Question

The temperature \(T\) inside a certain furnace varies with the time \(t\) according to the function \(T=(t+3) \sqrt{t+1}\) ' \(\mathrm{F},\) where \(t\) is in hours. Find the rate of change of \(T\) when \(t=2.35 \mathrm{h}\)

Short answer

The rate of change of temperature when \(t = 2.35\) hours is approximately \(3.9\mathrm{\,F/hr}\).

Step-by-step solution

Identify the derivative

The rate of change of the temperature with respect to time can be found by differentiating the function of temperature, T, with respect to time, t. The function is given as \(T=(t+3) \sqrt{t+1}\). Use the product rule to differentiate \(T\) where \(T\) is the product of two functions: \(u(t) = t + 3\) and \(v(t) = \sqrt{t + 1}\).

Differentiate the function

Apply the product rule for differentiation which states that \(d(uv)/dt = u(dv/dt) + v(du/dt)\). Here, \(du(t)/dt = 1\) and \(dv(t)/dt = 1/(2\sqrt{t+1})\). By applying the rule: \(dT/dt = (t+3) \cdot (1/(2\sqrt{t+1})) + \sqrt{t+1} \cdot 1\).

Simplify and find the rate of change at \(t=2.35\) hours

Substitute \(t = 2.35\) into the derived formula to obtain the value of \(dT/dt\). Thus, \(dT/dt|_{t=2.35} = (2.35+3) \cdot (1/(2\sqrt{2.35+1})) + \sqrt{2.35+1}\). Compute the value to find the rate at which temperature changes with time when \(t = 2.35\) hours.

Key concepts

Product Rule Differentiation

In calculus, the product rule is a vital tool for finding the derivative of a function that is the product of two or more functions. To recall, the derivative of a function represents the rate at which its values are changing at any given point. When you come across a function like \( f(x) = g(x) \cdot h(x) \), you cannot simply differentiate each function separately and multiply the results. The product rule comes to the rescue and instructs us to apply the following formula: \( (f(x))' = g(x) \cdot h'(x) + g'(x) \cdot h(x) \).

Let's consider our exercise, \( T=(t+3) \cdot \sqrt{t+1} \). Here, the temperature \( T \) is a product of two functions, \( u(t) = t + 3 \) and \( v(t) = \sqrt{t + 1} \). Following the product rule, we would first take the derivative of \( u(t) \), which gives us \( u'(t) = 1 \), as the derivative of a constant added to a variable is 1. Then, we differentiate \( v(t) \), which gives us \( v'(t) = \frac{1}{2\sqrt{t+1}} \), coming from the derivative of the square root of \(t+1\).

By combining these derivatives in the format provided by the product rule, we get the rate of change of the temperature function as \( \frac{dT}{dt} = (t+3) \cdot \left(\frac{1}{2\sqrt{t+1}}\right) + \sqrt{t+1} \cdot 1 \). This form of differentiation allows us to tackle more complex functions with ease, showing the versatility and power of calculus in managing even products of functions.

Derivative of a Function

The derivative of a function at any point gives us the slope of the tangent line to the function at that point. It is a measure of how the function value changes as its input changes. In other words, derivatives can be thought of as the instantaneous rate of change or the velocity of the function’s output with respect to its input. For a function \( y = f(x) \), the derivative is typically denoted by \( f'(x) \) or \( \frac{dy}{dx} \).

For simple functions, such as polynomials, the power rule tells us that for \( f(x) = x^n \), the derivative is \( f'(x) = nx^{n-1} \). For the function in our example, \( T=(t+3) \cdot \sqrt{t+1} \), we needed to use the product rule because we were dealing with a product of two functions, as explained in the previous section. By understanding the concept and technique of differentiation, we can apply this tool to find rates of change in all sorts of physical phenomena, such as temperature, velocity, or even economic functions.

Applying Calculus to Physical Problems

Calculus is frequently applied to solve physical problems where quantities vary continuously. By using derivatives, we can model and study the rates at which these physical quantities change. For example, in the context of our exercise, the derivative helps us understand how the temperature inside a furnace changes at a specific moment in time.

After using the product rule to differentiate the function representing the temperature over time, we can evaluate the derivative at a particular time to find the rate of change at that instant. This practically means that if we substitute \( t = 2.35 \) hours into our derivative, we can compute the exact rate at which the temperature is changing inside the furnace at 2.35 hours. In general, this method can be applied to a wide array of scenarios in physics, chemistry, biology, economics, and engineering to figure out how quickly something is changing - be it a temperature, population size, or the cost of goods - giving us valuable insights into the nature of change in our physical world.