Problem 25 - Calculus III

Question

Find \(\mathbf{r}^{\prime}(t) \cdot \mathbf{r}^{\prime \prime}(t)\) for \(\mathbf{r}(t)=-3 t^{5} \mathbf{i}+5 t \mathbf{j}+2 t^{2} \mathbf{k}\)

Step-by-Step Solution

Answer

The dot product \( \mathbf{r}'(t) \cdot \mathbf{r}''(t) \) is \( 900t^7 + 16t \).

Step 1: Find the First Derivative

First, we need to find the first derivative of the vector function \( \mathbf{r}(t) = -3t^5\mathbf{i} + 5t\mathbf{j} + 2t^2\mathbf{k} \). To do this, we differentiate each component with respect to \( t \).\[\mathbf{r}'(t) = \frac{d}{dt}(-3t^5)\mathbf{i} + \frac{d}{dt}(5t)\mathbf{j} + \frac{d}{dt}(2t^2)\mathbf{k} = -15t^4\mathbf{i} + 5\mathbf{j} + 4t\mathbf{k}.\]

Step 2: Find the Second Derivative

Next, we find the second derivative, \( \mathbf{r}''(t) \), by differentiating \( \mathbf{r}'(t) = -15t^4\mathbf{i} + 5\mathbf{j} + 4t\mathbf{k} \) again.\[\mathbf{r}''(t) = \frac{d}{dt}(-15t^4)\mathbf{i} + \frac{d}{dt}(5)\mathbf{j} + \frac{d}{dt}(4t)\mathbf{k} = -60t^3\mathbf{i} + 0 \mathbf{j} + 4\mathbf{k}.\]

Step 3: Compute the Dot Product

Now, we need to find the dot product \( \mathbf{r}'(t) \cdot \mathbf{r}''(t) \). The dot product of two vectors \( \mathbf{a} \) and \( \mathbf{b} \) is given by \( a_1b_1 + a_2b_2 + a_3b_3 \).\[\mathbf{r}'(t) \cdot \mathbf{r}''(t) = (-15t^4)(-60t^3) + (5)(0) + (4t)(4) = 900t^7 + 0 + 16t.\]

Step 4: Combine and Simplify the Expression

Add the terms together and simplify the expression.\[900t^7 + 16t.\]

Step 5: Present the Final Answer

The dot product \( \mathbf{r}'(t) \cdot \mathbf{r}''(t) \) is equal to \( 900t^7 + 16t \). This represents the result of the computation.

Key Concepts

Vector Differentiation

Vector differentiation involves taking the derivative of a vector function with respect to a certain variable, often time denoted as \( t \). This process is much like differentiating scalar functions, except we do it component-wise. In our exerciset, the vector function \( \mathbf{r}(t) = -3t^5\mathbf{i} + 5t\mathbf{j} + 2t^2\mathbf{k} \) needs differentiation.

• First, differentiate each component separately: * \( -3t^5 \) becomes \( -15t^4 \).
• Next, \( 5t \) gets reduced to just \( 5 \).
• Lastly, \( 2t^2 \) turns into \( 4t \).

By applying these rules, we get the first derivative \( \mathbf{r}'(t) = -15t^4\mathbf{i} + 5\mathbf{j} + 4t\mathbf{k} \). It's important to remember that differentiation simplifies each component as per standard calculus rules.

Dot Product

The dot product is a crucial operation in vector calculus that combines two vectors to yield a scalar value. It's computed by multiplying corresponding components of the vectors and adding the results together. For example, given two vectors \( \mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k} \) and \( \mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k} \), the dot product \( \mathbf{a} \cdot \mathbf{b} \) would be calculated as:\[a_1b_1 + a_2b_2 + a_3b_3.\]In this exercise, we apply the dot product to \( \mathbf{r}'(t) \) and \( \mathbf{r}''(t) \). This involves multiplying each corresponding component of the two derived vectors and summing the results:

• \( (-15t^4)(-60t^3) = 900t^7 \)
• \( (5)(0) = 0 \)
• \( (4t)(4) = 16t \)

Ultimately, the sum \( 900t^7 + 16t \) represents the dot product, translating two vectors into a singular scalar quantity.

Derivative of Vector Functions

Taking derivatives of vector functions incorporates understanding individual component differentiation and vector calculus rules. For multi-variable vector functions, this involves differentiating each element separately. When dealing with higher derivatives, such as second derivatives, follow the same process again for the result from the first differentiation.This exercise presents a demonstration:

• Starting with \( \mathbf{r}'(t) = -15t^4\mathbf{i} + 5\mathbf{j} + 4t\mathbf{k} \), we derive again to find the second derivative.
• Differentiate once more:
  \( -15t^4 \) becomes \( -60t^3 \).
• \( 5 \) derivative goes to zero, as constants vanish upon differentiation.
• Finally, \( 4t \) becomes \( 4 \), delivering the second derivative \( \mathbf{r}''(t) = -60t^3\mathbf{i} + 0\mathbf{j} + 4\mathbf{k} \).

Utilizing these steps helps fully understand behavior changes of vector functions over time, providing insights into both motion and dynamic systems in physics or engineering.