How do you apply the product rule to find the derivative of a vector function like R_i(t) = (t, t^3, 8t)?

Since R_i(t) is a vector with components, you differentiate each component separately using standard rules. The product rule is typically used when components are products of functions, but in this case, each component is a simple function of t, so their derivatives are straightforward: d/dt(t) = 1, d/dt(t^3) = 3t^2, and d/dt(8t) = 8.

What is the derivative R_i'(t) of the vector function R_i(t) = (t, t^3, 8t)?

The derivative R_i'(t) is (1, 3t^2, 8), obtained by differentiating each component separately.

How do you evaluate the derivative R_i'(t) at t = 2 for R_i(t) = (t, t^3, 8t)?

Substitute t = 2 into R_i'(t): R_i'(2) = (1, 3(2)^2, 8) = (1, 34, 8) = (1, 12, 8).

Is the product rule necessary for differentiating R_i(t) = (t, t^3, 8t)?

No, because each component is a simple function of t. The product rule is only needed if components are products of functions. Here, direct differentiation suffices.

What is the significance of the point R(2) = (2, 1, 0) in the context of differentiating R_i(t)?

The point R(2) = (2, 1, 0) indicates the value of the vector function at t=2, but when differentiating, we focus on R_i'(t). To evaluate the derivative at t=2, we use the derivative formula, which yields R_i'(2) = (1, 12, 8).

Evaluate The Derivative By Using The Appropriate Product Rule Where Ri(t) = (t,t3, 8t), R(2) = (2,1,0),

Evaluate The Derivative By Using The Appropriate Product Rule Where Ri(t) = (t,t3, 8t), R(2) = (2,1,0),

Understanding how to evaluate derivatives effectively is a fundamental skill in calculus, especially when dealing with vector-valued functions. In this article, we will explore the process of differentiating a vector function, specifically Ri(t) = (t, t³, 8t), using the appropriate product rule. Additionally, we will evaluate the derivative at a particular point, t = 2, and understand how initial conditions like R(2) = (2, 1, 0) relate to the differentiation process. This comprehensive guide aims to clarify the concepts, demonstrate detailed steps, and enhance your calculus problem-solving skills.

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Understanding Vector-Valued Functions and Their Derivatives

Before diving into the differentiation process, it is crucial to understand what vector-valued functions are and how their derivatives are computed.

What Is a Vector-Valued Function?

A vector-valued function assigns a vector to each point in its domain. It can be written as:

\[ \mathbf{R}(t) = \langle R1(t), R2(t), R_3(t) \rangle \]

where each component function \( R_i(t) \) is a real-valued function of t.

In our case:

\[ \mathbf{R}_i(t) = (t, t^3, 8t) \]

which means:

\( R_{i1}(t) = t \)

\( R_{i2}(t) = t^3 \)

\( R_{i3}(t) = 8t \)

Derivative of Vector-Valued Functions

The derivative of a vector function is obtained by differentiating each component function separately:

\[ \mathbf{R}'(t) = \left( \frac{d}{dt} R1(t), \frac{d}{dt} R2(t), \frac{d}{dt} R_3(t) \right) \]

This process is straightforward when the components are simple functions, but complexity arises when the components involve products or compositions requiring the product rule or chain rule.

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Applying the Product Rule to Vector Functions

The product rule in calculus is essential when differentiating products of functions. For scalar functions u(t) and v(t), the product rule states:

\[ \frac{d}{dt} [u(t) v(t)] = u'(t) v(t) + u(t) v'(t) \]

When dealing with vector functions, if the components involve products, the same rule applies component-wise.

Specifics of the Product Rule for Vector Components

Suppose a component of a vector function is a product of two functions, say \( R(t) = u(t) \cdot v(t) \). Its derivative is:

\[ R'(t) = u'(t) v(t) + u(t) v'(t) \]

This rule extends naturally to each component of a vector function.

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Differentiating Ri(t) = (t, t³, 8t)

In our case, the function Ri(t) = (t, t³, 8t) consists of three separate component functions:

\( R_1(t) = t \)

\( R_2(t) = t^3 \)

\( R_3(t) = 8t \)

Since these are simple functions, their derivatives are straightforward:

\( R_1'(t) = \frac{d}{dt} t = 1 \)

\( R_2'(t) = \frac{d}{dt} t^3 = 3t^2 \)

\( R_3'(t) = \frac{d}{dt} 8t = 8 \)

However, to illustrate the use of the product rule, consider how the derivative would look if the components involved products.

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Example: Derivative of a Component Using the Product Rule

Suppose a component is \( R(t) = t \cdot t^2 \). Its derivative would be:

\[ R'(t) = \frac{d}{dt} (t \cdot t^2) \]

Applying the product rule:

\[ R'(t) = (1) \cdot t^2 + t \cdot 2t = t^2 + 2t^2 = 3t^2 \]

Similarly, for our original components, if they involved products, the derivatives would involve applying the product rule to each.

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Calculating the Derivative at t = 2

Having differentiated each component, we now evaluate the derivative \( \mathbf{R}'(t) \) at \( t = 2 \).

From earlier, the derivatives are:

\( R_1'(t) = 1 \)

\( R_2'(t) = 3t^2 \)

\( R_3'(t) = 8 \)

Plugging in \( t = 2 \):

\( R_1'(2) = 1 \)

\( R_2'(2) = 3 \times (2)^2 = 3 \times 4 = 12 \)

\( R_3'(2) = 8 \)

Thus,

\[ \mathbf{R}'(2) = (1, 12, 8) \]

This vector represents the rate of change of the function at \( t = 2 \), giving insights into the velocity if the function models a physical path, or the slope if the function models a rate.

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Understanding the Initial Condition R(2) = (2, 1, 0)

The initial condition states that at \( t = 2 \), the vector function yields:

\[ \mathbf{R}(2) = (2, 1, 0) \]

This information is essential for solving problems such as initial value problems in differential equations or confirming the correctness of the function.

In our case, the initial condition confirms the specific point on the curve at \( t = 2 \), but it does not affect the derivative directly. The derivative depends on the functional form, which we've already differentiated.

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Summary of the Differentiation Process

To summarize, the steps to evaluate the derivative of a vector function like Ri(t) = (t, t³, 8t) using the appropriate rules are:

Identify each component function: Recognize the structure of each component.

Differentiate each component: Use basic differentiation rules; apply the product rule if components involve products.

Evaluate at the given point: Substitute the specific value of t into the derivatives.

Interpret the result: Understand what the derivative vector signifies in context.

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Additional Tips for Differentiating Vector Functions

Always differentiate component-wise; the derivative of a vector function is the vector of derivatives.
When components involve products, explicitly apply the product rule.
For composite functions, use the chain rule as necessary.
Remember that initial conditions are critical for solving differential equations but do not influence the differentiation process directly.

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Conclusion

Differentiating vector-valued functions like Ri(t) = (t, t³, 8t) is a fundamental skill in calculus, especially when analyzing motion, growth processes, or other phenomena modeled by vectors. By understanding how to apply the product rule appropriately, you can accurately compute derivatives, interpret the rate of change, and solve related problems efficiently. Evaluating the derivative at specific points, such as \( t = 2 \), provides meaningful insights into the behavior of the function at those moments. Mastery of these concepts enhances your overall calculus proficiency and prepares you for more advanced mathematical applications.

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