix}

When

N = 1 / (a (b - 1))

, the value of the function is

R / (a \cdot (b - 1) \cdot (1 + 1 / (b - 1 {))}^{b})

(for

b = 2

this simplifies to

R / (4 a)

> a = 1
> b = 2
> R = 1
> curve(shep(x, R, a, b), bty = "l", ylim = c(0, 0.3), from = 0, 
+     to = 5)
> abline(v = 1/(a * (b - 1)), lty = 2)
> abline(h = R/(a * (b - 1) * (1 + 1/(b - 1))^b), lty = 2)

There's actually another answer that we've missed by focusing on the numerator. As

N \to \infty

, the limit of the derivative is

\frac{R (aN)^{b - 1} (a (1 - b) N)}{(aN)^{2 b}} = \frac{R (1 - b)}{(aN)^{b}};

R > 0

(1 - b) < 0

for

b > 1

aN > 0

, so the whole thing is negative and decreasing in magnitude toward zero.

Exercise 4*: Holling type III functional response, standard parameterization:

f (x) = {ax}^{2} / (1 + {bx}^{2})

Asymptote: as

x \to \infty

{bx}^{2} + 1 \approx {bx}^{2}

and the function approaches

a / b

Half-maximum:

\begin{matrix} {ax}^{2} / (1 + {bx}^{2}) & = & (a / b) / 2 \\ {ax}^{2} & = & (a / b) / 2 (1 + {bx}^{2}) \\ {ax}^{2} & = & (a / b) / 2 (1 + {bx}^{2}) \\ (a - a / 2) x^{2} & = & (a / b) / 2 \\ x^{2} & = & (2 / a) (a / b) / 2 = 1 / b \\ x & = & \sqrt{1 / b} \end{matrix}

So, if we have asymptote

A = a / b

and half-max

H = \sqrt{1 / b}

, then

b = 1 / H^{2}

and

a = Ab = A / H^{2}

f (x) = \frac{(A / H^{2}) x^{2}}{1 + x^{2} / H^{2}}

which might be more simply written as

A (x / H)^{2} / (1 + (x / H)^{2})

Check with a plot:

> holling3 = function(x, A = 1, H = 1) {
+     A * (x/H)^2/(1 + (x/H)^2)
+ }
> curve(holling3(x, A = 2, H = 3), from = 0, to = 20, ylim = c(0, 
+     2.1))
> abline(h = c(1, 2), lty = 2)
> abline(v = 3, lty = 2)

Exercise 5*:

Population-dynamic:

n (t) = \frac{K}{1 + (\frac{K}{n (0)} - 1) \exp (- r t)}

Asymptote

K

, initial exponential slope

r

, value at

t = 0

n (0)

, derivative at

t = 0

r n (0) (1 - n (0) / K)

Statistical:

f (x) = \frac{e^{a + bx}}{1 + e^{a + bx}}

Asymptote 1, value at

x = 0

\exp (a) / (1 + \exp (a))

The easiest way to figure this out is first to set

K = 1

and multiply the population-dynamic version by

\exp (rt) / \exp (rt)

n (t) = \frac{\exp (rt)}{\exp (rt) + (\frac{1}{n (0)} - 1)}

and multiply the statistical version by

\exp (- a) / \exp (- a)

f (x) = \frac{\exp (bx)}{\exp (- a) + \exp (bx)}

This manipulation makes it clear (I hope) that

b = r

x = t

, and

(1 / n (0) - 1) = \exp (- a)

, or

a = - \log (1 / n (0) - 1)

, or

n (0) = 1 / (1 + \exp (- a))

Set up parameters and equivalents:

> a = -5
> b = 2
> n0 = 1/(1 + exp(-a))
> n0

[1] 0.006692851

> K = 1
> r = b

Draw the curves:

> curve(exp(a + b * x)/(1 + exp(a + b * x)), from = 0, to = 5, 
+     ylab = "")
> curve(K/(1 + (K/n0 - 1) * exp(-r * x)), add = TRUE, type = "p")
> legend(0, 1, c("statistical", "pop-dyn"), pch = c(NA, 1), lty = c(1, 
+     NA), merge = TRUE)

The merge=TRUE statement in the legend() command makes R plot the point and line types in a single column.

File translated from T_EX by T_TM, version 3.70.
On 14 Sep 2005, 16:37.