Problem 60 - Calculus 3

Question

Let $$\mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{j}-t^{4} \mathbf{k}$$ and $$s(t)=\sin (t) \mathbf{i}+e^{t} \mathbf{j}+\cos (t) \mathbf{k}$$ Here is the graph of the function: Find the following. $$\frac{d}{d t}\left[t^{2} \cdot s(t)\right]$$

Step-by-Step Solution

Answer

The derivative is \((2t \sin(t) + t^2 \cos(t)) \mathbf{i} + (2t e^t + t^2 e^t) \mathbf{j} + (2t \cos(t) - t^2 \sin(t)) \mathbf{k}\)."}

Step 1: Understand the Function

We need to find the derivative of the product of the scalar function \(t^2\) and the vector function \(s(t)\). The function given is \(t^2 \cdot s(t)\), which is a vector function.

Step 2: Recall the Product Rule for Derivatives

To differentiate a product \(u(t)\cdot v(t)\), we use the product rule: \(\frac{d}{dt}[u(t)\cdot v(t)] = u'(t)\cdot v(t) + u(t)\cdot v'(t)\). Here, \(u(t) = t^2\) and \(v(t) = s(t)\).

Step 3: Differentiate the Scalar Function

The scalar function \(u(t) = t^2\) is simply differentiated using the power rule: \(u'(t) = 2t\).

Step 4: Differentiate the Vector Function

The vector function \(s(t) = \sin (t) \mathbf{i} + e^{t} \mathbf{j} + \cos (t) \mathbf{k}\) is differentiated component-wise: - \(\frac{d}{dt}[\sin(t)] = \cos(t)\)- \(\frac{d}{dt}[e^t] = e^t\)- \(\frac{d}{dt}[\cos(t)] = -\sin(t)\)Thus, \(s'(t) = \cos(t) \mathbf{i} + e^{t} \mathbf{j} - \sin(t) \mathbf{k}\).

Step 5: Apply the Product Rule

Now apply the product rule: \[\frac{d}{dt}[t^2 \cdot s(t)] = 2t \cdot s(t) + t^2 \cdot s'(t)\]

Step 6: Calculate Each Part

Substitute back \(s(t)\) and \(s'(t)\) into the product rule from Step 5:1. \(2t \cdot s(t) = 2t(\sin(t) \mathbf{i} + e^t \mathbf{j} + \cos(t) \mathbf{k}) = 2t \sin(t) \mathbf{i} + 2t e^t \mathbf{j} + 2t \cos(t) \mathbf{k}\)2. \(t^2 \cdot s'(t) = t^2 (\cos(t) \mathbf{i} + e^t \mathbf{j} - \sin(t) \mathbf{k}) = t^2 \cos(t) \mathbf{i} + t^2 e^t \mathbf{j} - t^2 \sin(t) \mathbf{k}\)

Key Concepts

Derivative of Vector Functions

When working with vector functions, it's essential to understand how derivatives are applied. Vector functions are expressions combining multiple components—typically along the x, y, and z axes. For instance, in our given problem, the vector function is \(s(t) = \sin(t)\mathbf{i} + e^{t}\mathbf{j} + \cos(t)\mathbf{k}\). Each component function corresponds to a different vector axis.

Taking the derivative of a vector function requires differentiating each component individually. This is similar to scalar functions but applied to each separate part of the vector.

• For \(\sin(t)\mathbf{i}\), the derivative is \(\cos(t)\mathbf{i}\).
• For \(e^{t}\mathbf{j}\), it stays the same, \(e^{t}\mathbf{j}\).
• Lastly, \(\cos(t)\mathbf{k}\) simplifies to \(-\sin(t)\mathbf{k}\).

This results in the new vector function \(s'(t) = \cos(t)\mathbf{i} + e^{t}\mathbf{j} - \sin(t)\mathbf{k}\), which is vital for understanding how the function changes at any given point \(t\).

Product Rule

The product rule is a fundamental concept in calculus, used when differentiating the product of two functions. When you multiply two functions, like \(u(t) = t^2\) and \(v(t) = s(t)\), simply differentiating each and multiplying will not yield the correct derivative of the product. The product rule formula is:
\[\frac{d}{dt} [u(t) \cdot v(t)] = u'(t) \cdot v(t) + u(t) \cdot v'(t)\]
In applying this rule to our function, we first need to find \(u'(t)\) and \(v'(t)\). Using the power rule, \(u(t) = t^2\) becomes \(u'(t) = 2t\). We've already found \(s'(t)\) as the derivative of the vector function.
The \(u'(t)\cdot v(t)\) term is \(2t \cdot (\sin(t)\mathbf{i} + e^{t}\mathbf{j} + \cos(t)\mathbf{k})\), and the \(u(t)\cdot v'(t)\) is \(t^2 \cdot (\cos(t)\mathbf{i} + e^{t}\mathbf{j} - \sin(t)\mathbf{k})\). Combining these gives us the full derivative of the original product.

Differentiation

Differentiation is a cornerstone of calculus, allowing us to determine the rate at which a function is changing at any point. In the context of our exercise, we're differentiating to find the derivative of the product of a scalar function and a vector function.
This requires careful attention to how each component of the function behaves under the action of a derivative. Using basic rules like the product rule and the power rule simplifies this process. Here's a quick recap:

• Power Rule: For any function \(t^n\), its derivative is \(nt^{n-1}\). This was used to find \(u'(t) = 2t\) for \(t^2\).
• Component-wise Derivative: For vector functions, differentiate each function inside the vector individually.

By combining these differentiation techniques, we evaluated the complex functions and found their derivatives, giving us insight into how these functions change with each variable or parameter, enhancing our ability to analyze vector fields and systems.