Q2(999 points)

Let u(t)=(cost,sint,t)u(t)=(cos⁡t,sin⁡t,t), and suppose v(t)v(t) is a vector-valued function such that

v(π/3)=(1,−2,2),v′(π/3)=(1.1,0.6,0.2).v(π/3)=(1,−2,2),v′(π/3)=(1.1,0.6,0.2).

Using the identities for differentiating vector-valued functions, calculate to four significant figures the following quantities, at t=π/3t=π/3:

(**A**) (2u+t2v)′,(2u+t2v)′, (**B**)v⋅(u×v)′,v⋅(u×v)′, (**C**)(|v||u|2)′(|v||u|2)′.. i only need the answer for part c

To compute part **(C)**, we need to differentiate the expression \((|v| |u|^2)'\) at \(t = \frac{\pi}{3}\), where \(v(t)\) and \(u(t)\) are vector-valued functions.

### Step 1: Break down the expression
We want to differentiate:
\[
(|v| |u|^2)' = \frac{d}{dt} \left( |v| \cdot |u|^2 \right)
\]
We can apply the product rule of differentiation here:
\[
(|v| |u|^2)' = \frac{d}{dt} |v| \cdot |u|^2 + |v| \cdot \frac{d}{dt} |u|^2
\]

### Step 2: Compute \(|v|\)
The magnitude of \(v(t)\) is given by:
\[
|v| = \sqrt{v_1^2 + v_2^2 + v_3^2}
\]
At \(t = \frac{\pi}{3}\), we know \(v\left(\frac{\pi}{3}\right) = (1, -2, 2)\). Therefore,
\[
|v| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3
\]

### Step 3: Compute \(\frac{d}{dt} |v|\)
We can use the formula for differentiating the magnitude of a vector:
\[
\frac{d}{dt} |v| = \frac{v \cdot v'}{|v|}
\]
At \(t = \frac{\pi}{3}\), we know \(v\left(\frac{\pi}{3}\right) = (1, -2, 2)\) and \(v'\left(\frac{\pi}{3}\right) = (1.1, 0.6, 0.2)\). Therefore,
\[
v \cdot v' = 1(1.1) + (-2)(0.6) + 2(0.2) = 1.1 - 1.2 + 0.4 = 0.3
\]
Thus,
\[
\frac{d}{dt} |v| = \frac{0.3}{3} = 0.1
\]

### Step 4: Compute \(|u|\) and \(|u|^2\)
The magnitude of \(u(t) = (\cos t, \sin t, t)\) is:
\[
|u| = \sqrt{\cos^2 t + \sin^2 t + t^2}
\]
Since \(\cos^2 t + \sin^2 t = 1\), this simplifies to:
\[
|u| = \sqrt{1 + t^2}
\]
At \(t = \frac{\pi}{3}\),
\[
|u| = \sqrt{1 + \left(\frac{\pi}{3}\right)^2} = \sqrt{1 + \frac{\pi^2}{9}} = \sqrt{\frac{9 + \pi^2}{9}} = \frac{\sqrt{9 + \pi^2}}{3}
\]
Therefore,
\[
|u|^2 = 1 + \frac{\pi^2}{9}
\]

### Step 5: Compute \(\frac{d}{dt} |u|^2\)
Differentiating \(|u|^2 = 1 + t^2\), we get:
\[
\frac{d}{dt} |u|^2 = 2t
\]
At \(t = \frac{\pi}{3}\),
\[
\frac{d}{dt} |u|^2 = 2 \times \frac{\pi}{3} = \frac{2\pi}{3}
\]

### Step 6: Combine everything
We now combine all the components:
\[
(|v| |u|^2)' = \frac{d}{dt} |v| \cdot |u|^2 + |v| \cdot \frac{d}{dt} |u|^2
\]
At \(t = \frac{\pi}{3}\), this becomes:
\[
(|v| |u|^2)' = (0.1)(1 + \frac{\pi^2}{9}) + 3 \cdot \frac{2\pi}{3}
\]
Simplifying:
\[
(|v| |u|^2)' = 0.1 \left(1 + \frac{\pi^2}{9}\right) + 2\pi
\]
Substitute \(\pi \approx 3.1416\):
\[
(|v| |u|^2)' = 0.1 \left(1 + \frac{(3.1416)^2}{9}\right) + 2 \times 3.1416
\]
\[
(|v| |u|^2)' = 0.1 \left(1 + \frac{9.8696}{9}\right) + 6.2832
\]
\[
(|v| |u|^2)' = 0.1 \times 2.0966 + 6.2832 = 0.2097 + 6.2832 = 6.4929
\]

Thus, the value of \((|v| |u|^2)'\) at \(t = \frac{\pi}{3}\) is approximately **6.493** (to four significant figures).

Would you like further details or have any questions?

---

**Related Questions:**
1. How do you compute the derivative of vector-valued functions?
2. What are the steps to find the magnitude of a vector?
3. How does the product rule apply to vector magnitudes?
4. Can you explain the formula for differentiating the magnitude of a vector?
5. How does the chain rule apply to functions involving vectors?

**Tip:** Always verify that the dot products and magnitudes are computed accurately, as small errors can lead to significant deviations in final answers.

Let u(t) = (cos t, sin t, t), and suppose v(t) is a vector-valued function such that v(π/3) = (1, −2, 2), v′(π/3) = (1.1, 0.6, 0.2). Using the identities for differentiating vector-valued functions, calculate to four significant figures the following quantities, at t = π/3: (C) (|v||u|^2)' .

Explore the differentiation of vector-valued functions with an example problem calculating (|v| |u|^2)' at t = π/3. This problem covers key concepts in vector calculus, including the product rule and vector magnitude, providing a step-by-step solution. Suitable for university-level students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation