Math Problem Statement

A curve is described by the vector equation r(t) = t*i + t^2*j, t > 0. Calculate r'(t). Calculate the gradient of the curve at the point where t = 1. Determine a unit vector s parallel to the tangent to the curve where t = 1.

Solution

The image contains a vector calculus problem, which describes a curve by the vector equation r(t)=ti+t2j\mathbf{r}(t) = t\mathbf{i} + t^2\mathbf{j}, where t>0t > 0. There are three parts to the question:

  1. Calculate r(t)\mathbf{r}'(t), which is the derivative of the vector equation.
  2. Calculate the gradient of the curve at the point where t=1t = 1.
  3. Determine a unit vector s^\hat{s} parallel to the tangent to the curve where t=1t = 1.

Let’s solve each part step by step:

1. Calculate r(t)\mathbf{r}'(t)

The vector equation is: r(t)=ti+t2j\mathbf{r}(t) = t\mathbf{i} + t^2\mathbf{j}

To find r(t)\mathbf{r}'(t), differentiate each component with respect to tt: r(t)=ddt(ti)+ddt(t2j)\mathbf{r}'(t) = \frac{d}{dt}(t\mathbf{i}) + \frac{d}{dt}(t^2\mathbf{j}) r(t)=i+2tj\mathbf{r}'(t) = \mathbf{i} + 2t\mathbf{j}

So, r(t)=i+2tj\mathbf{r}'(t) = \mathbf{i} + 2t\mathbf{j}.

2. Calculate the gradient at t=1t = 1

The gradient of the curve is given by r(t)\mathbf{r}'(t), so we substitute t=1t = 1 into r(t)\mathbf{r}'(t): r(1)=i+2(1)j=i+2j\mathbf{r}'(1) = \mathbf{i} + 2(1)\mathbf{j} = \mathbf{i} + 2\mathbf{j}

Thus, the gradient at t=1t = 1 is i+2j\mathbf{i} + 2\mathbf{j}.

3. Determine a unit vector s^\hat{s} parallel to the tangent at t=1t = 1

To find a unit vector, we need to normalize r(1)\mathbf{r}'(1). The formula for the magnitude of a vector v=ai+bj\mathbf{v} = a\mathbf{i} + b\mathbf{j} is: v=a2+b2|\mathbf{v}| = \sqrt{a^2 + b^2}

For r(1)=i+2j\mathbf{r}'(1) = \mathbf{i} + 2\mathbf{j}, we have: r(1)=12+22=1+4=5|\mathbf{r}'(1)| = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}

Now, the unit vector s^\hat{s} is: s^=r(1)r(1)=i+2j5=15i+25j\hat{s} = \frac{\mathbf{r}'(1)}{|\mathbf{r}'(1)|} = \frac{\mathbf{i} + 2\mathbf{j}}{\sqrt{5}} = \frac{1}{\sqrt{5}}\mathbf{i} + \frac{2}{\sqrt{5}}\mathbf{j}

Thus, the unit vector s^\hat{s} parallel to the tangent to the curve at t=1t = 1 is: s^=15i+25j\hat{s} = \frac{1}{\sqrt{5}}\mathbf{i} + \frac{2}{\sqrt{5}}\mathbf{j}

Would you like further details or clarification on any step? Here are five follow-up questions to consider:

  1. What is the geometric interpretation of r(t)\mathbf{r}'(t)?
  2. How would you calculate the tangent vector at any arbitrary point t=t0t = t_0?
  3. Can you explain the significance of the unit vector s^\hat{s}?
  4. How does the magnitude of r(t)\mathbf{r}'(t) relate to the speed of motion along the curve?
  5. What would change if the curve were described in three dimensions?

Tip: The unit vector always has a magnitude of 1, which makes it useful in defining direction without scaling effects.

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Math Problem Analysis

Mathematical Concepts

Vector Calculus
Differentiation
Gradient
Tangent Vectors

Formulas

r'(t) = i + 2t*j (Derivative of vector function)
|v| = sqrt(a^2 + b^2) (Magnitude of vector)
Unit vector = v / |v|

Theorems

Differentiation of vector functions
Normalization of vectors

Suitable Grade Level

University Level (Calculus 2 or Vector Calculus)