Math Problem Statement

For the given position vectors r(t), compute the (tangent) velocity vector r'(t) for the given value of t: A) Let r(t) = ⟨cos(5t), sin(5t)⟩. Then r'(π/4) = ? B) Let r(t) = ⟨t^2, t^3⟩. Then r'(7) = ? C) Let r(t) = e^(5t)i + e^(-7t)j + tk. Then r'(-1) = ?

Solution

Let’s solve the problems step by step based on the given image.


Problem A:

Position vector:
r(t)=cos(5t),sin(5t)\mathbf{r}(t) = \langle \cos(5t), \sin(5t) \rangle
Find: r(π4)\mathbf{r}'\left(\frac{\pi}{4}\right).

  1. Compute r(t)\mathbf{r}'(t):
    Differentiate each component:
    r(t)=5sin(5t),5cos(5t)\mathbf{r}'(t) = \langle -5\sin(5t), 5\cos(5t) \rangle

  2. Substitute t=π4t = \frac{\pi}{4}:
    r(π4)=5sin(5π4),5cos(5π4)\mathbf{r}'\left(\frac{\pi}{4}\right) = \langle -5\sin\left(\frac{5\pi}{4}\right), 5\cos\left(\frac{5\pi}{4}\right) \rangle
    Since sin(5π4)=22\sin\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2} and cos(5π4)=22\cos\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2}, we get:
    r(π4)=5(22),5(22)=522,522\mathbf{r}'\left(\frac{\pi}{4}\right) = \langle -5\left(-\frac{\sqrt{2}}{2}\right), 5\left(-\frac{\sqrt{2}}{2}\right) \rangle = \langle \frac{5\sqrt{2}}{2}, -\frac{5\sqrt{2}}{2} \rangle


Problem B:

Position vector:
r(t)=t2,t3\mathbf{r}(t) = \langle t^2, t^3 \rangle
Find: r(7)\mathbf{r}'(7).

  1. Compute r(t)\mathbf{r}'(t):
    Differentiate each component:
    r(t)=2t,3t2\mathbf{r}'(t) = \langle 2t, 3t^2 \rangle

  2. Substitute t=7t = 7:
    r(7)=2(7),3(72)=14,147\mathbf{r}'(7) = \langle 2(7), 3(7^2) \rangle = \langle 14, 147 \rangle


Problem C:

Position vector:
r(t)=e5ti+e7tj+tk\mathbf{r}(t) = e^{5t}\mathbf{i} + e^{-7t}\mathbf{j} + t\mathbf{k}
Find: r(1)\mathbf{r}'(-1).

  1. Compute r(t)\mathbf{r}'(t):
    Differentiate each component:
    r(t)=5e5t,7e7t,1\mathbf{r}'(t) = \langle 5e^{5t}, -7e^{-7t}, 1 \rangle

  2. Substitute t=1t = -1:
    r(1)=5e5,7e7,1\mathbf{r}'(-1) = \langle 5e^{-5}, -7e^{7}, 1 \rangle


Final Answers:

A. r(π4)=522,522\mathbf{r}'\left(\frac{\pi}{4}\right) = \langle \frac{5\sqrt{2}}{2}, -\frac{5\sqrt{2}}{2} \rangle
B. r(7)=14,147\mathbf{r}'(7) = \langle 14, 147 \rangle
C. r(1)=5e5,7e7,1\mathbf{r}'(-1) = \langle 5e^{-5}, -7e^{7}, 1 \rangle

Would you like more details or explanations?
Here are five related questions to expand your understanding:

  1. How do you compute derivatives for vector-valued functions?
  2. Why are the tangent vectors useful in physics and geometry?
  3. What are the properties of sine and cosine at angles like π4\frac{\pi}{4}?
  4. How do exponential functions behave under differentiation?
  5. What is the significance of evaluating a derivative at a specific value?

Tip: When differentiating trigonometric or exponential functions, ensure you account for the chain rule!

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Vector calculus
Differentiation of vector-valued functions
Trigonometric functions
Exponential functions

Formulas

Derivative of cos(kx): -k*sin(kx)
Derivative of sin(kx): k*cos(kx)
Power rule: d/dx[t^n] = n*t^(n-1)
Exponential function derivative: d/dt[e^(kt)] = k*e^(kt)

Theorems

Chain Rule
Differentiation of vector functions

Suitable Grade Level

College-level Calculus