Problem 3 - Applied Calculus

Question

Differentiate the functions in Problems 1-20. Assume that \(A\) and \(B\) are constants. $$ y=t^{2}+5 \cos t $$

Step-by-Step Solution

Answer

The derivative is \( y' = 2t - 5 \sin t \).

Step 1: Differentiate the First Term

The function given is \( y = t^2 + 5 \cos t \). Start by differentiating the first term \( t^2 \) with respect to \( t \). Using the power rule, \( \frac{d}{dt}(t^n) = n \cdot t^{n-1} \), we find the derivative of \( t^2 \) as \( 2t \).

Step 2: Differentiate the Second Term

Next, differentiate the second term \( 5 \cos t \) with respect to \( t \). The derivative of \( \cos t \) is \( -\sin t \), and since we have a constant multiplier 5, by the constant multiple rule, the derivative is \( 5 \cdot (-\sin t) = -5 \sin t \).

Step 3: Combine the Derivatives

Combine the derivatives of both terms to get the derivative of the entire function. Thus, the derivative of \( y = t^2 + 5 \cos t \) is \( y' = 2t - 5 \sin t \).

Key Concepts

Power Rule

The power rule is a fundamental technique used in differentiation. It applies when you're working with terms where a variable is raised to a constant power. The power rule states that if you have a function of the form \(f(t) = t^n\), then the derivative \(f'(t)\) is calculated by multiplying the power \(n\) by \(t\) and reducing the power by one, resulting in \(n \cdot t^{n-1}\).

In the given exercise, the term \(t^2\) is differentiated using the power rule. Here, \(n = 2\), so applying the rule, we have:

• Bring down the power: \(2\).
• Maintain the base \(t\).
• Decrease the power by one, resulting in \(2\cdot t^{2-1} = 2t\).

This yields a derivative of \(2t\).

The power rule is incredibly useful because it simplifies the process of finding derivatives of polynomial terms.

Trigonometric Functions

Trigonometric functions, such as sine and cosine, have well-defined derivatives. These rules apply universally and are essential when differentiating functions involving these trigonometric elements.

For example, the derivative of \(\cos(t)\) is \(-\sin(t)\). This means whenever you differentiate a cosine function with respect to its angle, the result is the negative sine of that angle.

In the problem at hand, we differentiate \(\cos(t)\) in the term \(5 \cos(t)\). By recognizing that the derivative of \(\cos(t)\) is \(-\sin(t)\), this part of the solution becomes straightforward: simply apply the derivative rule for cosine and adjust for any constants.

Understanding these derivatives allows you to swiftly handle trigonometric parts in more complex functions, empowering you to compute their derivatives effectively.

Constant Multiplier Rule

The constant multiplier rule simplifies derivative calculations when a term includes a constant factor. If a constant \(k\) is multiplied by a function, the derivative of this term is the constant times the derivative of the function itself.

Formally, it can be expressed as: if \(f(t) = c \cdot g(t)\) where \(c\) is a constant, then the derivative \(f'(t)\) is \(c \cdot g'(t)\).

In our exercise, we have the term \(5 \cos(t)\). Here, \(5\) is the constant multiplier. Using the constant multiplier rule, we take the derivative of the inner function, \(\cos(t)\), which is \(-\sin(t)\), and multiply it by the constant \(5\).

• Find the derivative of the function: \(-\sin(t)\).
• Multiply by the constant: \(5\).

This results in \(-5 \sin(t)\). Therefore, the rule provides a quick way to handle constant factors during differentiation.