Tangent Planes & Normal Vectors Quiz

1) Given f(x,y) = x² + 2xy - 3y², what is the gradient vector ∇f at the point (1,2)?

⟨2x + 2y, 2x - 6y⟩

⟨6, -10⟩

⟨2, -6⟩

⟨4, -10⟩

To find the gradient vector, we compute the partial derivatives of f with respect to x and y. The partial derivative with respect to x is ∂f/∂x = 2x + 2y, and the partial derivative with respect to y is ∂f/∂y = 2x - 6y. Therefore, the gradient is ∇f = ⟨2x + 2y, 2x - 6y⟩. At the point (1,2), we substitute x = 1 and y = 2 into both components. For the first component: 2(1) + 2(2) = 2 + 4 = 6. For the second component: 2(1) - 6(2) = 2 - 12 = -10. Therefore, the gradient at (1,2) is ⟨6, -10⟩.

Explanation

To find the gradient vector, we compute the partial derivatives of f with respect to x and y. The partial derivative with respect to x is ∂f/∂x = 2x + 2y, and the partial derivative with respect to y is ∂f/∂y = 2x - 6y. Therefore, the gradient is ∇f = ⟨2x + 2y, 2x - 6y⟩. At the point (1,2), we substitute x = 1 and y = 2 into both components. For the first component: 2(1) + 2(2) = 2 + 4 = 6. For the second component: 2(1) - 6(2) = 2 - 12 = -10. Therefore, the gradient at (1,2) is ⟨6, -10⟩.

2) For the function f(x,y,z) = xyz + x² + y² - z², what is the directional derivative in the direction of v = ⟨1,1,1⟩ at (1,1,1)?

√3

5√3

5/√3

3√3

The directional derivative in the direction of a unit vector u is given by D_u f = ∇f · u. First, we find the gradient of f by computing the partial derivatives: ∂f/∂x = yz + 2x, ∂f/∂y = xz + 2y, ∂f/∂z = xy - 2z. At the point (1,1,1), these become ∂f/∂x = (1)(1) + 2(1) = 3, ∂f/∂y = (1)(1) + 2(1) = 3, ∂f/∂z = (1)(1) - 2(1) = -1. So ∇f = ⟨3,3,-1⟩. The direction vector v = ⟨1,1,1⟩ has magnitude √3, so the unit vector is u = ⟨1/√3, 1/√3, 1/√3⟩. The directional derivative is ∇f · u = ⟨3,3,-1⟩ · ⟨1/√3, 1/√3, 1/√3⟩ = (3/√3) + (3/√3) + (-1/√3) = (5/√3).

Explanation

The directional derivative in the direction of a unit vector u is given by D_u f = ∇f · u. First, we find the gradient of f by computing the partial derivatives: ∂f/∂x = yz + 2x, ∂f/∂y = xz + 2y, ∂f/∂z = xy - 2z. At the point (1,1,1), these become ∂f/∂x = (1)(1) + 2(1) = 3, ∂f/∂y = (1)(1) + 2(1) = 3, ∂f/∂z = (1)(1) - 2(1) = -1. So ∇f = ⟨3,3,-1⟩. The direction vector v = ⟨1,1,1⟩ has magnitude √3, so the unit vector is u = ⟨1/√3, 1/√3, 1/√3⟩. The directional derivative is ∇f · u = ⟨3,3,-1⟩ · ⟨1/√3, 1/√3, 1/√3⟩ = (3/√3) + (3/√3) + (-1/√3) = (5/√3).

3) Which of the following vector identities correctly describes the gradient of the product of two scalar functions f and g?

∇(fg) = ∇f · ∇g

∇(fg) = f∇g + g∇f

∇(fg) = ∇f + ∇g

∇(fg) = f∇f + g∇g

The gradient operator follows a product rule similar to the derivative product rule in single-variable calculus. The gradient of a product ∇(fg) is the first function times the gradient of the second plus the second function times the gradient of the first: f∇g + g∇f.

Explanation

The gradient operator follows a product rule similar to the derivative product rule in single-variable calculus. The gradient of a product ∇(fg) is the first function times the gradient of the second plus the second function times the gradient of the first: f∇g + g∇f.

4) Find the gradient of the function f(x,y) = x / y.

(1/y, -x/y²)

(1/y, 1/x)

(y, -x)

(1, -1)

We compute the partial derivatives. For ∂f/∂x, we treat y as a constant coefficient (1/y). The derivative of x(1/y) with respect to x is 1/y. For ∂f/∂y, we treat x as a constant. We are differentiating x(y⁻¹). Using the power rule, the derivative is x(-1)y⁻² = -x/y². Therefore, the gradient is (1/y, -x/y²).

Explanation

We compute the partial derivatives. For ∂f/∂x, we treat y as a constant coefficient (1/y). The derivative of x(1/y) with respect to x is 1/y. For ∂f/∂y, we treat x as a constant. We are differentiating x(y⁻¹). Using the power rule, the derivative is x(-1)y⁻² = -x/y². Therefore, the gradient is (1/y, -x/y²).

5) If ∇f(1,2) = (4, -2) and u = (3/5, 4/5) is a unit vector, what is the directional derivative in the direction of u?

2

4

4/5

20

The directional derivative is the dot product ∇f · u. Compute (4, -2) · (3/5, 4/5) = 4·(3/5) + (-2)·(4/5) = 12/5 - 8/5 = 4/5.

Explanation

The directional derivative is the dot product ∇f · u. Compute (4, -2) · (3/5, 4/5) = 4·(3/5) + (-2)·(4/5) = 12/5 - 8/5 = 4/5.

6) At which point is the gradient ∇f equal to the zero vector for f(x,y) = x² + y²?

(1,1)

(0,0)

(-1,2)

(2,0)

Compute partial derivatives: ∂f/∂x = 2x, ∂f/∂y = 2y. Set both to zero: 2x = 0 and 2y = 0, so x = 0, y = 0. Thus ∇f = (0,0) only at the origin (0,0).

Explanation

Compute partial derivatives: ∂f/∂x = 2x, ∂f/∂y = 2y. Set both to zero: 2x = 0 and 2y = 0, so x = 0, y = 0. Thus ∇f = (0,0) only at the origin (0,0).

7) Find the equation of the tangent plane to the sphere x² + y² + z² = 3 at the point (1, 1, 1).

X+y+z=3

X+y+z=1

2x+2y+2z=3

X+y+z=0

The sphere is a level surface of F(x,y,z) = x² + y² + z². The normal vector to the tangent plane is the gradient ∇F = (2x, 2y, 2z). At (1,1,1), the normal vector is (2, 2, 2). The equation of the plane is 2(x-1) + 2(y-1) + 2(z-1) = 0. Dividing by 2, we get (x-1) + (y-1) + (z-1) = 0, which simplifies to x + y + z - 3 = 0, or x + y + z = 3.

Explanation

The sphere is a level surface of F(x,y,z) = x² + y² + z². The normal vector to the tangent plane is the gradient ∇F = (2x, 2y, 2z). At (1,1,1), the normal vector is (2, 2, 2). The equation of the plane is 2(x-1) + 2(y-1) + 2(z-1) = 0. Dividing by 2, we get (x-1) + (y-1) + (z-1) = 0, which simplifies to x + y + z - 3 = 0, or x + y + z = 3.

8) Let r(t) be a path in the xy-plane and f(x,y) be a differentiable function. According to the multivariable Chain Rule, the rate of change of f with respect to t along the path, d/dt[f(r(t))], is equal to:

∇f(r(t)) · r'(t)

∇f(r(t)) × r'(t)

|∇f(r(t))| * |r'(t)|

∇f(r(t)) + r'(t)

The Chain Rule for a function of several variables states that if z = f(x,y) and x and y are functions of t, then dz/dt = (∂f/∂x)(dx/dt) + (∂f/∂y)(dy/dt). In vector notation, (∂f/∂x, ∂f/∂y) is the gradient ∇f, and (dx/dt, dy/dt) is the velocity vector r'(t). Therefore, the rate of change is the dot product ∇f · r'(t).

Explanation

The Chain Rule for a function of several variables states that if z = f(x,y) and x and y are functions of t, then dz/dt = (∂f/∂x)(dx/dt) + (∂f/∂y)(dy/dt). In vector notation, (∂f/∂x, ∂f/∂y) is the gradient ∇f, and (dx/dt, dy/dt) is the velocity vector r'(t). Therefore, the rate of change is the dot product ∇f · r'(t).

9) For f(x,y) = x² + y² at the point (1,1), what is the directional derivative in the direction of the vector <1,1>?

2

2√2

4

√2

∇f = (2x,2y) = (2,2) at (1,1). The vector <1,1> has magnitude √2, so the unit vector is <1/√2, 1/√2>. Directional derivative = (2)(1/√2) + (2)(1/√2) = 2/√2 + 2/√2 = 4/√2 = (4√2)/2 = 2√2.

Explanation

∇f = (2x,2y) = (2,2) at (1,1). The vector <1,1> has magnitude √2, so the unit vector is <1/√2, 1/√2>. Directional derivative = (2)(1/√2) + (2)(1/√2) = 2/√2 + 2/√2 = 4/√2 = (4√2)/2 = 2√2.

10) If ∇f(P) = 0 at a point P, then

Point is maximum

Directional derivative is zero in all directions

Gradient is perpendicular to level curve

F has vertical tangent plane

When the gradient is the zero vector, the dot product ∇f · u = 0 for any unit vector u. This means the directional derivative in every possible direction is zero, so the function is flat (stationary) at that point in all directions. This occurs at critical points (possible max, min, or saddle).

Explanation

When the gradient is the zero vector, the dot product ∇f · u = 0 for any unit vector u. This means the directional derivative in every possible direction is zero, so the function is flat (stationary) at that point in all directions. This occurs at critical points (possible max, min, or saddle).

11) The gradient of f(x,y) = 4x + 5y + 7 is

(4,5)

(0,0)

(4x,5y)

(4+7x,5+7y)

For a linear function f(x,y) = 4x + 5y + 7, the partial with respect to x is 4, and with respect to y is 5. The constant 7 disappears. So ∇f = (4,5) everywhere; it is a constant vector field.

Explanation

For a linear function f(x,y) = 4x + 5y + 7, the partial with respect to x is 4, and with respect to y is 5. The constant 7 disappears. So ∇f = (4,5) everywhere; it is a constant vector field.

12) At the point (2,1) on the level curve f(x,y) = k, the gradient ∇f(2,1) = (3,-4). Which of the following vectors could be tangent to the level curve at (2,1)?

(3,-4)

(-3,4)

(4,3)

(8,-6)

A tangent vector to the level curve must be perpendicular to the gradient (normal vector). So its dot product with ∇f must be zero. Check (4,3) · (3,-4) = 4·3 + 3·(-4) = 12 - 12 = 0. The others are not: (3,-4) is the gradient itself, (-3,4) is opposite gradient, (8,-6) = 2(4,-3), dot 3·8 + (-4)·(-6) = 24 + 24 = 48 ≠ 0.

Explanation

A tangent vector to the level curve must be perpendicular to the gradient (normal vector). So its dot product with ∇f must be zero. Check (4,3) · (3,-4) = 4·3 + 3·(-4) = 12 - 12 = 0. The others are not: (3,-4) is the gradient itself, (-3,4) is opposite gradient, (8,-6) = 2(4,-3), dot 3·8 + (-4)·(-6) = 24 + 24 = 48 ≠ 0.

13) The tangent line to the level curve f(x,y) = 9 at the point where ∇f = (2, -6) has normal vector

(2,-6)

(-6,2)

(1,-3)

Any scalar multiple of (-6,2)

The gradient is always normal to the level curve, so the normal vector to the tangent line at any point on the level curve is the gradient vector itself (or any scalar multiple). Thus (2, -6) is a normal vector to the tangent line.

Explanation

The gradient is always normal to the level curve, so the normal vector to the tangent line at any point on the level curve is the gradient vector itself (or any scalar multiple). Thus (2, -6) is a normal vector to the tangent line.

14) Given ∇f(3,1) = (6, -3), which vector gives directional derivative zero?

(1,2)

(-1,2)

(2,1)

(3,6)

We need v such that ∇f · v = 0. Compute (6, -3) · (1, 2) = 6·1 + (-3)·2 = 6 - 6 = 0. The other options do not give zero dot product.

Explanation

We need v such that ∇f · v = 0. Compute (6, -3) · (1, 2) = 6·1 + (-3)·2 = 6 - 6 = 0. The other options do not give zero dot product.

15) A heat-seeking particle is located at a point on a metal plate. If the particle moves in the direction of the gradient of the temperature function, how is it moving relative to the isotherms (lines of constant temperature)?

It is moving along an isotherm.

It is moving perpendicular to the isotherms.

It is moving at a 45-degree angle to the isotherms.

It is not moving.

Isotherms are the level curves of the temperature function. Since the gradient vector is always perpendicular (normal) to the level curves, a particle moving in the direction of the gradient is moving perpendicular to the isotherms.

Explanation

Isotherms are the level curves of the temperature function. Since the gradient vector is always perpendicular (normal) to the level curves, a particle moving in the direction of the gradient is moving perpendicular to the isotherms.

Tangent Lines, Tangent Planes & Normal Vectors