Problem 66 - University Calculus: Early Transcendentals

Question

In Exercises \(51-70,\) find \(d y / d t\). $$y=4 \sin (\sqrt{1+\sqrt{t}})$$

Step-by-Step Solution

Answer

The derivative is \(\frac{\cos(\sqrt{1+\sqrt{t}})}{\sqrt{t(1+\sqrt{t})}}\)."

Step 1: Identify the Function Structure

The function given is \( y = 4 \sin(\sqrt{1 + \sqrt{t}}) \). Notice that it is a composition of different functions: a sine function, a square root, and another square root function nested inside. This suggests the use of the chain rule for differentiation.

Step 2: Differentiate the Outer Function

Apply the chain rule to differentiate the outermost function. The derivative of \( \sin(u) \) with respect to \( u \) is \( \cos(u) \). Thus, the derivative of \( 4 \sin(u) \) is \( 4 \cos(u) \), where \( u = \sqrt{1 + \sqrt{t}} \).

Step 3: Differentiate the Middle Function

Now differentiate the inside function, \( \sqrt{1 + \sqrt{t}} \), with respect to \( t \). Set \( v = 1 + \sqrt{t} \), so that the outer function here is \( \sqrt{v} \). The derivative of \( \sqrt{v} \) with respect to \( v \) is \( \frac{1}{2\sqrt{v}} \).

Step 4: Differentiate the Innermost Function

Differentiate the innermost function, \( \sqrt{t} \), with respect to \( t \). The derivative is \( \frac{1}{2\sqrt{t}} \).

Step 5: Apply the Chain Rule

Combine all the derivatives using the chain rule. Multiply the derivatives together: \( \frac{dy}{dt} = 4 \cos(\sqrt{1 + \sqrt{t}}) \cdot \frac{1}{2\sqrt{1 + \sqrt{t}}} \cdot \frac{1}{2\sqrt{t}} \).

Step 6: Simplify the Derivative

Simplify the expression: \( \frac{dy}{dt} = \frac{4 \cos(\sqrt{1 + \sqrt{t}})}{4\sqrt{t} \sqrt{1 + \sqrt{t}}} \). This simplifies further to \( \frac{\cos(\sqrt{1 + \sqrt{t}})}{\sqrt{t(1 + \sqrt{t})}} \).

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus. It involves finding the derivative of a function, which is essentially the function's rate of change. When differentiating, one determines how a function's output value changes as the input value changes. This is particularly useful in problems involving motion, where you want to know how speed varies over time.For the function given in the exercise, differentiation helps us find how quickly the sine of a composed square root function changes with respect to time, denoted by \( t \). By following specific rules, like the chain rule in our case, we systematically break down complex functions into manageable parts, allowing us to compute their derivatives step-by-step.Differentiation is a key tool in mathematics and science used to solve real-world problems such as optimization and modeling natural phenomena.

Composition of Functions

The composition of functions is an essential concept when dealing with complex functions that are made up of simpler functions. In this exercise, the original function \( y = 4 \sin(\sqrt{1 + \sqrt{t}}) \) is a great example of a composite function. It combines a trigonometric sine function, an outer square root, and an inner square root. To understand the composition:

• The innermost part, \( \sqrt{t} \), is nested within the next function, \( 1 + \sqrt{t} \).
• This resulting expression is then placed inside another square root, \( \sqrt{1+\sqrt{t}} \).
• Finally, this whole expression becomes the argument of the sine function.

Using functions in this layered manner allows for powerful modeling in calculus. Breaking down compositions via differentiation involves carefully applying the chain rule, thereby ensuring each function's derivative is accounted for in the overall derivative calculation.

Trigonometric Functions

Trigonometric functions, such as sine, cosine, and tangent, are pivotal in mathematics, particularly in calculus and geometry. They relate the angles of triangles to the lengths of their sides and have applications in wave theory, oscillations, and circular motion.In our exercise, the function includes the sine function, noted as \( \sin(u) \). Sine oscillates, meaning it goes through cycles of values over its domain, typically angles. Understanding these cycles is crucial when analyzing periodic phenomena.When differentiating trigonometric functions, each has a specific derivative:

• The derivative of \( \sin(u) \) with respect to \( u \) is \( \cos(u) \).
• Likewise, the derivative of \( \cos(u) \) is \(-\sin(u) \).

These rules help us understand how trigonometric functions change, which, in turn, assists in solving a variety of real-world mathematical problems. Understanding these basic patterns enables one to handle more intricate functions, like those in our exercise, with greater confidence.