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suppose that the rate at which body weight w changes with age t is \\( …

Question

suppose that the rate at which body weight w changes with age t is \\( \frac { d w } { d t } \propto w ^ { a } \\), where \\( a > 0 \\) is a coefficient that takes different values for different species of animal. (a) the relative growth rate (percentage weight gained per unit of time) is defined as \\( g ( w ) = \frac { 1 } { w } \frac { d w } { d t } \\). write down a formula for \\( g ( w ) \\). for which values of a is the relative growth rate increasing, and for which values is it decreasing? (b) as fish grow larger, their weight increases each day but the relative growth rate decreases. if the rate of growth is described by \\( \frac { d w } { d t } \propto w ^ { a } \\), explain what constraints must be imposed on a. (b) in order for the relative growth rate to decrease as fish grow larger, the value of a has to be

Explanation:

Step1: Substitute \(\frac{dW}{dt}\) into \(G(W)\)

Since \(\frac{dW}{dt}\propto W^{a}\), we can write \(\frac{dW}{dt}=kW^{a}\) (where \(k>0\) is a proportionality constant). Then \(G(W)=\frac{1}{W}\frac{dW}{dt}=\frac{1}{W}\times kW^{a}=kW^{a - 1}\).

Step2: Find the derivative of \(G(W)\) with respect to \(W\)

Using the power rule \((x^{n})^\prime=nx^{n - 1}\), if \(G(W)=kW^{a - 1}\), then \(G^\prime(W)=k(a - 1)W^{a - 2}\).

Step3: Determine when \(G(W)\) is increasing or decreasing

For \(G(W)\) to be increasing, \(G^\prime(W)>0\). Since \(k>0\) and \(W>0\) (weight is non - negative), we need \(a-1>0\), i.e., \(a > 1\). For \(G(W)\) to be decreasing, \(G^\prime(W)<0\). Since \(k>0\) and \(W>0\), we need \(a - 1<0\), i.e., \(0 < a<1\).

Step4: Solve part (b)

We know from part (a) that for the relative growth rate \(G(W)\) to decrease as \(W\) (weight) increases, we need \(G^\prime(W)<0\). Since \(G(W)=kW^{a - 1}\) and \(G^\prime(W)=k(a - 1)W^{a - 2}\), and \(k>0\), \(W>0\), we must have \(a-1<0\) (because \(W^{a - 2}>0\) for \(W>0\) and \(a>0\)). So \(a<1\). Also, given \(a>0\) (from the problem statement \(\frac{dW}{dt}\propto W^{a},a > 0\)), so \(0 < a<1\).

Answer:

(a) \(G(W)=kW^{a - 1}\), increasing for \(a>1\), decreasing for \(0 < a<1\). So the answer for part (a) is B. (b) \(0 < a<1\)