Problem 3 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

In Chapter 7 we show that \([\cos (t)]^{\prime}=-\sin (t) .\) Use this formula and \([\sin t]^{\prime}=\cos t\) written earlier in this chapter to differentiate: $$ \begin{array}{ll} \text { a. } y(t)=\cos (2 t) & \text { b. } y(t)=\sin \left(\frac{\pi}{2} t\right) \\ \text { c. } y(t)=e^{\cos t} & \text { d. } y(t)=\cos \left(e^{t}\right) \\ \text { e. } y(t)=\sin (\cos t) & \text { f. } y(t)=\sin \left(\cos e^{t}\right) \\ \text { g. } y(t)=\ln (\sin t) & \text { h. } y(t)=\sec t=(\cos t)^{-1} \\ \text {i. } y(t)=\ln \left(\cos \left(e^{t}\right)\right) & \text { j. } y(t)=\ln \left(\cos \left(e^{\sin t}\right)\right) \end{array} $$

Step-by-Step Solution

Answer

a. -2sin(2t), b. (π/2)cos(π/2 t), c. e^{cos t} (-sin t), d. -sin(e^t) e^t, e. cos(cos t)(-sin t), f. cos(cos(e^t))(-sin(e^t)e^t), g. cot(t), h. sec(t)tan(t), i. -tan(e^t) e^t, j. -tan(e^{sin t})e^{sin t}cos t.

Step 1: Differentiate y(t) = cos(2t)

Apply the chain rule. Let u = 2t, then y(t) = cos(u). The derivative is given by \ \( y'(t) = -\sin(u) \cdot u'(t) \). Since \( u = 2t \), we have \( u'(t) = 2 \). Therefore, the derivative is \( y'(t) = -2 \sin(2t) \).

Step 2: Differentiate y(t) = sin(π/2 * t)

Use the chain rule. Let v = \( \frac{\pi}{2}t \), then \( y(t) = \sin(v) \). The derivative is given by \( y'(t) = \cos(v) \cdot v'(t) \). Since \( v = \frac{\pi}{2}t \), we have \( v'(t) = \frac{\pi}{2} \). Therefore, the derivative is \( y'(t) = \frac{\pi}{2} \cos\left(\frac{\pi}{2}t\right) \).

Step 3: Differentiate y(t) = e^{cos t}

Apply the chain rule. Let w = cos(t), then \( y(t) = e^w \). The derivative is \( y'(t) = e^w \cdot w'(t) \). Since \( w = \cos(t) \), we have \( w'(t) = -\sin(t) \). Therefore, the derivative is \( y'(t) = e^{\cos t} (-\sin t) \).

Step 4: Differentiate y(t) = cos(e^t)

Apply the chain rule. Let x = e^t, then \( y(t) = \cos(x) \). The derivative is \( y'(t) = -\sin(x) \cdot x'(t) \). Since \( x = e^t \), we have \( x'(t) = e^t \). Therefore, the derivative is \( y'(t) = -\sin(e^t) e^t \).

Step 5: Differentiate y(t) = sin(cos t)

Use the chain rule. Let y = cos(t), then \( y(t) = \sin(y) \). The derivative is \( y'(t) = \cos(y) \cdot y'(t) \). Since \( y = \cos(t) \), we have \( y'(t) = -\sin(t) \). Therefore, the derivative is \( y'(t) = \cos(\cos t) (-\sin t) \).

Step 6: Differentiate y(t) = sin(cos(e^t))

Apply the chain rule twice. Let z = cos(e^t), then \( y(t) = \sin(z) \). First derivative with respect to z is \( \cos(z) \cdot z'(t) \). Let a = e^t, then \( z = \cos(a) \). So, \( z'(t) = -\sin(a)\cdot a'(t) = -\sin(e^t) e^t \).Therefore, the derivative is \( y'(t) = \cos(\cos(e^t))[-\sin(e^t) e^t] \).

Step 7: Differentiate y(t) = ln(sin t)

Use the chain rule. The derivative of \( \ln u \) is \( \frac{1}{u} \cdot u'(t) \). Here, \( u = \sin(t) \), so \( u'(t) = \cos(t) \).Thus, \( y'(t) = \frac{1}{\sin(t)} \cdot \cos(t) = \cot(t) \).

Step 8: Differentiate y(t) = sec(t) = (cos t)^{-1}

Use the quotient rule or chain rule. The derivative is \( y'(t) = \big(-\sin(t)(\cos(t))^{-2} \big) \). It simplifies to \( y'(t) = \sec(t)\tan(t) \).

Step 9: Differentiate y(t) = ln(cos(e^t))

Apply the chain rule twice. Let b = cos(e^t), then \( y(t) = \ln(b) \). Derivative is \( \frac{1}{b} \cdot b'(t) \). Since \( b = \cos(e^t) \), derive as \( b'(t) = -\sin(e^t) e^t \).So, \( y'(t) = \frac{-\sin(e^t) e^t}{\cos(e^t)} = -\tan(e^t) e^t \).

Step 10: Differentiate y(t) = ln(cos(e^{sin t}))

Apply chain rule multiple times. Let c = e^{sin(t)}, then b = cos(c), and \( y(t) = \ln(b) \).First, \( y'(t) = \frac{1}{b} \cdot b'(t) \).Then, \( b'(t) = -\sin(c) c' \).Finally, \( c' = \cos(t) \cdot e^{\sin(t)} \). The derivative is \( y'(t) = -\tan(c) e^{\sin(t)} \cos(t) \).

Key Concepts

Chain Rule

The chain rule is an essential tool in calculus for finding the derivative of composite functions. When you have a function inside another function, the chain rule helps you differentiate it by breaking it down into simpler parts. Think of it like peeling an onion layer by layer.

The basic idea is to first differentiate the outer function, while keeping the inner function unchanged, and then multiply it by the derivative of the inner function. This concept can feel complex at first, but with practice, it becomes a powerful skill in calculus.

For example, consider the function \( y(t) = \cos(2t) \). By using the chain rule:

• Identify the outer function as \( \cos(u) \), where \( u = 2t \).
• Find the derivative of the outer function: \( -\sin(u) \).
• Now calculate the derivative of the inner function: \( 2 \), because \( u = 2t \).
• Combine them: \( y'(t) = -2\sin(2t) \).

This step-by-step application of the chain rule quickly turns complex functions into something manageable.

Derivatives

Derivatives are a cornerstone of calculus and represent the rate at which a function changes at any given point. Simply put, they are about finding the slope of a curve or the rate of change.

Imagining derivatives in a practical sense helps; think of driving a car. If your speedometer reads how fast you're going at a specific instant, this is akin to finding the derivative of your position over time. In the world of mathematics, derivatives are used everywhere to find slopes of tangents to curves.

Let's consider the example \( y(t) = \sin(\frac{\pi}{2} t) \):

• Identify the function as \( \sin(v) \) with \( v = \frac{\pi}{2} t \).
• Differentiate \( \sin(v) \) to get \( \cos(v) \).
• Next, find the derivative of the inner function: \( \frac{\pi}{2} \).
• Combine them: \( y'(t) = \frac{\pi}{2} \cos\left(\frac{\pi}{2} t\right) \).

Understanding derivatives is crucial because they tell you how a function behaves, revealing lots of useful insights.

Calculus

Calculus is a vast field in mathematics focused on change and motion, comprising mainly of differentiation and integration.
Differentiation, the process we've been discussing, lets us calculate slopes and rate of changes. It provides a method for finding the instantaneous rate of change, which has applications in various scientific fields, such as physics, biology, and economics.
In understanding calculus, the conceptual shift from algebra to calculus is all about moving from static equations to dynamic systems. Think of calculus as the mathematics of change.
For beginners, consider these basic principles:

• Functions can be differentiated to determine how they change and grow.
• Rules like the chain rule make dealing with complicated functions simpler.
• Every derivative finds a slope - how a tiny change in input affects the output.

Calculus might seem challenging, but once you grasp the core principles, you'll have a powerful tool for understanding and analyzing a variety of real-world problems.