Question

{Use of Tech} Growth rate of spotted owlets The rate of growth (in g/week) of the body mass of Indian spotted owlets is modeled by the function r(t) = 10,147.9e⁻²·²ᵗ/(37.98e⁻²·² + 1), where t is the age (in weeks) of the owlets. What value of t \u003e 0 maximizes r? What is the physical meaning of the maximum value?

Accepted Answer

To find the value of t that maximizes the growth rate function r(t), we need to find the critical points of the function. This involves taking the derivative of r(t) with respect to t and setting it equal to zero. The function given is r(t) = $$ \frac{10,147.9e^{-2.2t}}{37.98e^{-2.2} + 1} . We $$will use the quotient rule to differentiate this function. The quotient rule states that if you have a function $$ \frac{u(t)}{v(t)} $$, its derivative is $$ \frac{u'(t)v(t) - u(t)v'(t)}{(v(t))^2} . $$Identify u(t) and v(t) from the function: $$ u(t) = 10,147.9e^{-2.2t} $$ and $$ v(t) = 37.98e^{-2.2} + 1 . $$Differentiate both: $$ u'(t) = -22,325.38e^{-2.2t} $$ and $$ v'(t) = 0 $$ since v(t) is a constant with respect to t. Substitute u(t), u'(t), v(t), and v'(t) into the quotient rule formula to find r'(t). Simplify the expression to find the critical points by setting r'(t) = 0. Solve the equation obtained from setting r'(t) = 0 to find the value of t that maximizes r(t). The physical meaning of the maximum value of r(t) is the age at which the growth rate of the owlets' body mass is at its highest.

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Key Concepts

Differentiation

Critical Points

Second Derivative Test