Question

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

Short answer

Question: Find the derivative of the dot product of the vector-valued functions $$f(t) = t^2(\mathbf{i}+2 \mathbf{j}-2 t \mathbf{k})$$ and $$g(t) = \left(e^{t} \mathbf{i}+2 e^{t} \mathbf{j}-3 e^{-t} \mathbf{k}\right)$$ with respect to the variable t. Answer: The derivative of the dot product with respect to t is $$\frac{d}{dt}(f(t) \cdot g(t)) = 10t e^t + 10t^2 e^t + 6t^3 e^{-t} - 18t^2 e^{-t}$$.

Step-by-step solution

Find the derivatives of f(t) and g(t) with respect to t

To find the derivatives of the component functions, we differentiate each component of the vector with respect to t. For f(t), we have: $$\frac{d}{dt}(t^2 \mathbf{i}) = 2t \mathbf{i}$$ $$\frac{d}{dt}(2t^2 \mathbf{j}) = 4t \mathbf{j}$$ $$\frac{d}{dt}(-2t^3 \mathbf{k}) = -6t^2 \mathbf{k}$$ So, the derivative of f'(t) is given by: $$f'(t) = 2t \mathbf{i} + 4t \mathbf{j} - 6t^2 \mathbf{k}$$ Similarly, for g(t), we have: $$\frac{d}{dt}(e^t \mathbf{i}) = e^t \mathbf{i}$$ $$\frac{d}{dt}(2e^t \mathbf{j}) = 2e^t \mathbf{j}$$ $$\frac{d}{dt}(-3e^{-t} \mathbf{k}) = 3e^{-t} \mathbf{k}$$ So, the derivative of g'(t) is given by: $$g'(t) = e^t \mathbf{i} + 2e^t \mathbf{j} + 3e^{-t} \mathbf{k}$$

Apply the product rule for derivatives

Now we will apply the product rule for derivatives to calculate the required derivative: $$\frac{d}{dt}(f(t) \cdot g(t)) = f'(t) \cdot g(t) + f(t) \cdot g'(t)$$ The dot products will be: $$f'(t) \cdot g(t) = (2t \mathbf{i} + 4t \mathbf{j} - 6t^2 \mathbf{k}) \cdot (e^t \mathbf{i} + 2e^t \mathbf{j} - 3e^{-t} \mathbf{k})$$ and $$f(t) \cdot g'(t) = (t^2(\mathbf{i}+2 \mathbf{j}-2 t \mathbf{k})) \cdot (e^t \mathbf{i} + 2e^t \mathbf{j} + 3e^{-t} \mathbf{k})$$

Compute the dot products

Now compute both dot products: $$f'(t) \cdot g(t) = 2te^t + 8te^t - 18t^2 e^{-t}$$ $$f(t) \cdot g'(t) = t^2e^t + 4t^2e^t + 6t^3 e^{-t}$$ And finally, add both dot products to find the derivative: $$\frac{d}{dt}(f(t) \cdot g(t)) = (2te^t + 8te^t - 18t^2 e^{-t}) + (t^2e^t + 4t^2e^t + 6t^3 e^{-t})$$ $$\frac{d}{dt}(f(t) \cdot g(t)) = 10t e^t + 10t^2 e^t + 6t^3 e^{-t} - 18t^2 e^{-t}$$

Key concepts

Derivatives

Understanding the concept of derivatives is essential in vector calculus. When dealing with vector functions, derivatives help us see how each component of the vector changes with respect to a variable. Think of derivatives as a measure of how fast something is changing at any given point.

In this problem, we start by computing the derivative of each component of the vector function separately. For example:

• The derivative of the \(t^2 \, \mathbf{i}\) component with respect to \(t\) is \(2t \, \mathbf{i}\), indicating that this part grows linearly as \(t\) increases.
• Similarly, by differentiating other terms like \(2t^2 \, \mathbf{j}\) and \(-2t^3 \, \mathbf{k}\), we capture the rate of change for these parts too, resulting in the vectors \(4t \, \mathbf{j}\) and \(-6t^2 \, \mathbf{k}\) respectively.

This breakdown allows us to understand each dimensional change when a vector is differentiated. A critical step in vector calculus that builds the foundation for accurately calculating more complex expressions.

Dot Product

The dot product is a key operation involving two vectors, often producing a scalar result. It combines corresponding components of the vectors in a special way. Think of it as a method to measure how much two vectors point in the same direction.

In this exercise, the dot product is crucial to combining the differentiated vector functions. When you compute the dot product, you multiply each corresponding component of two vectors and then sum these results.

• For example, in the expression \(f'(t) \cdot g(t)\), you would multiply \(2t \, \mathbf{i}\) with \(e^t \, \mathbf{i}\), yielding \(2t e^t\).
• Similarly, obtaining the dot product for \(4t \, \mathbf{j}\) and \(2e^t \, \mathbf{j}\) results in \(8t e^t\).

The dot product simplifies the expression by reducing the combination of multi-dimensional components into a single resultant number, making it easier to manage and solve in calculus problems.

Product Rule

In calculus, the product rule is a fundamental principle used to find the derivative of a product of two functions. This rule is essential in vector calculus when dealing with vector functions.

It states that if you have two functions, say \(u(t)\) and \(v(t)\), their derivative \((uv)'\) is given by \(u'v + uv'\). This can be easily remembered by thinking of it as "the derivative of the first times the second plus the first times the derivative of the second."

• In this exercise, we use this rule to differentiate the product of two vector functions \(f(t)\) and \(g(t)\).
• By applying the product rule, their derivative becomes \(\frac{d}{dt}(f(t) \cdot g(t)) = f'(t) \cdot g(t) + f(t) \cdot g'(t)\).

Using the product rule correctly helps break down the calculation into manageable parts, ultimately leading to a simpler and more accurate outcome.

Vector Functions

Vector functions are a way to articulate mathematical relationships involving vectors. Rather than treating each component separately, vector functions let you work with all components as a single entity.

They are particularly useful in sciences and engineering, where quantities like force and velocity are naturally expressed as vectors.

• For example, a vector function might be expressed as \(\mathbf{r}(t) = x(t) \, \mathbf{i} + y(t) \, \mathbf{j} + z(t) \, \mathbf{k}\), where each letter represents a component that can change as \(t\) changes.
• These functions make it straightforward to perform operations like differentiation and integration, which are crucial to understanding movements and changes in physical systems.

When solving calculus problems, vector functions simplify the process by providing a unifying approach to manage the direction-dependent variables seamlessly.