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Chapter 2: Problem 12

Compute the tangent vectors to the given path. $$\mathbf{c}(t)=\left(3 t^{2}, t^{3}\right)$$

Short Answer

Expert verified
The tangent vector is \( \mathbf{c}'(t) = (6t, 3t^2) \).

Step by step solution

01

Understand the Path Equation

The path is defined by the vector function \( \mathbf{c}(t) = (3t^2, t^3) \), where \( t \) is a parameter. Our task is to find the tangent vectors along this path for various values of \( t \).
02

Find the Derivative of the Path

To find the tangent vector, we differentiate \( \mathbf{c}(t) \) with respect to \( t \). This gives us the velocity vector, which is tangent to the path.The derivative is obtained component-wise:1. Differentiate the first component: \( \frac{d}{dt}(3t^2) = 6t \).2. Differentiate the second component: \( \frac{d}{dt}(t^3) = 3t^2 \).
03

Write the Tangent Vector

Combining the results from Step 2, the tangent vector to the path \( \mathbf{c}(t) \) is given by:\[ \mathbf{c}'(t) = (6t, 3t^2). \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vector Functions
A vector function allows us to describe a path in space by giving a vector for each point along the path. Think of it like a map that tells us exactly where we are at any given time. If we have a function like \( \mathbf{c}(t) = (3t^2, t^3) \), the vector is described in terms of a parameter \( t \). Here, each component of the vector depends on \( t \), which means as \( t \) changes, our position along the path changes too.
  • The first component, \( 3t^2 \), controls movement in the horizontal direction.
  • The second component, \( t^3 \), controls movement in the vertical direction.
Understanding vector functions is crucial because they help us model paths, shapes, or motions through space. They serve as the foundation for studying more dynamic behaviors like motion and rates of change in calculus.
Differentiation
Differentiation is a key concept in calculus that lets us find the rate at which something changes. When we differentiate a vector function like \( \mathbf{c}(t) = (3t^2, t^3) \), we're looking to see how the path changes over time.
  • For \( 3t^2 \), differentiation gives \( 6t \). This tells us how quickly the horizontal position is changing as \( t \) increases.
  • For \( t^3 \), differentiation gives \( 3t^2 \). This tells us how quickly the vertical position is changing with respect to \( t \).
By finding these derivatives, we're not just seeing the current position, but the direction and speed at which the point is moving. This process provides us with a new function, called the tangent or velocity vector, which describes this change.
Velocity Vectors
When we say "velocity vectors," we're referring to vectors that show the speed and direction of an object's motion. In the context of our path \( \mathbf{c}(t) = (3t^2, t^3) \), the derivative \( \mathbf{c}'(t) = (6t, 3t^2) \) becomes our velocity vector.
The components of this vector tell us:
  • \( 6t \) indicates how fast the horizontal movement happens as \( t \) changes.
  • \( 3t^2 \) shows the rate of vertical movement.
Velocity vectors are essential in physics and engineering when analyzing motions like car trajectories or particle paths. They help us understand not just where an object is, but where it's headed and how fast, serving as a blueprint for dynamic systems in motion.

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