Problem 24 - University Calculus: Early Transcendentals

Question

Find the derivative of \(y\) with respect to the appropriate variable. $$y=\sin ^{-1}(1-t)$$

Step-by-Step Solution

Answer

The derivative is \( \frac{dy}{dt} = -\frac{1}{\sqrt{t(2-t)}} \).

Step 1: Understand the Function

The function given is \( y = \sin^{-1}(1-t) \). This represents the inverse sine function, also known as arcsine, for the expression \( 1-t \). Our goal is to find \( \frac{dy}{dt} \).

Step 2: Use the Derivative Formula for Inverse Sine

Recall that the derivative of \( \sin^{-1}(u) \) with respect to \( u \) is \( \frac{1}{\sqrt{1-u^2}} \). Here, \( u = 1-t \). The derivative \( \frac{dy}{du} = \frac{1}{\sqrt{1-(1-t)^2}} \).

Step 3: Differentiate the Inner Function

We need \( \frac{du}{dt} \) where \( u = 1-t \). Differentiating with respect to \( t \), we get \( \frac{du}{dt} = -1 \).

Step 4: Apply the Chain Rule

Using the chain rule, \( \frac{dy}{dt} = \frac{dy}{du} \times \frac{du}{dt} \). Substituting from previous steps, we get \( \frac{dy}{dt} = \frac{1}{\sqrt{1-(1-t)^2}} \times (-1) \).

Step 5: Simplify the Expression

Simplify the expression under the square root: \( 1-(1-t)^2 = 1 - (1 - 2t + t^2) = 2t - t^2 = t(2-t) \). So, \( \frac{1}{\sqrt{t(2-t)}} \). Thus, \( \frac{dy}{dt} = -\frac{1}{\sqrt{t(2-t)}} \).

Key Concepts

Chain Rule

The chain rule is essential in differentiating composite functions. It allows us to relate the derivative of a function to the derivatives of its inner and outer functions. When dealing with a composition like \( y = \sin^{-1}(1-t) \), the chain rule is crucial.

In this expression, \( 1-t \) is the inner function, and the inverse sine is the outer function. The chain rule states \( \frac{dy}{dt} = \frac{dy}{du} \times \frac{du}{dt} \). This simply means, first, you take the derivative of the outer function with respect to the inner function \( u \), and then multiply it by the derivative of the inner function \( u \) with respect to \( t \).

For \( y = \sin^{-1}(1-t) \), this process helps unravel the layers of the composition, leading to the derivative of the original expression.

Inverse Sine Function

The inverse sine function, written as \( \sin^{-1}(x) \) or \( \arcsin(x) \), returns the angle whose sine is \( x \). It’s an essential concept in calculus, which helps in finding dimensions like angles or when working backward from known sine values.

When differentiating \( \sin^{-1}(u) \), use the specific derivative rule \( \frac{d}{du}\sin^{-1}(u) = \frac{1}{\sqrt{1-u^2}} \). This formula arises from its trigonometric identity. It's crucial, especially when the input \( u \) is another function, like \( 1-t \). Knowing this derivative helps in tackling problems involving inverse trigonometric functions effectively.

Understanding how to apply these rules allows one to solve for derivatives easily, simplifying complex calculus problems involving inverse trigonometric expressions.

Calculus

Calculus, the mathematical study of change, plays a significant role in understanding and calculating derivatives. Derivatives represent the rate at which a quantity changes, and are foundational in calculus.

In problems like the derivative of \( y = \sin^{-1}(1-t) \), calculus provides the tools and rules such as the derivative of inverse sine and the chain rule. These concepts allow us to differentiate complex expressions.

• Limits: Understanding limits is crucial as they form the basis of the derivative definition.
• Continuity: Checking if a function is continuous ensures we can apply calculus techniques correctly.
• Derivatives: They are used to find slopes and rates of change, helping in solving practical problems in physics or engineering.

Calculus offers methods to derive formulas, solve equations, and understand the behavior of functions in detailed ways, making it indispensable for analyzing dynamic systems.