Find The Derivative Of The Vector Function

Finding the derivative of a vector function is a fundamental concept in calculus, especially when dealing with motion in space, curves, and vector fields. It extends the familiar idea of differentiation from scalar functions to functions that output vectors. This article aims to provide a comprehensive guide on how to find the derivative of a vector function, complete with examples and explanations.

Understanding Vector Functions

A vector function is a function that takes one or more scalar variables (often time, denoted as t) and returns a vector. In three-dimensional space, a vector function can be represented as:

r(t) = <f(t), g(t), h(t)> = f(t)i + g(t)j + h(t)k

Where:

• r(t) is the vector function.
• f(t), g(t), and h(t) are scalar functions representing the components of the vector in the x, y, and z directions, respectively.
• i, j, and k are the unit vectors along the x, y, and z axes.

The Derivative of a Vector Function

The derivative of a vector function r(t) is another vector function, denoted as r'(t) or dr/dt, which represents the instantaneous rate of change of the vector r(t) with respect to the scalar variable t. It is found by differentiating each component function separately:

r'(t) = <f'(t), g'(t), h'(t)> = f'(t)i + g'(t)j + h'(t)k

Where:

• f'(t), g'(t), and h'(t) are the derivatives of the scalar component functions f(t), g(t), and h(t), respectively.

Geometric Interpretation

Geometrically, the derivative r'(t) of a vector function r(t) represents a tangent vector to the curve traced by r(t) at the point corresponding to t. The tangent vector points in the direction of the curve at that point and its magnitude indicates the speed at which the curve is being traced.

Steps to Find the Derivative of a Vector Function

To find the derivative of a vector function, follow these straightforward steps:

1. Identify the Component Functions: Begin by identifying the scalar component functions f(t), g(t), and h(t) of the vector function r(t).
2. Differentiate Each Component: Differentiate each component function with respect to t to find f'(t), g'(t), and h'(t). Use standard differentiation rules for scalar functions (power rule, product rule, chain rule, etc.).
3. Form the Derivative Vector Function: Combine the derivatives of the component functions to form the derivative vector function r'(t) = <f'(t), g'(t), h'(t)>.
4. Simplify (if possible): Simplify the resulting vector function if possible.

Examples

Let’s illustrate the process with several examples.

Example 1: Basic Polynomial Vector Function

Find the derivative of the vector function:

r(t) = <t^2, 3t, t^3 + 1>

Solution:

1. Identify Component Functions:

   • f(t) = t^2
   • g(t) = 3t
   • h(t) = t^3 + 1

2. Differentiate Each Component:

   • f'(t) = 2t
   • g'(t) = 3
   • h'(t) = 3t^2

3. Form the Derivative Vector Function:

   r'(t) = <2t, 3, 3t^2>

Example 2: Trigonometric Vector Function

Find the derivative of the vector function:

r(t) = <cos(t), sin(t), t>

Solution:

1. Identify Component Functions:

   • f(t) = cos(t)
   • g(t) = sin(t)
   • h(t) = t

2. Differentiate Each Component:

   • f'(t) = -sin(t)
   • g'(t) = cos(t)
   • h'(t) = 1

3. Form the Derivative Vector Function:

   r'(t) = <-sin(t), cos(t), 1>

This vector function represents the tangent vector to a helix.

Example 3: Exponential and Polynomial Vector Function

Find the derivative of the vector function:

r(t) = <e^(2t), t^2 + 2t, 5>

Solution:

1. Identify Component Functions:

   • f(t) = e^(2t)
   • g(t) = t^2 + 2t
   • h(t) = 5

2. Differentiate Each Component:

   • f'(t) = 2e^(2t)
   • g'(t) = 2t + 2
   • h'(t) = 0

3. Form the Derivative Vector Function:

   r'(t) = <2e^(2t), 2t + 2, 0>

Example 4: Vector Function with Product Rule

Find the derivative of the vector function:

r(t) = <t*cos(t), t*sin(t), t^2>

Solution:

1. Identify Component Functions:

   • f(t) = t*cos(t)
   • g(t) = t*sin(t)
   • h(t) = t^2

2. Differentiate Each Component:

   • For f(t) = t*cos(t), use the product rule: (uv)' = u'v + uv'

     f'(t) = (1)*cos(t) + t*(-sin(t)) = cos(t) - t*sin(t)

   • For g(t) = t*sin(t), use the product rule: (uv)' = u'v + uv'

     g'(t) = (1)*sin(t) + t*(cos(t)) = sin(t) + t*cos(t)

   • h'(t) = 2t

3. Form the Derivative Vector Function:

   r'(t) = <cos(t) - t*sin(t), sin(t) + t*cos(t), 2t>

Higher-Order Derivatives

Just as with scalar functions, you can find higher-order derivatives of vector functions. The second derivative, denoted as r''(t), is the derivative of the first derivative r'(t). Similarly, the third derivative r'''(t) is the derivative of the second derivative, and so on.

To find higher-order derivatives, simply repeat the process of differentiating each component function.

Example 5: Second Derivative

Given r(t) = <t^2, sin(t), cos(t)>, find r''(t).

Solution:

1. Find the First Derivative:

   r'(t) = <2t, cos(t), -sin(t)>

2. Find the Second Derivative:

   r''(t) = <2, -sin(t), -cos(t)>

Properties of Vector Function Derivatives

The derivatives of vector functions follow several useful properties, analogous to those of scalar functions.

1. Constant Vector Rule: If c is a constant vector, then d/dt [c] = 0.
2. Constant Multiple Rule: If c is a scalar constant, then d/dt [c*r(t)] = c*r'(t).
3. Sum/Difference Rule: d/dt [r(t) ± s(t)] = r'(t) ± s'(t).
4. Scalar Product Rule: If f(t) is a scalar function, then d/dt [f(t)*r(t)] = f'(t)*r(t) + f(t)*r'(t).
5. Dot Product Rule: d/dt [r(t) · s(t)] = r'(t) · s(t) + r(t) · s'(t).
6. Cross Product Rule: d/dt [r(t) x s(t)] = r'(t) x s(t) + r(t) x s'(t).
7. Chain Rule: If r(t) = u(f(t)), then d/dt [r(t)] = u'(f(t)) * f'(t).

Applications of Vector Function Derivatives

The derivative of a vector function has many important applications in physics and engineering.

Velocity and Acceleration

In physics, if r(t) represents the position vector of a particle at time t, then:

• The velocity vector v(t) is the first derivative of the position vector: v(t) = r'(t).
• The acceleration vector a(t) is the second derivative of the position vector: a(t) = v'(t) = r''(t).

The speed of the particle is the magnitude of the velocity vector: |v(t)|.

Tangent Lines and Curves

As mentioned earlier, r'(t) represents a tangent vector to the curve traced by r(t) at the point corresponding to t. This is useful for finding tangent lines to curves in space.

Arc Length

The arc length s of a curve traced by r(t) from t = a to t = b is given by the integral:

s = ∫[a to b] |r'(t)| dt

Where |r'(t)| is the magnitude of the derivative vector, representing the speed along the curve.

Curvature

The curvature κ (kappa) of a curve at a point is a measure of how much the curve bends at that point. It can be calculated using the formula:

κ = |r'(t) x r''(t)| / |r'(t)|^3

Common Mistakes to Avoid

1. Incorrect Differentiation: Make sure to correctly differentiate each component function. Double-check your calculus rules, especially the chain rule, product rule, and quotient rule.
2. Forgetting the Vector Notation: The derivative r'(t) is a vector function, so remember to express your answer in vector form.
3. Misapplying Differentiation Rules: Ensure that you apply the correct differentiation rules for dot products, cross products, and scalar products.
4. Algebraic Errors: Be careful with algebraic manipulations, especially when simplifying the derivative vector function.

Advanced Techniques

Parametric Equations

Vector functions are often used to represent parametric equations of curves and surfaces. Understanding how to differentiate vector functions is crucial for analyzing these curves and surfaces.

Vector Fields

In advanced calculus, vector functions are also used to define vector fields, which assign a vector to each point in space. Differentiating vector fields involves more complex operations like the gradient, divergence, and curl, which are essential in physics and engineering.

Practical Examples in Physics

Projectile Motion

Consider a projectile launched with an initial velocity v0 at an angle θ with the horizontal. The position vector r(t) can be expressed as:

r(t) = <(v0*cos(θ))*t, (v0*sin(θ))*t - (1/2)*g*t^2>

Where g is the acceleration due to gravity.

The velocity vector v(t) is:

v(t) = <v0*cos(θ), v0*sin(θ) - g*t>

The acceleration vector a(t) is:

a(t) = <0, -g>

Circular Motion

For an object moving in a circle of radius R with constant angular velocity ω, the position vector r(t) can be expressed as:

r(t) = <R*cos(ωt), R*sin(ωt)>

The velocity vector v(t) is:

v(t) = <-Rω*sin(ωt), Rω*cos(ωt)>

The acceleration vector a(t) is:

a(t) = <-Rω^2*cos(ωt), -Rω^2*sin(ωt)> = -ω^2*r(t)

Conclusion

Finding the derivative of a vector function is a vital skill in calculus and has extensive applications in physics, engineering, and computer graphics. By understanding the basic concepts, mastering the differentiation rules, and practicing with examples, you can effectively analyze and manipulate vector functions. Whether you're calculating velocities and accelerations, finding tangent lines, or determining arc lengths, the ability to differentiate vector functions is indispensable.