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Derivatives are a fundamental concept in calculus and are crucial for understanding how functions change. In this exercise, we start by finding the derivative of the function that models oxygen consumption of a bird embryo. The function given is a cubic polynomial, and when we differentiate it, we get its derivative: \[c'(t)=-0.195t^{2}+6.8t-22\]The derivative tells us the rate at which the original function changes at any given point. You can think of it as the slope of the tangent line to the curve at any point \(t\). Differentiation is the process used to find this instantaneous rate of change. In practical terms, for this function, the derivative \(c'(t)\) represents how quickly the oxygen consumption rate is changing per day. Understanding this concept helps us predict and analyze the behavior of dynamic processes, just like seeing how a car's speedometer needle moves tells you about your acceleration on a highway.

The quadratic formula is a powerful tool used to find the roots of a quadratic equation. In our exercise, we encountered a quadratic equation after finding the first derivative:\[-0.195t^{2}+6.8t-22=0\]To solve for \(t\), we apply the quadratic formula:\[t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\]where \(a = -0.195\), \(b = 6.8\), and \(c = -22\). Solving the quadratic equation gives us the critical points:

• \(t_1 \approx 13.37\)
• \(t_2 \approx 8.63\)

These solutions tell us where the derivative is zero, signifying potential local maxima or minima of the original function. The quadratic formula works for any quadratic equation and ensures we can always find potential turning points or important values like peaks and troughs in applied problems, such as tracking the maximum rate of biological processes.

The second derivative test is a method to determine whether a critical point is a maximum or minimum by using the second derivative of the function. After finding critical points, we continue by calculating the second derivative from our equation:\[c''(t)=-0.39t+6.8\]We then evaluate the second derivative at each critical point. For the first critical point \(t_1 \approx 13.37\), we find:\[c''(t_1) \,\text{is negative}\], indicating a local maximum because the curve is concave down at this point. Evaluating the second derivative at \(t_2 \approx 8.63\), we find:\[c''(t_2) \,\text{is positive}\],telling us there's a local minimum as the curve is concave up. Thus, the point where the function's rate of change is fastest is when \(t\) is around 13 days. By applying the second derivative test, we can classify critical points, aiding in interpretation, and decide the nature of turning points, which is vital for practical decision-making in scientific research and engineering. The simplest way to remember is: if the second derivative is positive, the function is curving upwards, and if it's negative, it curves downwards.