Problem 111 - 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. $$\mathbf{v}(t)=\left\langle t, t^{2}\right\rangle$$

Step-by-Step Solution

Answer

The difference quotient is \(\langle 1, 2t + h \rangle\).

Step 1: Calculate v(t+h)

First, substitute \(t + h\) into the vector-valued function \(\mathbf{v}(t) = \langle t, t^2 \rangle\). This results in \(\mathbf{v}(t + h) = \langle t + h, (t + h)^2 \rangle\).

Step 2: Expand (t + h)^2

Expand the expression \((t + h)^2\) to get \(t^2 + 2th + h^2\). Therefore, \(\mathbf{v}(t + h) = \langle t + h, t^2 + 2th + h^2 \rangle\).

Step 3: Compute v(t + h) - v(t)

Subtract \(\mathbf{v}(t) = \langle t, t^2 \rangle\) from \(\mathbf{v}(t + h) = \langle t + h, t^2 + 2th + h^2 \rangle\). This results in \(\langle t + h - t, t^2 + 2th + h^2 - t^2 \rangle = \langle h, 2th + h^2 \rangle\).

Step 4: Divide by h

Divide each component of the vector \(\langle h, 2th + h^2 \rangle\) by \(h\). This results in \(\langle \frac{h}{h}, \frac{2th + h^2}{h} \rangle = \langle 1, 2t + h \rangle\).

Step 5: Conclusion: Write the Final Expression

The difference quotient of the vector-valued function is \(\langle 1, 2t + h \rangle\).

Key Concepts

Vector-Valued Function

In mathematics, a vector-valued function is a function that takes one or more variables and returns a vector. A simple example of a vector-valued function in two dimensions could be written as \( \mathbf{v}(t) = \langle x(t), y(t) \rangle \). This means that for each input \( t \), the function outputs a vector with components \( x(t) \) and \( y(t) \).
Vector-valued functions are crucial because they can describe physical phenomena such as velocity or force, which have both magnitude and direction. When working with vector-valued functions, it's essential to manage each vector's components separately.

• The process involves finding expressions for each component.
• For example, in the exercise, \( \mathbf{v}(t) = \langle t, t^2 \rangle \), the first component is the independent variable \( t \) and the second is its square \( t^2 \).

Understanding vector-valued functions is foundational for studying more advanced topics like vector calculus, which involves operations on vector fields.

Expansion of Binomials

One of the steps involved in finding a difference quotient is expanding binomials. A binomial is a polynomial with two terms, such as \( (t+h) \). When we need to expand \( (t+h)^2 \), we can use the binomial expansion method, which involves:

• Applying the formula \( (a+b)^2 = a^2 + 2ab + b^2 \).
• In the context of the exercise, \( a = t \) and \( b = h \), leading to \( (t+h)^2 = t^2 + 2th + h^2 \).

Expanding binomials is a fundamental algebraic skill needed in calculus and other mathematical applications.
This skill enables us to simplify expressions and solve equations more generally.
Mastery of binomial expansion allows you to handle more complex expressions easily, especially when working with derivatives and integrals where such expansions frequently occur.

Calculus Concepts

Calculus introduces several concepts, one of which is the difference quotient. This concept is foundational for understanding derivatives because it measures the average rate of change of a function over an interval. The formula for the difference quotient is \( \frac{v(t+h)-v(t)}{h} \).
The difference quotient is the stepping stone to calculus, driving us towards the concept of instantaneous rate of change. As \( h \) approaches zero, the difference quotient becomes the derivative, which provides the slope of the tangent line at a specific point on the function.

• In our exercise, this is manifested through the difference quotient of \( \mathbf{v}(t) = \langle t, t^2 \rangle \), resulting in \( \langle 1, 2t + h \rangle \).
• The final expression highlights the importance of processing each component of a vector separately.

Understanding these fundamental calculus concepts paves the way for exploring more advanced topics like integration and multi-variable calculus, where vectors and limits play critical roles.