Math 111 Introduction to the Calculus Fall 2011 Final Exam Solutions

Solution:

The expression simplifies to (x+3) for all x 3, and since we're taking the limit as x approaches 3 (not evaluating at x=3), we can substitute directly.

Solution:

To find the derivative, we apply both the product rule and chain rule:

Solution:

We use the trigonometric identity sin(x) = 1 - cos(x) to rewrite the integral:

Let u = cos(x), then du = -sin(x) dx:

Now evaluating from 0 to /2:

Solution:

Let's define our variables:

• V = volume of water in the cone
• h = water height at any time
• r = radius of the water surface at any time

For similar triangles: r/h = 3/6, so r = h/2.

The volume of a cone is V = rh/3. Substituting r = h/2:

Differentiating with respect to time:

Solving for dh/dt:

Therefore, when the water is 2 m deep, the water level is rising at a rate of 5/ m/min.

Solution:

Let's define our variables:

• x = length and width of the square base
• y = height of the box

Volume constraint: V = xy = 78,088

Surface area: S = x + 4xy (base + four sides)

From volume constraint: y = 78,088/x

Substituting y into surface area equation:

To find minimum surface area, set dS/dx = 0:

Finding y:

Verifying this is a minimum: dS/dx = 2 + 624,704/x > 0 at x 53.84

Therefore, the dimensions that minimize material are: Base 53.84 cm 53.84 cm, Height 26.92 cm

Solution:

First, find intersection point of y = x and y = 2:

The region is bounded by x = 0 on the left, x = 4 on the right, y = 2 on top, and y = x on the bottom.

Integrating with respect to y:

Solution:

Using the Fundamental Theorem of Calculus and Chain Rule:

Therefore, the derivative of F(x) is e sin(e).

Solution:

Let's evaluate the improper integral:

Case 1: p = 1

The integral diverges when p = 1.

Case 2: p 1

This limit exists only when 1-p < 0, or p > 1. In this case:

Conclusion: The improper integral converges when p > 1 and diverges when p 1.

Solution:

The Maclaurin series is the Taylor series centered at x = 0.

Derivatives of f(x) = e^2x:

The Maclaurin series is:

Therefore, the Maclaurin series for e^2x is _n=0 (2x)/n! = 1 + 2x + 2x/2! + 2x/3! + ...

Solution:

First and second derivatives:

Possible inflection point: Set f''(x) = 0

Determine sign of f''(x) on each interval:

Since concavity changes at x = 1, we have an inflection point there.

The y-coordinate:

Conclusion:

• f(x) is concave downward on (-, 1)
• f(x) is concave upward on (1, )
• The inflection point is at (1, -1)

Solution:

This is a separable differential equation. Separating variables:

Integrating both sides:

Applying the initial condition y(0) = 1:

Substituting C back:

Therefore, the solution to the differential equation is y = 2/(2 - x).