Problem 112 - Precalculus

Question

The difference quotient for the vector-valued function \(v(i)\) is defined as \(\frac{v(t+h)-v(t)}{h}\). In Exercises 111 and 112 , find the difference quotient of the vector-valued function. $$\mathbf{v}(t)=\left\langle t^{2}+1, t^{3}\right\rangle$$

Step-by-Step Solution

Answer

The difference quotient is \( \left\langle 2t + h, 3t^2 + 3th + h^2 \right\rangle \).

Step 1: Expression for the Function at t+h

Start with the vector-valued function \( \mathbf{v}(t) = \left\langle t^2 + 1, t^3 \right\rangle \). Compute \( \mathbf{v}(t+h) \). Substitute \( t+h \) into the function: \[ \mathbf{v}(t+h) = \left\langle (t+h)^2 + 1, (t+h)^3 \right\rangle. \]

Step 2: Expand the Expression

Now, expand the expressions for \( (t+h)^2 \) and \( (t+h)^3 \):- \( (t+h)^2 = t^2 + 2th + h^2 \)- \( (t+h)^3 = t^3 + 3t^2h + 3th^2 + h^3 \)

Step 3: Substitute Expanded Terms into v(t+h)

Substitute the expanded terms from Step 2 back into \( \mathbf{v}(t+h) \): \[ \mathbf{v}(t+h) = \left\langle t^2 + 2th + h^2 + 1, t^3 + 3t^2h + 3th^2 + h^3 \right\rangle \]

Step 4: Compute the Difference Quotient

Write the difference quotient \( \frac{\mathbf{v}(t+h) - \mathbf{v}(t)}{h} \). Subtract \( \mathbf{v}(t) \) from \( \mathbf{v}(t+h) \):\[ \mathbf{v}(t+h) - \mathbf{v}(t) = \left\langle t^2 + 2th + h^2 + 1 - t^2 - 1, t^3 + 3t^2h + 3th^2 + h^3 - t^3 \right\rangle \]

Step 5: Simplification

Simplify the expression from Step 4: \[ \mathbf{v}(t+h) - \mathbf{v}(t) = \left\langle 2th + h^2, 3t^2h + 3th^2 + h^3 \right\rangle \]

Step 6: Divide by h and Simplify Further

Now divide each component by \( h \) to complete the difference quotient:\[ \frac{\mathbf{v}(t+h) - \mathbf{v}(t)}{h} = \left\langle \frac{2th + h^2}{h}, \frac{3t^2h + 3th^2 + h^3}{h} \right\rangle \]Simplify this to get:\[ = \left\langle 2t + h, 3t^2 + 3th + h^2 \right\rangle \]

Key Concepts

Vector-valued Function

In calculus, vector-valued functions are fascinating entities. Instead of mapping a single value, they map values to vectors. Consider our function \( \mathbf{v}(t) = \left\langle t^2 + 1, t^3 \right\rangle \). This function outputs a two-dimensional vector for every value of \( t \). Here, each component of the vector is a function itself: one is a quadratic \( t^2 + 1 \), and the other is a cubic \( t^3 \).
Understanding how these functions behave as \( t \) changes can unravel the path or trajectory represented by the vector. For example, in physics, such functions can depict the position or velocity of an object in space over time. This breaks complex multi-dimensional problems into manageable pieces.

Expansion of Expressions

Expression expansion is a key skill in algebra and calculus, especially when dealing with difference quotients. Expand expressions in order to simplify complex problems. Our task was to expand \( (t+h)^2 \) and \( (t+h)^3 \).
The expansion of \( (t+h)^2 \) is achieved by multiplying \( (t+h)(t+h) \), giving us \( t^2 + 2th + h^2 \). Similarly, expanding \( (t+h)^3 \) uses \((t+h)^2(t+h)\), resulting in \( t^3 + 3t^2h + 3th^2 + h^3 \).

• This expansion helps in subsequent steps like subtraction and division.
• It breaks down complex polynomials into manageable parts.

Such expansions are foundational for deriving the difference quotient efficiently.

Simplification

Simplification refines mathematical expressions to make them more manageable, especially when calculating the difference quotient. This process involves reducing complexity by combining like terms.
In our exercise, the difference \( \mathbf{v}(t+h) - \mathbf{v}(t) \) initially presented itself as:

• \( \left\langle t^2 + 2th + h^2 + 1 - t^2 - 1, t^3 + 3t^2h + 3th^2 + h^3 - t^3 \right\rangle \)

The expression simplifies by cancelling out terms like \( t^2 - t^2 \) and \( 1 - 1 \), leaving:

• \( \left\langle 2th + h^2, 3t^2h + 3th^2 + h^3 \right\rangle \)

This simplification exposes the core elements that need further calculation, like dividing by \( h \). Simplification is crucial in reducing errors and understanding relationships between terms.

Calculus

Calculus fundamentally deals with change and motion. It breaks real-world situations into mathematical problems. One of its pivotal operations is the difference quotient, a bridge to derivatives.
In our vector-valued function \( \mathbf{v}(t) \), the difference quotient \( \frac{\mathbf{v}(t+h) - \mathbf{v}(t)}{h} \) is the foundation for understanding the rate of change or the slope of vectors at any point \( t \). By dividing each component by \( h \) and simplifying:

• \( \left\langle \frac{2th + h^2}{h}, \frac{3t^2h + 3th^2 + h^3}{h} \right\rangle \)

Simplifying gives us the form \( \left\langle 2t + h, 3t^2 + 3th + h^2 \right\rangle \), indicating how each component changes as \( h \) approaches zero. Understanding these components is vital for progressing into the derivatives and integrals central to calculus.