Problem 47 - Thomas Calculus

Question

In Exercises \(39-48,\) find \(d y / d t\) $$ y=\sqrt{1+\cos \left(t^{2}\right)} $$

Step-by-Step Solution

Answer

\( \frac{dy}{dt} = -\frac{t \sin(t^2)}{\sqrt{1 + \cos(t^2)}} \)

Step 1: Identify the Function

The function given is \(y = \sqrt{1 + \cos(t^2)}\). We need to find \( \frac{dy}{dt} \), which represents the derivative of \(y\) with respect to \(t\).

Step 2: Apply the Chain Rule

To differentiate \(y = \sqrt{1 + \cos(t^2)}\), recognize it as \((1 + \cos(t^2))^{1/2}\). Use the chain rule: if \(y = u^{1/2}\), then \( \frac{dy}{dt} = \frac{1}{2}u^{-1/2} \cdot \frac{du}{dt} \). Here, \(u = 1 + \cos(t^2)\).

Step 3: Differentiate the Inner Function

Now differentiate the inner function \(u = 1 + \cos(t^2)\). First, find the derivative of \(\cos(t^2)\), which is \(-\sin(t^2) \cdot 2t\) due to the chain rule. So, \(\frac{du}{dt} = 0 - 2t \sin(t^2) = -2t \sin(t^2)\).

Step 4: Combine Derivatives

Substitute \(\frac{du}{dt}\) into the equation from Step 2. We have:\[ \frac{dy}{dt} = \frac{1}{2}(1 + \cos(t^2))^{-1/2} \cdot (-2t \sin(t^2)) \]Simplifying this gives:\[ \frac{dy}{dt} = -\frac{t \sin(t^2)}{\sqrt{1 + \cos(t^2)}} \]

Step 5: Final Answer

After simplification and applying the chain rule, we found that:\[ \frac{dy}{dt} = -\frac{t \sin(t^2)}{\sqrt{1 + \cos(t^2)}} \] This is the derivative of \(y\) with respect to \(t\).

Key Concepts

Derivative

The concept of a derivative is fundamental in calculus and represents how a function changes as its input changes. In simple terms, it's the rate at which the function's value is changing at any given point. For a function of one variable like our example, the derivative gives us the slope of the tangent line to the curve represented by the function at any point.
In our exercise, we need to find the derivative of the function with respect to the variable, which in this case is time variable \(t\). We are looking for \( \frac{dy}{dt} \), which represents the instantaneous rate of change of \(y\) as \(t\) changes. This concept is crucial because derivatives allow us to predict the behavior of functions and to understand changes in various systems modeled by those functions.
In practical applications, derivatives help in tracking things like velocity in physics, rates in finance, and efficiency in engineering. They play an indispensable role in optimization problems as well.

Trigonometric Functions

Trigonometric functions such as sine and cosine are periodic functions that are crucial in modeling oscillations and waves, among other phenomena. In calculus, they often appear within other functions and need to be differentiated just like in our exercise.
Our original function involves \(\cos(t^2)\), a typical trigonometric function. The differentiation of trigonometric functions requires understanding their derivatives. For instance:

• The derivative of \(\sin(x)\) is \(\cos(x)\).
• The derivative of \(\cos(x)\) is \(-\sin(x)\).

Given the function \(\cos(t^2)\), we have to apply the chain rule to find its derivative because \(t^2\) is not just \(t\), making the function more complex.
Trigonometric functions are essential in various fields, from physics and engineering to computer graphics, where they help in modeling reality and solving practical problems.

Differentiation Techniques

Differentiation techniques are strategies used to find the derivative of a function efficiently. The chain rule is one of the most important methods in differentiation, particularly useful when dealing with composite functions like \(y = \sqrt{1 + \cos(t^2)}\).
The chain rule states that if a function \(y = f(g(t))\), where \(f\) and \(g\) are functions of \(t\), the derivative \(\frac{dy}{dt}\) can be found by multiplying \(\frac{df}{dg}\) by \(\frac{dg}{dt}\).
For our function, we express the square root as \((1 + \cos(t^2))^{1/2}\) and set \(u = 1 + \cos(t^2)\). Applying the chain rule requires us to:

• Differentiating the outer function \(u^{1/2}\) resulting in \(\frac{1}{2}u^{-1/2}\).
• Then differentiating the inner function \(u = 1 + \cos(t^2)\), applying the chain rule again for \(\cos(t^2)\) as \(-2t \sin(t^2)\).

Finally, these are combined to complete the differentiation process, showcasing how differentiation techniques simplify the process of finding derivatives in complex expressions.