Calves The daily milk consumption (in kilograms) for calves...

Calves The daily milk consumption (in kilograms) for calves can be approximated by the function $$y=b_{0} w^{b_{1}} e^{-b_{2} w}$$ where $w$ is the age of the calf (in weeks) and $b_{0}, b_{1},$ and $b_{2}$ are constants. Source: Animal Production. $$\begin{array}{l}{\text { (a) The age in days is given by } t=7 w . \text { Use this fact to convert }} \\ {\text { the function above to a function in terms of } t \text { . }}\end{array}$$ $$\begin{array}{c}{\text { (b) For a group of Angus calves, } b_{0}=5.955, b_{1}=0.233, \text { and }} \\ {b_{2}=0.027 . \text { Use the trapezoidal rule with } n=10 \text { , and then }} \\ {\text { Simpson's rule with } n=10, \text { to find the total amount of milk }} \\ {\text { consumed by one of these calves over the first } 25 \text { weeks of life. }}\end{array}$$ $$ \begin{array}{l}{\text { (c) For a group of Nelore calves, } b_{0}=8.409, b_{1}=0.143, \text { and }} \\ {b_{2}=0.037 . \text { Use the trapezoidal rule with } n=10 \text { , and then }} \\ {\text { Simpson's rule with } n=10, \text { to find the total amount of milk }} \\ {\text { consumed by one of these calves over the first } 25 \text { weeks of life. }}\end{array} $$

Calves The daily milk consumption (in kilograms) for calves can be approximated by the function
$$y=b_{0} w^{b_{1}} e^{-b_{2} w}$$
where $w$ is the age of the calf (in weeks) and $b_{0}, b_{1},$ and $b_{2}$ are constants. Source: Animal Production.
$$\begin{array}{l}{\text { (a) The age in days is given by } t=7 w . \text { Use this fact to convert }} \\ {\text { the function above to a function in terms of } t \text { . }}\end{array}$$
$$\begin{array}{c}{\text { (b) For a group of Angus calves, } b_{0}=5.955, b_{1}=0.233, \text { and }} \\ {b_{2}=0.027 . \text { Use the trapezoidal rule with } n=10 \text { , and then }} \\ {\text { Simpson's rule with } n=10, \text { to find the total amount of milk }} \\ {\text { consumed by one of these calves over the first } 25 \text { weeks of life. }}\end{array}$$
$$
\begin{array}{l}{\text { (c) For a group of Nelore calves, } b_{0}=8.409, b_{1}=0.143, \text { and }} \\ {b_{2}=0.037 . \text { Use the trapezoidal rule with } n=10 \text { , and then }} \\ {\text { Simpson's rule with } n=10, \text { to find the total amount of milk }} \\ {\text { consumed by one of these calves over the first } 25 \text { weeks of life. }}\end{array}
$$

Key Concepts

Change of Variables

This concept involves substituting one variable for another, typically to change the domain or units of a function. In many applications, especially when integrating with respect to a different variable, a relationship is established between the original variable and the new variable, and the function is rewritten accordingly. This is essential for matching the domain of integration or to simplify the problem.

Trapezoidal Rule

The trapezoidal rule is a numerical integration method that approximates the area under a curve by dividing the region into trapezoids rather than using exact analytical integration. The area of each trapezoid is calculated and summed to estimate the total integral. This method is straightforward and effective for functions that are relatively linear on small intervals.

Simpson's Rule

Simpson's rule is another numerical integration technique which approximates the area under a curve by fitting parabolic segments through sets of three adjacent points. By using quadratic approximations instead of linear ones, Simpson’s rule generally provides a higher accuracy than the trapezoidal rule, especially when the function being integrated is smooth and well-behaved.

Transcript

00:01 So in this question, we're given this equation here, and this represents the daily milk consumption in kilograms for calves.

00:09 Okay, so even though it says it's a daily consumption, the variable we're given here is w, which is actually weeks.

00:17 So which is why in part a, the first thing we need to do is we need to actually convert this equation and use days instead of weeks.

00:25 So let us go ahead and do that.

00:27 So let t be the number of days.

00:30 Then that's going to be 7 times w.

00:33 So you're going to isolate w, and that's going to be t over 7, and then you're going to take w equals t7, you're going to do a substitution.

00:43 So everywhere that you see w, you're going to plug in t over 7.

00:46 So this means our equation is actually y equals b not times t over 7 to the power of b1, and then e to the power of negative b2 times t over 7.

01:00 And that's all you have to do.

01:02 You can just leave it like this.

01:05 So now part b, we are asked to use trapezoidal rule and simpson's rule to approximate the total amount of milk consumed by one of these calves over the first 25 weeks of life.

01:20 Okay.

01:21 So note that, again, they're giving us weeks, so which means we need to convert it to days.

01:27 So let's start with that.

01:30 So the first 25 weeks, which means we're starting when they're just born.

01:34 So that would be zero days.

01:38 And then 25 weeks, so we multiply that by 7, so that would be 175 days.

01:43 And we're told to use n equals 10 in our approximations.

01:49 And for this question, we're given particular values for b not, which is 5 .955.

02:00 And b1 is 0 .233, and b2 is 0 .027.

02:10 So you're going to take these values and you're going to create your table.

02:15 So your table, you're going to have your tis, which starts at zero, and then goes up in an increment of b minus a over n all the way until you hit b...

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