Math Problem Statement

Find r'(t)*r"(t) given the following vector function. r(t)=(5t^2+5t)i+(5t^2+5t)j

Solution

Given the vector function r(t)=(5t2+5t)i^+(5t2+5t)j^\mathbf{r}(t) = (5t^2 + 5t) \hat{i} + (5t^2 + 5t) \hat{j}, we are tasked with finding r(t)r(t)\mathbf{r}'(t) \cdot \mathbf{r}''(t), which is the dot product of the first and second derivatives of the vector function.

Step 1: Differentiate r(t)\mathbf{r}(t) to find r(t)\mathbf{r}'(t)

The vector function is given as: r(t)=(5t2+5t)i^+(5t2+5t)j^\mathbf{r}(t) = (5t^2 + 5t) \hat{i} + (5t^2 + 5t) \hat{j}

Differentiate r(t)\mathbf{r}(t) with respect to tt: r(t)=ddt((5t2+5t)i^+(5t2+5t)j^)\mathbf{r}'(t) = \frac{d}{dt} \left( (5t^2 + 5t) \hat{i} + (5t^2 + 5t) \hat{j} \right)

For the i^\hat{i}-component: ddt(5t2+5t)=10t+5\frac{d}{dt}(5t^2 + 5t) = 10t + 5

For the j^\hat{j}-component: ddt(5t2+5t)=10t+5\frac{d}{dt}(5t^2 + 5t) = 10t + 5

Thus, we have: r(t)=(10t+5)i^+(10t+5)j^\mathbf{r}'(t) = (10t + 5) \hat{i} + (10t + 5) \hat{j}

Step 2: Differentiate r(t)\mathbf{r}'(t) to find r(t)\mathbf{r}''(t)

Now, differentiate r(t)\mathbf{r}'(t) with respect to tt: r(t)=ddt((10t+5)i^+(10t+5)j^)\mathbf{r}''(t) = \frac{d}{dt} \left( (10t + 5) \hat{i} + (10t + 5) \hat{j} \right)

For the i^\hat{i}-component: ddt(10t+5)=10\frac{d}{dt}(10t + 5) = 10

For the j^\hat{j}-component: ddt(10t+5)=10\frac{d}{dt}(10t + 5) = 10

Thus, we have: r(t)=10i^+10j^\mathbf{r}''(t) = 10 \hat{i} + 10 \hat{j}

Step 3: Compute the dot product r(t)r(t)\mathbf{r}'(t) \cdot \mathbf{r}''(t)

The dot product of two vectors A=a1i^+a2j^\mathbf{A} = a_1 \hat{i} + a_2 \hat{j} and B=b1i^+b2j^\mathbf{B} = b_1 \hat{i} + b_2 \hat{j} is given by: AB=a1b1+a2b2\mathbf{A} \cdot \mathbf{B} = a_1 b_1 + a_2 b_2

For r(t)=(10t+5)i^+(10t+5)j^\mathbf{r}'(t) = (10t + 5) \hat{i} + (10t + 5) \hat{j} and r(t)=10i^+10j^\mathbf{r}''(t) = 10 \hat{i} + 10 \hat{j}, we have: r(t)r(t)=(10t+5)(10)+(10t+5)(10)\mathbf{r}'(t) \cdot \mathbf{r}''(t) = (10t + 5)(10) + (10t + 5)(10) r(t)r(t)=100t+50+100t+50\mathbf{r}'(t) \cdot \mathbf{r}''(t) = 100t + 50 + 100t + 50 r(t)r(t)=200t+100\mathbf{r}'(t) \cdot \mathbf{r}''(t) = 200t + 100

Final Answer:

r(t)r(t)=200t+100\mathbf{r}'(t) \cdot \mathbf{r}''(t) = 200t + 100

Would you like further details or have any questions?

Here are 5 related questions to explore further:

  1. How do you compute the cross product between two vector functions?
  2. How would the result change if the vector function had a third component k^\hat{k}?
  3. What is the physical significance of r(t)r(t)\mathbf{r}'(t) \cdot \mathbf{r}''(t) in the context of motion?
  4. Can you compute r(t)×r(t)\mathbf{r}(t) \times \mathbf{r}'(t) for the same vector function?
  5. How do you interpret r(t)\mathbf{r}'(t) and r(t)\mathbf{r}''(t) geometrically in terms of motion?

Tip: The dot product of two vectors is maximized when the vectors are parallel and minimized when they are perpendicular.

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Math Problem Analysis

Mathematical Concepts

Vector Calculus
Dot Product
Derivatives of Vector Functions

Formulas

Dot product: A · B = a1*b1 + a2*b2
First derivative: r'(t) = d/dt[(5t^2 + 5t)i + (5t^2 + 5t)j]
Second derivative: r''(t) = d/dt[r'(t)]

Theorems

Dot product of vectors
Rules of differentiation (product and chain rule)

Suitable Grade Level

Undergraduate