Please provide the answer in a neat format.2. Suppose that Vf(x, y) is always parallel to the position vector (x,y). Convince
me that f(3,0) = f(-3,0). You will need to first compute the normal vecto

Question

Please provide the answer in a neat format.2. Suppose that Vf(x, y) is always parallel to the position vector (x,y). Convince
me that f(3,0) = f(-3,0). You will need to first compute the normal vector to the
circle that is centered at the origin and has radius 3.

4. Consider the following information: for the function f(x,y) = x2 + y "2, we
have that f(2,0) = 4 and the directional derivative of f at (2,0) along the vector
(3,2) is equal to 12. Explain the geometric significance of this value of the
directional derivative, as follows:
Parametrize a linear curve C1 in 2D that goes through (2,0) and has velocity
(3,2).
Lift this curve to a new curve C2 along the graph of f(x,y), so that the (x,y)
components of C2 match up with C1. What is the parametrization of C2?
Without computing anything, provide a good estimate of f(2.03, 0.02). What does
it tell us about C2?

2] Parametrize a linear curve C3 that is tangent to C2 at (2,0,4). [

VIPSolutions · Accepted Answer

# Solutions

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## Problem 2

**Given:**  
- ∇f(x, y) is always parallel to the vector (x, y).
- Show that f(3,) = f(-3,).
- Use the normal vector to the circle centered at the origin with radius 3.

### Step 1: Understanding the Gradient Condition

- If ∇f(x, y) is parallel to (x, y), then ∇f(x, y) = λ(x, y) for some scalar λ(x, y).

### Step 2: Circle and Normal Vector

- The equation for a circle centered at the origin with radius 3 is:  
  $ x^2 + y^2 = 9 $
- The normal vector to the circle at (x, y) is (x, y), which matches the position vector.

### Step 3: Implication for f

- Since the gradient is normal to level curves, the level curves of f must be the circles centered at the origin (because the gradient points radially outwards).
- Therefore, f(x, y) is constant on each circle:  
  $ f(x, y) = g(x^2 + y^2) $, for some function g.

### Step 4: Evaluate f at (3,) and (-3,)

- $ f(3,) = g(3^2 + ^2) = g(9) $
- $ f(-3,) = g((-3)^2 + ^2) = g(9) $
- **Thus, $ f(3,) = f(-3,) $.**

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## Problem 4

**Given:**  
- $ f(x, y) = x^2 + y^2 $
- $ f(2, ) = 4 $
- Directional derivative at (2, ) along (3, 2) is 12.

### Step 1: Geometric Significance of the Directional Derivative

- The directional derivative gives the rate of change of f in the direction (3, 2) at (2, ).
- A value of 12 means f is increasing most rapidly in that direction with rate 12.

### Step 2: Parametrize Curve $ C_1 $

- Curve $ C_1 $ through (2, ) with velocity (3, 2):  
  $ C_1(t) = (2 + 3t,  + 2t) = (2 + 3t, 2t) $

### Step 3: Lift Curve to $ C_2 $ on the Graph of f

- $ C_2(t) = (2 + 3t, 2t, f(2 + 3t, 2t)) $
- $ f(2 + 3t, 2t) = (2 + 3t)^2 + (2t)^2 = 4 + 12t + 9t^2 + 4t^2 = 4 + 12t + 13t^2 $
- So,  
  $ C_2(t) = (2 + 3t, 2t, 4 + 12t + 13t^2) $

### Step 4: Estimate $ f(2.03, .02) $

- $ (2.03, .02) $ is close to $ (2, ) $.
- Use linear approximation:  
  $ f(2.03, .02) \approx f(2, ) + \nabla f(2, ) \cdot (.03, .02) $
- $ \nabla f(x, y) = (2x, 2y) \Rightarrow \nabla f(2, ) = (4, ) $
- $ \nabla f(2, ) \cdot (.03, .02) = 4 \times .03 = .12 $
- Estimate: $ 4 + .12 = 4.12 $
- **Conclusion:** $ f $ increases rapidly in the direction of (3, 2), matching the large directional derivative.

### Step 5: Parametrize Tangent Curve $ C_3 $ at $ (2, , 4) $

- Tangent vector to $ C_2 $ at $ t= $ is (3, 2, 12).
- $ C_3(s) = (2 + 3s,  + 2s, 4 + 12s) $

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## **Summary Table**

| Part | Key Result |
|------|------------|
| 2    | $ f(3,) = f(-3,) $ because f is constant on circles centered at the origin. |
| 4    | - $ C_1(t) = (2 + 3t, 2t) $  - $ C_2(t) = (2 + 3t, 2t, 4 + 12t + 13t^2) $  - $ f(2.03, .02) \approx 4.12 $  - $ C_3(s) = (2 + 3s, 2s, 4 + 12s) $ |

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