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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14.20 Please help me answer this math problem.
![To find the function \( \mathbf{r}(t) \) that satisfies the following conditions:
\[
\mathbf{r}'(t) = \langle 1, 3t^2, 2t \rangle ; \, \mathbf{r}(1) = \langle 6, -2, 5 \rangle
\]
We need to integrate the components of the derivative vector \( \mathbf{r}'(t) \) to find \( \mathbf{r}(t) \).
1. Integrate the first component:
\[
\int 1 \, dt = t + C_1
\]
2. Integrate the second component:
\[
\int 3t^2 \, dt = t^3 + C_2
\]
3. Integrate the third component:
\[
\int 2t \, dt = t^2 + C_3
\]
Next, apply the initial condition \( \mathbf{r}(1) = \langle 6, -2, 5 \rangle \) to solve for the constants \( C_1, C_2, C_3 \).
Thus, \( \mathbf{r}(t) = \langle t + C_1, t^3 + C_2, t^2 + C_3 \rangle \).
By substituting \( t = 1 \):
\[
\langle 1 + C_1, 1^3 + C_2, 1^2 + C_3 \rangle = \langle 6, -2, 5 \rangle
\]
This yields the equations:
1. \( 1 + C_1 = 6 \) ⟹ \( C_1 = 5 \)
2. \( 1 + C_2 = -2 \) ⟹ \( C_2 = -3 \)
3. \( 1 + C_3 = 5 \) ⟹ \( C_3 = 4 \)
Finally, the function is:
\[
\mathbf{r}(t) = \langle t + 5, t^3 - 3, t^2 + 4 \rangle
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb3bd88a3-9d65-41d6-8e96-bcf2f13eb11b%2Facd2ff00-4c16-4b79-89fa-8912a94c0f06%2Febzn77w_processed.png&w=3840&q=75)
Transcribed Image Text:To find the function \( \mathbf{r}(t) \) that satisfies the following conditions:
\[
\mathbf{r}'(t) = \langle 1, 3t^2, 2t \rangle ; \, \mathbf{r}(1) = \langle 6, -2, 5 \rangle
\]
We need to integrate the components of the derivative vector \( \mathbf{r}'(t) \) to find \( \mathbf{r}(t) \).
1. Integrate the first component:
\[
\int 1 \, dt = t + C_1
\]
2. Integrate the second component:
\[
\int 3t^2 \, dt = t^3 + C_2
\]
3. Integrate the third component:
\[
\int 2t \, dt = t^2 + C_3
\]
Next, apply the initial condition \( \mathbf{r}(1) = \langle 6, -2, 5 \rangle \) to solve for the constants \( C_1, C_2, C_3 \).
Thus, \( \mathbf{r}(t) = \langle t + C_1, t^3 + C_2, t^2 + C_3 \rangle \).
By substituting \( t = 1 \):
\[
\langle 1 + C_1, 1^3 + C_2, 1^2 + C_3 \rangle = \langle 6, -2, 5 \rangle
\]
This yields the equations:
1. \( 1 + C_1 = 6 \) ⟹ \( C_1 = 5 \)
2. \( 1 + C_2 = -2 \) ⟹ \( C_2 = -3 \)
3. \( 1 + C_3 = 5 \) ⟹ \( C_3 = 4 \)
Finally, the function is:
\[
\mathbf{r}(t) = \langle t + 5, t^3 - 3, t^2 + 4 \rangle
\]
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