All questions of Calculus for Electrical Engineering (EE) Exam

Given Equation:
- x is real, find the maximum value of (-x^2 + 3x + 7)

Step 1: Find the vertex of the parabola
- The given equation is in the form of a quadratic equation, -x^2 + 3x + 7.
- To find the maximum value, we need to find the vertex of the parabola represented by this equation.
- The x-coordinate of the vertex is given by the formula: x = -b/2a, where a=-1 and b=3 in this case.
- Substituting the values of a and b, we get x = -3/(2*(-1)) = 3/2.
- Now, substitute x = 3/2 back into the equation to find the maximum value.

Step 2: Calculate the maximum value
- Substitute x = 3/2 into the equation: (-3/2)^2 + 3*(3/2) + 7
- Simplify the expression to find the maximum value: -9/4 + 9/2 + 7 = 37/4
Therefore, the maximum value of the given equation (-x^2 + 3x + 7) when x is real is 37/4. Hence, the correct answer is option 'C'.

Explain the questions

Divergence Theorem:
The divergence theorem relates a volume integral of a vector field to a surface integral of the vector field over the bounding surface of the volume. It states that the flux of a vector field across a closed surface is equal to the volume integral of the divergence of the vector field over the enclosed volume.

Given Vector Field:
The given vector field is F = (x^2, y^2, z^2).

Divergence of the Vector Field:
To calculate the divergence of the vector field, we need to find the partial derivatives of each component with respect to their corresponding variables and sum them up.

∇ · F = ∂(x^2)/∂x + ∂(y^2)/∂y + ∂(z^2)/∂z
= 2x + 2y + 2z

Application of Divergence Theorem:
Using the divergence theorem, we can evaluate the flux of the vector field across a closed surface by calculating the volume integral of the divergence over the enclosed volume.

In this case, the vector field is evaluated at a distance of one unit from the origin, which means we are considering a sphere centered at the origin with a radius of one unit.

Volume Integral:
To calculate the volume integral, we need to find the triple integral of the divergence of the vector field over the volume of the sphere.

∫∫∫ (2x + 2y + 2z) dV

Spherical Coordinates:
To evaluate the triple integral, it is convenient to switch to spherical coordinates since we are dealing with a sphere.

∫∫∫ (2ρsinφcosθ + 2ρsinφsinθ + 2ρcosφ) ρ^2sinφ dρ dθ dφ

Limits of Integration:
The limits of integration for ρ, θ, and φ are as follows:

ρ: 0 to 1 (since we are considering the sphere of radius one unit)
θ: 0 to 2π (full revolution around the z-axis)
φ: 0 to π (from the positive z-axis to the negative z-axis)

Integration:
Evaluating the triple integral, we get:

∫∫∫ (2ρsinφcosθ + 2ρsinφsinθ + 2ρcosφ) ρ^2sinφ dρ dθ dφ

= ∫[0 to 2π] ∫[0 to π] ∫[0 to 1] (2ρ^3sin^2φcosθ + 2ρ^3sin^2φsinθ + 2ρ^3sinφcosφ) dρ dθ dφ

= 2/4 * 2π * 2 * π * 1^4 * 1/4 + 0 + 0

= 2π/2

= π

Therefore, the divergence theorem value for the given vector field at a distance of one unit from the origin is π, which is approximately 3.

The given function is:
f(x) = (1 - cos(x))sin(x)

To find the maximum value of the function:
We can find the maximum value of the function by finding the critical points and determining whether they are maximum or minimum points.

Finding the critical points:
The critical points occur when the derivative of the function is zero or undefined. Let's find the derivative of the given function.

f'(x) = (1 - cos(x))cos(x) + sin(x)(-sin(x))
= cos(x) - cos^2(x) - sin^2(x)
= cos(x) - (1 - sin^2(x))
= cos(x) - 1 + sin^2(x)
= sin^2(x) + cos(x) - 1

Simplifying the derivative:
To find the critical points, we need to solve the equation f'(x) = 0.

sin^2(x) + cos(x) - 1 = 0

Using the identity sin^2(x) = 1 - cos^2(x):
1 - cos^2(x) + cos(x) - 1 = 0

Simplifying further:
-cos^2(x) + cos(x) = 0

Factoring out cos(x):
cos(x)(-cos(x) + 1) = 0

Setting each factor to zero:
cos(x) = 0 or -cos(x) + 1 = 0

Solving the first equation:
cos(x) = 0
This occurs when x = π/2 or x = 3π/2.

Solving the second equation:
-cos(x) + 1 = 0
cos(x) = 1
This occurs when x = 0 or x = 2π.

Therefore, the critical points are x = π/2, 3π/2, 0, and 2π.

Determining the nature of critical points:
To determine whether the critical points are maximum or minimum points, we can use the second derivative test. Let's find the second derivative of the function.

f''(x) = d/dx (sin^2(x) + cos(x) - 1)
= 2sin(x)cos(x) - sin(x)

Using the identity 2sin(x)cos(x) = sin(2x):
f''(x) = sin(2x) - sin(x)

Simplifying the second derivative:
f''(x) = 2sin(x)cos(x) - sin(x)
= sin(x)(2cos(x) - 1)

Evaluating the second derivative at the critical points:
f''(π/2) = sin(π/2)(2cos(π/2) - 1)
= 1(2(0) - 1)
= -1

f''(3π/2) = sin(3π/2)(2cos(3π/2) - 1)
= -1(2(0) - 1)
= 1

f''(0

To find the directional derivative of the function f(x,y,z) = xy + yz + zx, we need to find the gradient vector of f at the point P(1,1,1) and then take the dot product of the gradient vector with the tangent vector of the curve at P.

First, let's find the gradient vector of f:

∇f = (∂f/∂x, ∂f/∂y, ∂f/∂z)
= (y+z, x+z, x+y)

Now, let's find the tangent vector of the curve at P(1,1,1). We can do this by taking the derivative of each component of the curve with respect to t:

r(t) = (x(t), y(t), z(t))
= (t, t^2, t^3)

dr/dt = (dx/dt, dy/dt, dz/dt)
= (1, 2t, 3t^2)

At P(1,1,1), t = 1, so the tangent vector at P is:

dr/dt = (1, 2(1), 3(1)^2)
= (1, 2, 3)

Finally, we can find the directional derivative by taking the dot product of the gradient vector and the tangent vector:

Directional derivative = ∇f · dr/dt
= (y+z, x+z, x+y) · (1, 2, 3)
= (1+1, 1+1, 1+2)
= (2, 2, 3)

Therefore, the directional derivative of f = xy + yz + zx along the tangent to the curve at P(1,1,1) is (2, 2, 3).

The correct answer is b) Gauss divergence and Stokes.