Vector Function and Derivative

Oct 13, 2025 by recoveryshake.com 73 views

Introduction

In vector calculus, the derivative of a vector function and the unit tangent vector are fundamental concepts used to describe the behavior of curves and surfaces in space. In this article, we will explore how to calculate the derivative of a vector function, the unit tangent vector, and evaluate the unit tangent vector at a specific point.

Vector Function and Derivative

A vector function is a function that takes a scalar input and returns a vector output. In this case, we are given the vector function:

r(t) = \left\langle 4t^4, 3t \right\rangle

To find the derivative of this vector function, we will apply the derivative rules to each component separately.

Derivative of the First Component

The derivative of the first component is found by differentiating the function $4t^4$ with respect to $t$ . Using the power rule of differentiation, we get:

\frac{d}{dt} (4t^4) = 16t^3

Derivative of the Second Component

The derivative of the second component is found by differentiating the function $3t$ with respect to $t$ . Using the power rule of differentiation, we get:

\frac{d}{dt} (3t) = 3

Derivative of the Vector Function

Now that we have found the derivatives of each component, we can write the derivative of the vector function as:

r^{\prime}(t) = \langle 16t^3, 3 \rangle

Unit Tangent Vector

The unit tangent vector is a vector that points in the direction of the curve at a given point. It is defined as:

T(t) = \frac{r^{\prime}(t)}{\| r^{\prime}(t) \|}

where $\| r^{\prime}(t) \|$ is the magnitude of the derivative of the vector function.

Magnitude of the Derivative

To find the magnitude of the derivative, we will use the formula:

\| r^{\prime}(t) \| = \sqrt{(16t^3)^2 + 3^2}

Simplifying the expression, we get:

\| r^{\prime}(t) \| = \sqrt{256t^6 + 9}

Unit Tangent Vector

Now that we have found the magnitude of the derivative, we can write the unit tangent vector as:

T(t) = \frac{\langle 16t^3, 3 \rangle}{\sqrt{256t^6 + 9}}

Evaluating the Unit Tangent Vector at a Specific Point

To evaluate the unit tangent vector at a specific point, we will substitute the value of $t$ into the expression for the unit tangent vector.

Evaluating the Unit Tangent Vector at $t=1$

Substituting $t=1$ into the expression for the unit tangent vector, we get:

T(1) = \frac{\langle 16(1)^3, 3 \rangle}{\sqrt{256(1)^6 + 9}}

Simplifying the expression, we get:

T(1) = \frac{\langle 16, 3 \rangle}{\sqrt{265}}

Simplifying the Expression

To simplify the expression, we can divide the numerator and denominator by the square root of the denominator:

T(1) = \frac{\langle 16, 3 \rangle}{\sqrt{265}} \cdot \frac{\sqrt{265}}{\sqrt{265}}

Simplifying the expression, we get:

T(1) = \frac{\langle 16\sqrt{265}, 3\sqrt{265} \rangle}{265}

Conclusion

$In this article, we have calculated the derivative of a vector function, the unit tangent vector, and evaluated the unit tangent vector at a specific point. We have shown that the derivative of the vector function is $\langle 16t^3, 3 \rangle$, the unit tangent vector is $\frac{\langle 16t^3, 3 \rangle}{\sqrt{256t^6 + 9}}$, and the unit tangent vector at $t=1$ is $\frac{\langle 16\sqrt{265}, 3\sqrt{265} \rangle}{265}$. These results demonstrate the importance of vector calculus in describing the behavior of curves and surfaces in space.<br/> **Vector Calculus Q&A: Derivatives and Unit Tangents**$

Introduction

In our previous article, we explored the concepts of derivatives and unit tangents in vector calculus. We calculated the derivative of a vector function, the unit tangent vector, and evaluated the unit tangent vector at a specific point. In this article, we will answer some frequently asked questions about derivatives and unit tangents.

Q&A

Q: What is the derivative of a vector function?

A: The derivative of a vector function is a vector that represents the rate of change of the function with respect to the input variable.

Q: How do I calculate the derivative of a vector function?

A: To calculate the derivative of a vector function, you need to apply the derivative rules to each component of the vector separately.

Q: What is the unit tangent vector?

A: The unit tangent vector is a vector that points in the direction of the curve at a given point. It is defined as the derivative of the vector function divided by the magnitude of the derivative.

Q: How do I calculate the unit tangent vector?

A: To calculate the unit tangent vector, you need to find the magnitude of the derivative and then divide the derivative by the magnitude.

Q: What is the significance of the unit tangent vector?

A: The unit tangent vector is significant because it provides a way to describe the direction of a curve at a given point. It is used in many applications, including physics, engineering, and computer graphics.

Q: Can I use the unit tangent vector to find the curvature of a curve?

A: Yes, you can use the unit tangent vector to find the curvature of a curve. The curvature is defined as the rate of change of the unit tangent vector with respect to the arc length.

Q: How do I calculate the curvature of a curve?

A: To calculate the curvature of a curve, you need to find the rate of change of the unit tangent vector with respect to the arc length. This can be done using the formula:

\kappa = \frac{dT}{ds}

where $T$ is the unit tangent vector and $s$ is the arc length.

Q: What is the relationship between the unit tangent vector and the normal vector?

A: The unit tangent vector and the normal vector are related by the formula:

N = \frac{dT}{ds}

where $N$ is the normal vector and $T$ is the unit tangent vector.

Q: Can I use the unit tangent vector and the normal vector to find the curvature of a curve?

A: Yes, you can use the unit tangent vector and the normal vector to find the curvature of a curve. The curvature is defined as:

\kappa = \frac{dT}{ds} = \frac{N}{\| N \|}

Q: What is the significance of the curvature of a curve?

A: The curvature of a curve is significant because it provides a way to describe the shape of a curve. It is used in many applications, including physics, engineering, and computer graphics.

Conclusion

In this article, we have answered some frequently asked questions about derivatives and unit tangents in vector calculus. We have discussed the significance of the unit tangent vector and the curvature of a curve, and provided formulas for calculating these quantities. We hope that this article has been helpful in clarifying some of the concepts in vector calculus.