Table of contents

0. Functions7h 52m
1. Limits and Continuity2h 2m
2. Intro to Derivatives1h 33m
3. Techniques of Differentiation3h 18m
4. Applications of Derivatives2h 38m
5. Graphical Applications of Derivatives6h 2m
6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
7. Antiderivatives & Indefinite Integrals1h 26m
8. Definite Integrals4h 44m
9. Graphical Applications of Integrals2h 27m
- Area Between Curves
  1h 18m
- Introduction to Volume & Disk Method
  1h 8m
10. Physics Applications of Integrals 3h 16m
- Kinematics
  1h 13m
- Work
  2h 3m
11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 34m
12. Techniques of Integration7h 39m
13. Intro to Differential Equations2h 55m
14. Sequences & Series5h 36m
15. Power Series2h 19m
- Introduction to Power Series
  1h 9m
- Taylor Series & Taylor Polynomials
  1h 10m
16. Parametric Equations & Polar Coordinates7h 58m

5. Graphical Applications of Derivatives

Finding Global Extrema

Calculus

5. Graphical Applications of Derivatives

Finding Global Extrema

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Multiple Choice

Let $f (x, y) = x^{2} + 3 y^{2}$ . Find the maximum rate of change of $f$ at the point $(1, 2)$ and the direction in which it occurs.

The maximum rate of change is

8

, and it occurs in the direction of the vector

(1, 3)

The maximum rate of change is

5

, and it occurs in the direction of the vector

(1, 2)

The maximum rate of change is

10

, and it occurs in the direction of the vector

(2, 6)

The maximum rate of change is

13

, and it occurs in the direction of the vector

(2, 12)

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Verified step by step guidance

Step 1: Recall that the maximum rate of change of a function f(x, y) at a given point occurs in the direction of the gradient vector ∇f(x, y). The magnitude of the gradient vector gives the maximum rate of change.

Step 2: Compute the gradient of f(x, y). The gradient is a vector of partial derivatives: ∇f(x, y) = (∂f/∂x, ∂f/∂y). For f(x, y) = x^2 + 3y^2, calculate ∂f/∂x = 2x and ∂f/∂y = 6y.

Step 3: Evaluate the gradient at the given point (1, 2). Substitute x = 1 and y = 2 into the gradient components: ∇f(1, 2) = (2(1), 6(2)) = (2, 12).

Step 4: The magnitude of the gradient vector ∇f(1, 2) gives the maximum rate of change. Compute the magnitude using the formula ||∇f|| = √((∂f/∂x)^2 + (∂f/∂y)^2). Substitute the components of the gradient vector: ||∇f(1, 2)|| = √(2^2 + 12^2).

Step 5: The direction in which the maximum rate of change occurs is given by the unit vector in the direction of the gradient. Normalize the gradient vector ∇f(1, 2) = (2, 12) by dividing each component by the magnitude ||∇f(1, 2)||.