Question

practice another newborn blue whales are approximately 24 feet long and weigh 3 tons. young whales are nursed for 7 months, and by the time of weaning they often are 56 feet long and weigh 23 tons. let l and w denote the length (in feet) and the weight (in tons), respectively, of a whale that is t months of age. (a) if l and t are linearly related, express l in terms of t. (give your answer in slope - intercept form.) (b) what is the daily increase in the length of a young whale? (use 1 month = 30 days. round your answer to three decimal places.) ft/day (c) if w and t are linearly related, express w in terms of t. (give your answer in slope - intercept form.) (d) what is the daily increase in the weight of a young whale? (round your answer to three decimal places.) tons/day

Explanation:

Step1: Determine two - point data for length

We know that at birth, assume \(t = 0\), \(L_0\) (initial length) is not given precisely, but we can use the data at weaning. At \(t_1=7\) months (\(t_1 = 7\times30=210\) days), \(L_1 = 56\) feet and at \(t = 0\), assume \(L_0\) (new - born length) is \(24\) feet.

Step2: Calculate the daily increase in length

The slope \(m\) of the linear function \(L(t)=mt + b\) for length is given by \(m=\frac{L_1 - L_0}{t_1}\). Substituting \(L_0 = 24\), \(L_1 = 56\) and \(t_1=210\) days, we have \(m=\frac{56 - 24}{210}=\frac{32}{210}\approx0.152\) ft/day.

Step3: Write the linear equation for length

The linear equation for length \(L\) in terms of \(t\) (in months) with \(t\) in months and 1 month = 30 days. First, convert the slope to a monthly - based slope. Since \(m = \frac{32}{210}\) ft/day, and 1 month = 30 days, the slope for \(t\) in months is \(m_{month}=\frac{32}{210}\times30=\frac{32}{7}\approx4.571\). The initial length \(b = 24\), so \(L(t)=\frac{32}{7}t+24\).

Step4: Determine two - point data for weight

At birth (\(t = 0\)), \(W_0 = 3\) tons and at \(t_1 = 7\) months (\(t_1=210\) days), \(W_1 = 23\) tons.

Step5: Calculate the daily increase in weight

The slope \(n\) of the linear function \(W(t)=nt + c\) for weight is \(n=\frac{W_1 - W_0}{t_1}\). Substituting \(W_0 = 3\), \(W_1 = 23\) and \(t_1 = 210\) days, we get \(n=\frac{23 - 3}{210}=\frac{20}{210}\approx0.095\) tons/day.

Step6: Write the linear equation for weight

The linear equation for weight \(W\) in terms of \(t\) (in months) with 1 month = 30 days. First, convert the slope to a monthly - based slope. The slope for \(t\) in months is \(n_{month}=\frac{20}{210}\times30=\frac{20}{7}\approx2.857\). The initial weight \(c = 3\), so \(W(t)=\frac{20}{7}t + 3\).

Answer:

(a) \(L(t)=\frac{32}{7}t + 24\)
(b) \(0.152\) ft/day
(c) \(W(t)=\frac{20}{7}t+3\)
(d) \(0.095\) tons/day