Find the Parametric equation of the tangent vector of curve r(t) = t^2 i + 2t^3 j + 3t k at t = 1.
The Correct Answer and Explanation is :
To find the parametric equation of the tangent vector of the curve
[
\mathbf{r}(t) = t^2 \mathbf{i} + 2t^3 \mathbf{j} + 3t \mathbf{k}
]
at ( t = 1 ), follow these steps:
Step 1: Compute the Velocity Vector
The velocity vector is the derivative of (\mathbf{r}(t)):
[
\mathbf{v}(t) = \frac{d}{dt} (t^2 \mathbf{i} + 2t^3 \mathbf{j} + 3t \mathbf{k})
]
Differentiating component-wise:
- ( \frac{d}{dt} (t^2) = 2t ), so the i-component is ( 2t ).
- ( \frac{d}{dt} (2t^3) = 6t^2 ), so the j-component is ( 6t^2 ).
- ( \frac{d}{dt} (3t) = 3 ), so the k-component is ( 3 ).
Thus, the velocity (tangent) vector is:
[
\mathbf{v}(t) = (2t) \mathbf{i} + (6t^2) \mathbf{j} + 3 \mathbf{k}
]
Step 2: Evaluate at ( t = 1 )
Substituting ( t = 1 ):
[
\mathbf{v}(1) = (2(1)) \mathbf{i} + (6(1)^2) \mathbf{j} + 3 \mathbf{k}
]
[
= 2 \mathbf{i} + 6 \mathbf{j} + 3 \mathbf{k}
]
Step 3: Parametric Equations
The tangent line at ( t = 1 ) passes through the point ( \mathbf{r}(1) ) with direction ( \mathbf{v}(1) ).
First, find ( \mathbf{r}(1) ):
[
\mathbf{r}(1) = (1^2) \mathbf{i} + 2(1)^3 \mathbf{j} + 3(1) \mathbf{k} = 1 \mathbf{i} + 2 \mathbf{j} + 3 \mathbf{k}
]
The parametric equations of the tangent line:
[
x = 1 + 2s, \quad y = 2 + 6s, \quad z = 3 + 3s
]
where ( s ) is a parameter along the tangent direction.
Explanation (300 Words)
A parametric equation of the tangent vector provides a way to describe a tangent line at a specific point on a curve. In vector form, the curve is given by ( \mathbf{r}(t) ), and the tangent vector at any point is obtained by differentiating ( \mathbf{r}(t) ). This derivative, ( \mathbf{v}(t) ), represents the instantaneous rate of change of ( \mathbf{r}(t) ) and is thus tangent to the curve at each ( t ).
For this problem, we first computed ( \mathbf{v}(t) ) by differentiating each component of ( \mathbf{r}(t) ). This gives a vector function ( \mathbf{v}(t) = (2t, 6t^2, 3) ), which describes how the position changes over time.
To find the specific tangent vector at ( t = 1 ), we evaluated ( \mathbf{v}(1) ), yielding ( (2,6,3) ). This vector represents the direction of the tangent line.
Next, we found the position of the curve at ( t = 1 ), which is ( \mathbf{r}(1) = (1,2,3) ). Using the point-direction equation of a line, we wrote parametric equations for the tangent line using the form:
[
(x, y, z) = (x_0, y_0, z_0) + s(v_x, v_y, v_z)
]
Substituting ( (1,2,3) ) as the point and ( (2,6,3) ) as the direction, we derived the parametric equations ( x = 1 + 2s ), ( y = 2 + 6s ), ( z = 3 + 3s ). These describe all points along the tangent line at ( t = 1 ).