I submitted a picture of the question **Question:**Find a parametric equation for the tangent line to the curve defined by $\mathbf{r}(t) = \langle t^2, 4\sqrt{t}, e^{2t} \rangle$ when $t = 1$.**Solution:**To find the parametric equation for the tangent line, follow these steps:1. **Find $\mathbf{r}'(t)$, the derivative of the vector function:** $\mathbf{r}(t) = \langle t^2, 4\sqrt{t}, e^{2t} \rangle$. - Differentiate each component with respect to $t$: - $\frac{d}{dt}(t^2) = 2t$ - $\frac{d}{dt}(4\sqrt{t}) = \frac{d}{dt}(4t^{1/2}) = 2t^{-1/2}$ - $\frac{d}{dt}(e^{2t}) = 2e^{2t}$ Thus, $\mathbf{r}'(t) = \langle 2t, 2t^{-1/2}, 2e^{2t} \rangle$.2. **Evaluate $\mathbf{r}(t)$ and $\mathbf{r}'(t)$ at $t = 1$:** $\mathbf{r}(1) = \langle 1^2, 4\sqrt{1}, e^{2\cdot1} \rangle = \langle 1, 4, e^2 \rangle$. $\mathbf{r}'(1) = \langle 2\cdot1, 2\cdot1^{-1/2}, 2e^{2\cdot1} \rangle = \langle 2, 2, 2e^2 \rangle$.3. **Write the equation of the tangent line:** The parametric equation of the tangent line at $t = 1$ is given by: $\mathbf{T}(t) = \mathbf{r}(1) + \mathbf{r}'(1) \cdot (t - 1)$. Plug in the values:

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Description: I submitted a picture of the question **Question:**Find a parametric equation for the tangent line to the curve defined by $\mathbf{r}(t) = \langle t^2, 4\sqrt{t}, e^{2t} \rangle$ when $t = 1$.**Solution:**To find the parametric equation for the

Given is the equation of a 3-dimensional parametric equation, to find the equation of the tangent we…

Answered: Find a parametric equation for the…

Math

Calculus

Find a parametric equation for the tangent line to the curve defined by r(t) =< t?, 4\ſt, e2t> when t = 1.

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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I submitted a picture of the question

**Question:**

Find a parametric equation for the tangent line to the curve defined by $\mathbf{r}(t) = \langle t^2, 4\sqrt{t}, e^{2t} \rangle$ when $t = 1$.

**Solution:**

To find the parametric equation for the tangent line, follow these steps:

1. **Find $\mathbf{r}'(t)$, the derivative of the vector function:**

$\mathbf{r}(t) = \langle t^2, 4\sqrt{t}, e^{2t} \rangle$.

- Differentiate each component with respect to $t$:
- $\frac{d}{dt}(t^2) = 2t$
- $\frac{d}{dt}(4\sqrt{t}) = \frac{d}{dt}(4t^{1/2}) = 2t^{-1/2}$
- $\frac{d}{dt}(e^{2t}) = 2e^{2t}$

Thus, $\mathbf{r}'(t) = \langle 2t, 2t^{-1/2}, 2e^{2t} \rangle$.

2. **Evaluate $\mathbf{r}(t)$ and $\mathbf{r}'(t)$ at $t = 1$:**

$\mathbf{r}(1) = \langle 1^2, 4\sqrt{1}, e^{2\cdot1} \rangle = \langle 1, 4, e^2 \rangle$.

$\mathbf{r}'(1) = \langle 2\cdot1, 2\cdot1^{-1/2}, 2e^{2\cdot1} \rangle = \langle 2, 2, 2e^2 \rangle$.

3. **Write the equation of the tangent line:**

The parametric equation of the tangent line at $t = 1$ is given by:

$\mathbf{T}(t) = \mathbf{r}(1) + \mathbf{r}'(1) \cdot (t - 1)$.

Plug in the values:

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