Question

Compute the following derivatives. $$\frac{d}{d t}\left(\left(t^{3} \mathbf{i}+6 \mathbf{j}-2 \sqrt{t} \mathbf{k}\right) \times\left(3 t \mathbf{i}-12 t^{2} \mathbf{j}-6 t^{-2} \mathbf{k}\right)\right)$$

Short answer

For the given vector functions $$\mathbf{A}(t) = t^{3} \mathbf{i} + 6 \mathbf{j} - 2 \sqrt{t} \mathbf{k}$$ and $$\mathbf{B}(t) = 3t \mathbf{i} - 12t^{2} \mathbf{j} - 6t^{-2} \mathbf{k},$$ their cross product is $$\mathbf{A} \times \mathbf{B} = \mathbf{i}(-36t^{-2} - 24t^{\frac{3}{2}}) - \mathbf{j}(-6t + 6\sqrt{t}) + \mathbf{k}(-12t^5 - 18t).$$ The derivative of the cross product with respect to t is $$\frac{d}{dt} (\mathbf{A} \times \mathbf{B}) = \mathbf{i}(72t^{-3} - 36t^{\frac{1}{2}}) - \mathbf{j}(-6 + \frac{3}{\sqrt{t}}) + \mathbf{k}(-60t^4 - 18).$$

Step-by-step solution

Write down the given vector functions

We are given two vector functions: $$\mathbf{A}(t) = t^{3} \mathbf{i} + 6 \mathbf{j} - 2 \sqrt{t} \mathbf{k},$$ $$\mathbf{B}(t) = 3t \mathbf{i} - 12t^{2} \mathbf{j} - 6t^{-2} \mathbf{k}.$$

Compute the cross product of these vectors

Recall that the cross product of two vectors \(\mathbf{A}\) and \(\mathbf{B}\) is given by $$\mathbf{A} \times \mathbf{B} = \left|\begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ A_{x} & A_{y} & A_{z}\\ B_{x} & B_{y} & B_{z} \end{array}\right| = \mathbf{i} \left(A_{y} B_{z} - A_{z} B_{y}\right) - \mathbf{j} \left(A_{x} B_{z} - A_{z} B_{x}\right) + \mathbf{k} \left(A_{x} B_{y} - A_{y} B_{x}\right).$$ Applying this to vectors \(\mathbf{A}\) and \(\mathbf{B}\), we get: $$\mathbf{A} \times \mathbf{B} = \mathbf{i}(6(- 6t^{-2}) - (- 2\sqrt{t})(-12t^2)) - \mathbf{j}(t^3(-6t^{-2})-(- 2\sqrt{t})(3t)) + \mathbf{k}(t^3(-12t^2)-(6)(3t)).$$ After simplifying: $$\mathbf{A} \times \mathbf{B} = \mathbf{i}(-36t^{-2} - 24t^{\frac{3}{2}}) - \mathbf{j}(-6t + 6\sqrt{t}) + \mathbf{k}(-12t^5 - 18t).$$

Compute the derivative of the cross product vector function

Now that we have the cross product \(\mathbf{A} \times \mathbf{B}\), we can compute the derivative with respect to \(t\): $$\frac{d}{dt} (\mathbf{A} \times \mathbf{B}) = \frac{d}{dt}\left(\mathbf{i}(-36t^{-2} - 24t^{\frac{3}{2}}) - \mathbf{j} (-6t + 6\sqrt{t}) + \mathbf{k}(-12t^5 - 18t)\right).$$ Computing the derivative component-wise, we get: $$\frac{d}{dt} (\mathbf{A} \times \mathbf{B}) = \mathbf{i}\left(\frac{d}{dt}(-36t^{-2} - 24t^{\frac{3}{2}})\right) - \mathbf{j}\left(\frac{d}{dt}(-6t + 6\sqrt{t})\right) + \mathbf{k}\left(\frac{d}{dt}(-12t^5 - 18t)\right).$$ This gives us: $$\frac{d}{dt} (\mathbf{A} \times \mathbf{B}) = \mathbf{i}(72t^{-3} - 36t^{\frac{1}{2}}) - \mathbf{j}(-6 + \frac{3}{\sqrt{t}}) + \mathbf{k}(-60t^4 - 18).$$

Key concepts

Vector Functions

A vector function is a function that takes one or more variables and returns a vector. In vector calculus, we usually deal with vector functions that depend on a parameter, often time \(t\). These functions output vectors in three-dimensional space \(\mathbb{R}^3\) with components for each of the \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\) directions.

Consider the vector functions \(\mathbf{A}(t)\) and \(\mathbf{B}(t)\) given in the exercise:

• \(\mathbf{A}(t) = t^3 \mathbf{i} + 6 \mathbf{j} - 2 \sqrt{t} \mathbf{k}\)
• \(\mathbf{B}(t) = 3t \mathbf{i} - 12t^2 \mathbf{j} - 6t^{-2} \mathbf{k}\)

Each component is a scalar function of \(t\), which together define the direction and magnitude of the vector at any given \(t\).
This allows us to describe paths and other phenomena parametrically, as each \(t\) corresponds to a point in space defined by a vector.

Cross Product

The cross product is an operation on two vectors in three-dimensional space, producing another vector that is perpendicular to both of the original vectors.
The magnitude of this vector is proportional to the area of the parallelogram formed by the original vectors. The formula for the cross product \(\mathbf{A} \times \mathbf{B}\) is:

\[\mathbf{A} \times \mathbf{B} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \A_x & A_y & A_z \B_x & B_y & B_z\end{vmatrix}\]

By applying this determinant formula to the vectors \(\mathbf{A}(t)\) and \(\mathbf{B}(t)\) provided:

• Result in the \(\mathbf{i}\) direction: combines \(A_yB_z - A_zB_y\)
• Result in the \(\mathbf{j}\) direction: combines \(A_zB_x - A_xB_z\)
• Result in the \(\mathbf{k}\) direction: combines \(A_xB_y - A_yB_x\)

This operation is central in many physical applications where it is used to determine perpendicularity and directions such as torque, angular momentum, and the magnetic force on a moving charge.

Derivative of a Vector

The derivative of a vector function with respect to \(t\) measures its rate of change. Just like regular functions, vector functions can be differentiated by treating each component of the vector separately.

In the context of the exercise, the derivative of the cross product vector \(\mathbf{A} \times \mathbf{B}\) with respect to \(t\) involves:

• Differentiating each component of the vector product \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\).
• Ensuring rules of differentiation (like the product rule) are appropriately applied.

Consider for example:

\[\frac{d}{dt} \left( -36t^{-2} \right) = 72t^{-3}\]
It is possible to differentiate each term independently, which allows for the calculation of how one vector changes as \(t\) changes.

Component-wise Differentiation

Component-wise differentiation involves taking the derivative of each part of a vector function separately. Each component is handled as its own independent function. This method simplifies the process of finding derivatives in vector calculus.

Take the cross product result from the exercise for illustration:

• \(\mathbf{i}\) component: contains \(-36t^{-2} - 24t^{3/2}\)
• \(\mathbf{j}\) component: contains \(-6t + 6\sqrt{t}\)
• \(\mathbf{k}\) component: contains \(-12t^5 - 18t\)

Each of these components is differentiated using standard rules from calculus:

• Power Rule: \(d/dt[t^n] = nt^{n-1}\)
• Constant Multiple Rule: \(d/dt[cf(t)] = c\cdot d/dt[f(t)]\)

The outcome is the derivative of the vector across each direction, providing a comprehensive view of how the entire vector function's behavior evolves with time.