Gradient Applications: Chain Rule, Geometry & Physical Interpretation

1) For the function f(x,y) = e^(xy) + sin(x), what is the equation of the level curve that passes through the point (0,0)?

E^(xy) + sin(x) = 0

E^(xy) + sin(x) = 1

E^(xy) + sin(x) = 2

E^(xy) + sin(x) = 0.5

A level curve consists of all points (x,y) where f(x,y) = c for some constant c. To find the equation of the level curve that passes through (0,0), we first compute f(0,0) = e^(0·0) + sin(0) = e⁰+ 0 = 1 + 0 = 1. Therefore, the level curve passing through (0,0) is the set of points where f(x,y) = 1, which gives us e^(xy) + sin(x) = 1.

Explanation

A level curve consists of all points (x,y) where f(x,y) = c for some constant c. To find the equation of the level curve that passes through (0,0), we first compute f(0,0) = e^(0·0) + sin(0) = e⁰+ 0 = 1 + 0 = 1. Therefore, the level curve passing through (0,0) is the set of points where f(x,y) = 1, which gives us e^(xy) + sin(x) = 1.

2) The temperature function in a room is given by T(x,y) = 100 - 2x² - y². A person starts at (3,4) and wants to walk in the direction where the temperature increases most rapidly. What is the direction of steepest ascent?

⟨-6, -4⟩

⟨6, 4⟩

⟨-12, -8⟩

⟨12, 4⟩

The direction of steepest ascent is the direction of the gradient vector, ∇T. First, we compute the partial derivatives of the temperature function T(x,y) = 100 - 2x² - y². The partial derivative with respect to x is ∂T/∂x = -4x. The partial derivative with respect to y is ∂T/∂y = -2y. The gradient vector is ∇T = ⟨-4x, -2y⟩. We evaluate the gradient at the point (3,4): ∇T(3,4) = ⟨-4(3), -2(4)⟩ = ⟨-12, -8⟩. This vector points in the direction of the most rapid increase in temperature. Any vector that is a positive scalar multiple of ⟨-12, -8⟩ points in the same direction. Choice A, ⟨-6, -4⟩, is equal to (1/2) * ⟨-12, -8⟩ and is therefore a vector pointing in the correct direction. Choice B points in the direction of steepest descent.

Explanation

The direction of steepest ascent is the direction of the gradient vector, ∇T. First, we compute the partial derivatives of the temperature function T(x,y) = 100 - 2x² - y². The partial derivative with respect to x is ∂T/∂x = -4x. The partial derivative with respect to y is ∂T/∂y = -2y. The gradient vector is ∇T = ⟨-4x, -2y⟩. We evaluate the gradient at the point (3,4): ∇T(3,4) = ⟨-4(3), -2(4)⟩ = ⟨-12, -8⟩. This vector points in the direction of the most rapid increase in temperature. Any vector that is a positive scalar multiple of ⟨-12, -8⟩ points in the same direction. Choice A, ⟨-6, -4⟩, is equal to (1/2) * ⟨-12, -8⟩ and is therefore a vector pointing in the correct direction. Choice B points in the direction of steepest descent.

3) If ∇f = (2x, 2y) and ∇g = (y, x), what is the gradient of the sum function h(x,y) = f(x,y) + g(x,y)?

(2x + y, 2y + x)

(2xy,2xy)

(2x - y, 2y - x)

(2x,2y)

The gradient operator is linear. This means that the gradient of a sum is the sum of the gradients: ∇(f + g) = ∇f + ∇g. We simply add the corresponding components of the two vectors: (2x + y, 2y + x).

Explanation

The gradient operator is linear. This means that the gradient of a sum is the sum of the gradients: ∇(f + g) = ∇f + ∇g. We simply add the corresponding components of the two vectors: (2x + y, 2y + x).

4) For the function f(x,y) = x³ - 3xy², what is the angle between the gradient at (1,1) and the vector ⟨1,-1⟩?

0 degrees

45 degrees

90 degrees

135 degrees

The angle between two vectors can be found using the dot product formula. We first compute the gradient: ∂f/∂x = 3x² - 3y² and ∂f/∂y = -6xy. At (1,1), these are ∂f/∂x = 3(1)² - 3(1)² = 3 - 3 = 0 and ∂f/∂y = -6(1)(1) = -6. So ∇f = ⟨0, -6⟩. The dot product of ⟨0, -6⟩ and ⟨1, -1⟩ is (0)(1) + (-6)(-1) = 0 + 6 = 6. The vector ⟨1, -1⟩ has magnitude √2 and the magnitude of ∇f is √(0² + (-6)²) = 6. The angle θ satisfies cos(θ) = (∇f · u)/(|∇f||u|) = 6/(6√2) = 1/√2. Therefore, θ = 45 degrees.

Explanation

The angle between two vectors can be found using the dot product formula. We first compute the gradient: ∂f/∂x = 3x² - 3y² and ∂f/∂y = -6xy. At (1,1), these are ∂f/∂x = 3(1)² - 3(1)² = 3 - 3 = 0 and ∂f/∂y = -6(1)(1) = -6. So ∇f = ⟨0, -6⟩. The dot product of ⟨0, -6⟩ and ⟨1, -1⟩ is (0)(1) + (-6)(-1) = 0 + 6 = 6. The vector ⟨1, -1⟩ has magnitude √2 and the magnitude of ∇f is √(0² + (-6)²) = 6. The angle θ satisfies cos(θ) = (∇f · u)/(|∇f||u|) = 6/(6√2) = 1/√2. Therefore, θ = 45 degrees.

5) For the surface z = x² + y², what is the normal vector at the point (1,1,2)?

⟨1,1,-4⟩

⟨2,2,-1⟩

⟨1,1,-2⟩

⟨2,2,-2⟩

We can rewrite the surface equation as f(x,y,z) = x² + y² - z = 0. The gradient of f gives a normal vector to the surface. We compute ∇f = ⟨2x, 2y, -1⟩. At the point (1,1,2), this becomes ⟨2(1), 2(1), -1⟩ = ⟨2, 2, -1⟩. Therefore, the normal vector at (1,1,2) is ⟨2,2,-1⟩.

Explanation

We can rewrite the surface equation as f(x,y,z) = x² + y² - z = 0. The gradient of f gives a normal vector to the surface. We compute ∇f = ⟨2x, 2y, -1⟩. At the point (1,1,2), this becomes ⟨2(1), 2(1), -1⟩ = ⟨2, 2, -1⟩. Therefore, the normal vector at (1,1,2) is ⟨2,2,-1⟩.

6) A particle moves along the curve defined by f(x,y) = 5, where f(x,y) = x² + xy + y². At the point (1,2), what is the rate of change of f along the particle's path?

0

5

10

15

The particle moves along the level curve f(x,y) = 5. On a level curve, the function value is constant, so f(x,y) = 5 for all points on the curve. Therefore, the rate of change of f along the particle's path is zero. This is because the gradient is perpendicular to the level curve, so moving along the curve (which is tangent to the level curve) results in no change in the function value.

Explanation

The particle moves along the level curve f(x,y) = 5. On a level curve, the function value is constant, so f(x,y) = 5 for all points on the curve. Therefore, the rate of change of f along the particle's path is zero. This is because the gradient is perpendicular to the level curve, so moving along the curve (which is tangent to the level curve) results in no change in the function value.

7) What is the geometric interpretation of the gradient vector?

It points in direction of greatest decrease.

It is perpendicular to level curves and points toward increasing values.

It is tangent to level curves.

It has no geometric meaning.

The gradient vector has several important geometric interpretations. First, it is perpendicular (normal) to level curves (in 2D) or level surfaces (in 3D). Second, it points in the direction of greatest increase of the function. Third, its magnitude gives the rate of change in that direction. Therefore, the correct interpretation is that the gradient is perpendicular to level curves and points toward increasing function values.

Explanation

The gradient vector has several important geometric interpretations. First, it is perpendicular (normal) to level curves (in 2D) or level surfaces (in 3D). Second, it points in the direction of greatest increase of the function. Third, its magnitude gives the rate of change in that direction. Therefore, the correct interpretation is that the gradient is perpendicular to level curves and points toward increasing function values.

8) For the surface x² + y² + z² = 9, what is the equation of the tangent plane at the point (2,2,1)?

2x+2y+z=9

2x+2y+z=13

4x+4y+2z=18

X+y+z=5

We can write the surface as f(x,y,z) = x² + y² + z² - 9 = 0. The gradient gives a normal vector to the surface: ∇f = ⟨2x, 2y, 2z⟩. At the point (2,2,1), this becomes ⟨4, 4, 2⟩. The equation of the tangent plane at (x₀,y₀,z₀) with normal vector ⟨a,b,c⟩ is a(x - x₀) + b(y - y₀) + c(z - z₀) = 0. Substituting the values: 4(x - 2) + 4(y - 2) + 2(z - 1) = 0. Expanding: 4x - 8 + 4y - 8 + 2z - 2 = 0, which simplifies to 4x + 4y + 2z - 18 = 0, or 2x + 2y + z - 9 = 0. Therefore, the equation is 2x + 2y + z = 9.

Explanation

We can write the surface as f(x,y,z) = x² + y² + z² - 9 = 0. The gradient gives a normal vector to the surface: ∇f = ⟨2x, 2y, 2z⟩. At the point (2,2,1), this becomes ⟨4, 4, 2⟩. The equation of the tangent plane at (x₀,y₀,z₀) with normal vector ⟨a,b,c⟩ is a(x - x₀) + b(y - y₀) + c(z - z₀) = 0. Substituting the values: 4(x - 2) + 4(y - 2) + 2(z - 1) = 0. Expanding: 4x - 8 + 4y - 8 + 2z - 2 = 0, which simplifies to 4x + 4y + 2z - 18 = 0, or 2x + 2y + z - 9 = 0. Therefore, the equation is 2x + 2y + z = 9.

9) If f(x,y) = x²e^y, what is the magnitude of the gradient at the point (1,0)?

1

√2

2

√5

We compute the partial derivatives: ∂f/∂x = 2xe^y and ∂f/∂y = x²e^y. At the point (1,0), these become ∂f/∂x = 2(1)e⁰= 2(1)(1) = 2 and ∂f/∂y = (1)²e⁰= 1(1) = 1. So ∇f = ⟨2, 1⟩. The magnitude of the gradient is |∇f| = √(2² + 1²) = √(4 + 1) = √5.

Explanation

We compute the partial derivatives: ∂f/∂x = 2xe^y and ∂f/∂y = x²e^y. At the point (1,0), these become ∂f/∂x = 2(1)e⁰= 2(1)(1) = 2 and ∂f/∂y = (1)²e⁰= 1(1) = 1. So ∇f = ⟨2, 1⟩. The magnitude of the gradient is |∇f| = √(2² + 1²) = √(4 + 1) = √5.

10) Let L(x,y) be the linear approximation (tangent plane approximation) of the function f(x,y) at the point (a,b). If the gradient of f at (a,b) is the zero vector, what is the equation of the linear approximation?

L(x,y)=f(a,b)

L(x,y)=0

L(x,y)=x+y

L(x,y)=f(a,b)+x+y

The formula for linear approximation is L(x,y) = f(a,b) + fx(a,b)(x-a) + fy(a,b)(y-b). The terms fx(a,b) and fy(a,b) are the components of the gradient. If the gradient is the zero vector, then fx=0 and fy=0. The equation simplifies to L(x,y) = f(a,b) + 0 + 0, which means the tangent plane is horizontal and the height is constant at the value of the function at that point.

Explanation

The formula for linear approximation is L(x,y) = f(a,b) + fx(a,b)(x-a) + fy(a,b)(y-b). The terms fx(a,b) and fy(a,b) are the components of the gradient. If the gradient is the zero vector, then fx=0 and fy=0. The equation simplifies to L(x,y) = f(a,b) + 0 + 0, which means the tangent plane is horizontal and the height is constant at the value of the function at that point.

11) Two surfaces are defined by F(x,y,z) = 0 and G(x,y,z) = 0. If these two surfaces intersect orthogonally (at a 90-degree angle) at a point P, what must be true about their gradients at P?

∇F=∇G

∇F×∇G=0

∇F·∇G=0

∇F=−∇G

The angle between two surfaces at an intersection point is defined as the angle between their normal vectors at that point. The gradient vector ∇F is the normal vector to the surface F=0, and ∇G is the normal vector to G=0. For the surfaces to be orthogonal, their normal vectors must be perpendicular. Two vectors are perpendicular if and only if their dot product is zero.

Explanation

The angle between two surfaces at an intersection point is defined as the angle between their normal vectors at that point. The gradient vector ∇F is the normal vector to the surface F=0, and ∇G is the normal vector to G=0. For the surfaces to be orthogonal, their normal vectors must be perpendicular. Two vectors are perpendicular if and only if their dot product is zero.

12) Consider the function f(x,y) = ax² + by². If the gradient of f at the point (1, 1) is the vector <2, -4>, what are the values of the constants a and b?

A=1, b=−2

A=2, b=−4

A=1, b=−4

A=2, b=−2

We compute the gradient of f(x,y) = ax² + by², so ∂f/∂x = 2ax and ∂f/∂y = 2by. Evaluating the gradient at (1,1) gives us ∇f(1,1) = <2a(1), 2b(1)> = <2a, 2b>. We are given that ∇f(1,1) = <2, -4>. Equating components, we have 2a = 2 implying a = 1. Also, 2b = -4 implies b = -2.

Explanation

We compute the gradient of f(x,y) = ax² + by², so ∂f/∂x = 2ax and ∂f/∂y = 2by. Evaluating the gradient at (1,1) gives us ∇f(1,1) = <2a(1), 2b(1)> = <2a, 2b>. We are given that ∇f(1,1) = <2, -4>. Equating components, we have 2a = 2 implying a = 1. Also, 2b = -4 implies b = -2.

13) For a differentiable function f(x,y), if the gradient at a point P is <3, 4>, which of the following unit vectors represents a direction in which the function value does not change?

<3/5,4/5>

<4/5,-3/5>

<4/5,3/5>

<-3/5,-4/5>

The function value does not change in directions tangent to the level curve. These directions are perpendicular to the gradient. We need a vector u such that ∇f · u = 0.

Test the dot product with the gradient <3, 4>:

Option A: 3(3/5) + 4(4/5) = 25/5 ≠ 0.

Option B: 3(4/5) + 4(-3/5) = 12/5 - 12/5 = 0.

Option C: 3(4/5) + 4(3/5) = 24/5 ≠ 0.

Option D is opposite to the gradient (steepest descent).

Therefore, B is the correct direction.

Explanation

The function value does not change in directions tangent to the level curve. These directions are perpendicular to the gradient. We need a vector u such that ∇f · u = 0.

Test the dot product with the gradient <3, 4>:

Option A: 3(3/5) + 4(4/5) = 25/5 ≠ 0.

Option B: 3(4/5) + 4(-3/5) = 12/5 - 12/5 = 0.

Option C: 3(4/5) + 4(3/5) = 24/5 ≠ 0.

Option D is opposite to the gradient (steepest descent).

Therefore, B is the correct direction.

14) Find the cosine of the angle between the cylinder x² + y² = 2 and the plane x + z = 4 at the point (1, 1, 3).

1/2

1/√3

1/√6

0

First, find the gradient of the cylinder surface f(x,y,z) = x² + y². ∇f = <2x, 2y, 0>. At (1,1,3), ∇f = <2, 2, 0>.

Next, find the gradient of the plane g(x,y,z) = x + z. ∇g = <1, 0, 1>.

The angle θ is the angle between these normal vectors.

cos(θ) = |∇f · ∇g| / (|∇f| |∇g|).

Dot product: <2, 2, 0> · <1, 0, 1> = 2(1) + 2(0) + 0(1) = 2.

Magnitude of ∇f: √(2² + 2² + 0²) = √8 = 2√2.

Magnitude of ∇g: √(1² + 0² + 1²) = √2.

cos(θ) = 2 / (2√2 * √2) = 2 / (2 * 2) = 2/4 = 1/2.

Explanation

First, find the gradient of the cylinder surface f(x,y,z) = x² + y². ∇f = <2x, 2y, 0>. At (1,1,3), ∇f = <2, 2, 0>.

Next, find the gradient of the plane g(x,y,z) = x + z. ∇g = <1, 0, 1>.

The angle θ is the angle between these normal vectors.

cos(θ) = |∇f · ∇g| / (|∇f| |∇g|).

Dot product: <2, 2, 0> · <1, 0, 1> = 2(1) + 2(0) + 0(1) = 2.

Magnitude of ∇f: √(2² + 2² + 0²) = √8 = 2√2.

Magnitude of ∇g: √(1² + 0² + 1²) = √2.

cos(θ) = 2 / (2√2 * √2) = 2 / (2 * 2) = 2/4 = 1/2.

15) A bug is crawling on the surface z = x² - y². The bug is currently at the point (2, 1) in the xy-plane. In which direction should the bug move to maintain its current height (z-value) while increasing its x-coordinate?

<1,2>

<1,-2>

<2,1>

<1,1>

To maintain height, the bug must move tangent to the level curve. The level curve at (2,1) is x² - y² = 2² - 1² = 3.

The gradient is ∇f = <2x, -2y>. At (2,1), ∇f = <4, -2>.

The direction of motion v = <vx, vy> must be perpendicular to <4, -2>, so their dot product is zero: 4vx - 2vy = 0, which implies 2vx = vy.

The vector <1, 2> satisfies this (2(1) = 2).

We also need to check if the x-coordinate increases. In the vector <1, 2>, the x-component is positive (1), so x increases.

(Note: The vector <-1, -2> also maintains height but decreases x). Thus, <1, 2> is the correct direction.

Explanation

To maintain height, the bug must move tangent to the level curve. The level curve at (2,1) is x² - y² = 2² - 1² = 3.

The gradient is ∇f = <2x, -2y>. At (2,1), ∇f = <4, -2>.

The direction of motion v = <vx, vy> must be perpendicular to <4, -2>, so their dot product is zero: 4vx - 2vy = 0, which implies 2vx = vy.

The vector <1, 2> satisfies this (2(1) = 2).

We also need to check if the x-coordinate increases. In the vector <1, 2>, the x-component is positive (1), so x increases.

(Note: The vector <-1, -2> also maintains height but decreases x). Thus, <1, 2> is the correct direction.