You CAN Ace Calculus

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Single Variable Calculus |
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Multi-Variable Calculus |
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Acceleration Vector |

Arc Length (Vector Functions) |

Arc Length Function |

Arc Length Parameter |

Conservative Vector Fields |

Cross Product |

Curl |

Curvature |

Cylindrical Coordinates |

Lagrange Multipliers |

Line Integrals |

Partial Derivatives |

Partial Integrals |

Path Integrals |

Potential Functions |

Principal Unit Normal Vector |

Differential Equations |
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Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.

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This is the second exam for first semester single variable calculus.

Each exam page contains a full exam with detailed solutions. Most of these are actual exams from previous semesters used in college courses. You may use these as practice problems or as practice exams. Here are some suggestions on how to use these to help you prepare for your exams.

- Set aside a chunk of full, uninterrupted time, usually an hour to two, to work each exam.

- Go to a quiet place where you will not be interrupted that duplicates your exam situation as closely as possible.

- Use the same materials that you are allowed in your exam (unless the instructions with these exams are more strict).

- Use your calculator as little as possible except for graphing and checking your calculations.

- Work the entire exam before checking any solutions.

- After checking your work, rework any problems you missed and go to the 17calculus page discussing the material to perfect your skills.

- Work as many practice exams as you have time for. This will give you practice in important techniques, experience in different types of exam problems that you may see on your own exam and help you understand the material better by showing you what you need to study.

IMPORTANT -

Exams can cover only so much material. Instructors will sometimes change exams from one semester to the next to adapt an exam to each class depending on how the class performs during the semester while they are learning the material. So just because you do well (or not) on these practice exams, does not necessarily mean you will do the same on your exam. Your instructor may ask completely different questions from these. That is why working lots of practice problems will prepare you better than working just one or two practice exams.

Calculus is not something that can be learned by reading. You have to work problems on your own and struggle through the material to really know calculus, do well on your exam and be able to use it in the future.

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Time |
2.5 hours |

Questions |
5 |

Total Points |
70 |

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Calculator |
not allowed |

Formula Sheet(s) |
one page 8.5x11 or A4; both sides okay |

Other Tools |
ruler |

*Instructions:*

- Show all your work.

- For each problem, correct answers are worth 1 point. The remaining points are earned by showing calculations and giving reasoning that justify your conclusions.

- Correct notation counts (i.e. points will be taken off for incorrect notation).

- Give your answers in exact, simplified form.

[20 points] Water is pouring into an inverted cone (point up) at the rate of 3 cubic meters per minute. The height of the cone is 10 meters and the radius of its base is 5 meters. How fast is the water level rising when the water stands 4 meters above the base?

Problem Statement |
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[20 points] Water is pouring into an inverted cone (point up) at the rate of 3 cubic meters per minute. The height of the cone is 10 meters and the radius of its base is 5 meters. How fast is the water level rising when the water stands 4 meters above the base?

Hint |
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The equation for the volume of a cone with height *h* and radius *r* is \(V=\pi r^2h/3\).

Problem Statement |
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Final Answer |
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\(\displaystyle{ \frac{1}{3\pi} }\) meters/min |

Problem Statement |
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Hint |
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The equation for the volume of a cone with height *h* and radius *r* is \(V=\pi r^2h/3\).

Solution |
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First, let's assign some variables. Call *w* the height of the water and *y* the distance from the top of the water to the point of the cone. Since the cone is 10 meters tall, we can write \(y=10-w\).

The radius of the top of the water we will call *r*. Since the water is the larger part of the cone, the volume of the water is given by taking the volume of the entire cone and subtracting the empty space above the water. So the volume of water is given by \(\displaystyle{ V=\frac{\pi(5)^2 (10)}{3} - \frac{\pi r^2 y}{3} }\). We want to try to keep the variable *w* in the equation since we are given the height of the water is \(w=4\) meters. So substitute our equation for *y* into the volume equation to get
\(\displaystyle{ V=\frac{\pi(5)^2 (10)}{3} - \frac{\pi r^2 (10-w)}{3} }\).

Now we need to find an expression for *r* in terms of *w*. To get this, we use the idea of similar triangles. In this problem we have \(\displaystyle{ \frac{r}{5} = \frac{y}{10} }\). Solving for *r*, we get \(\displaystyle{ r = \frac{10-w}{2} }\). Now we are ready to put this expression into the equation for volume.

\(\displaystyle{ V = \frac{205\pi}{3} - \frac{\pi}{3}\left[ \frac{10-w}{2} \right]^2 (10-w) }\) |

\(\displaystyle{ V = \frac{250\pi}{3} - \frac{\pi}{12}(10-w)^3 }\) |

Take the derivative with respect to |

\(\displaystyle{ \frac{dV}{dt} = -\frac{\pi}{12}(3)(10-w)^2 (-1)\frac{dw}{dt} }\) |

\(\displaystyle{ \frac{dV}{dt} = \frac{\pi}{4}(10-w)^2 \frac{dw}{dt} }\) |

We were given that \(dV/dt = 3 \) and \(w=4\). |

\(\displaystyle{ 3 = \frac{\pi}{4}(10-4)^2 \frac{dw}{dt} }\) |

\(\displaystyle{ \frac{dw}{dt} = \frac{1}{3\pi} }\) meters/min |

Final Answer |
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\(\displaystyle{ \frac{1}{3\pi} }\) meters/min |

close solution |

[15 points] Calculate the derivative of \(\displaystyle{ f(x) = \frac{e^{2x}}{x^2+1} }\)

**(a)** using the product rule
**(b)** using the quotient rule

**(c)** Compare your answers to parts (a) and (b). Are they the same? Why or why not?

Problem Statement |
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[15 points] Calculate the derivative of \(\displaystyle{ f(x) = \frac{e^{2x}}{x^2+1} }\)

**(a)** using the product rule
**(b)** using the quotient rule

**(c)** Compare your answers to parts (a) and (b). Are they the same? Why or why not?

Final Answer |
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\(\displaystyle{ f'(x) = \frac{2e^{2x}(x^2-x+1)}{(x^2+1)^2} }\) |

Problem Statement |
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**(a)** using the product rule
**(b)** using the quotient rule

**(c)** Compare your answers to parts (a) and (b). Are they the same? Why or why not?

Solution |
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**(a)** product rule

Rewrite \(f(x)\) as \( f(x) = e^{2x} (x^2+1)^{-1} \) |

\(\displaystyle{ f'(x) = e^{2x}\frac{d}{dx}[(x^2+1)^{-1}] + (x^2+1)^{-1}\frac{d}{dx}[e^{2x}] }\) |

\(\displaystyle{ f'(x) = e^{2x}(-1)(x^2+1)^{-2}(2x) + (2e^{2x})(x^2+1)^{-1} }\) |

\(\displaystyle{ f'(x) = \frac{-2xe^{2x}}{(x^2+1)^2} + \frac{2e^{2x}}{x^2+1} }\) |

Multiply the numerator and denominator of the second term on the right by \(x^2+1\) to get a common denominator. |

\(\displaystyle{ f'(x) = \frac{-2xe^{2x}+2(x^2+1)e^{2x}}{(x^2+1)^2} }\) |

\(\displaystyle{ f'(x) = \frac{2e^{2x}(x^2-x+1)}{(x^2+1)^2} }\) |

**(b)** quotient rule

\(\displaystyle{ f'(x) = \frac{(x^2+1)2e^{2x} - e^{2x}(2x)}{(x^2+1)^2} }\) |

\(\displaystyle{ f'(x) = \frac{2e^{2x}(x^2-x+1)}{(x^2+1)^2} }\) |

**(c)** Yes, they are the same as they should always be. I have simplied both expressions for my answers in parts a and b to the same form to show that they are equal.

Final Answer |
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\(\displaystyle{ f'(x) = \frac{2e^{2x}(x^2-x+1)}{(x^2+1)^2} }\) |

close solution |

[10 points] Calculate \(dy/dx\) for \( y = \sqrt{x}e^{x^2}(x^2+1)^{10} \) using logarithmic differentiation.

Problem Statement |
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[10 points] Calculate \(dy/dx\) for \( y = \sqrt{x}e^{x^2}(x^2+1)^{10} \) using logarithmic differentiation.

Final Answer |
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\(\displaystyle{ \frac{dy}{dx} = \left[ \frac{1}{2x} + 2x + \frac{20x}{x^2+1} \right] \sqrt{x}e^{x^2}(x^2+1)^{10} }\) |

Problem Statement |
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Solution |
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\( y = \sqrt{x}e^{x^2}(x^2+1)^{10} \) |

\(\displaystyle{ \ln y = \frac{1}{2}\ln x + x^2 + 10\ln(x^2+1) }\) |

\(\displaystyle{ \frac{1}{y}\frac{dy}{dx} = \frac{1}{2x} + 2x + \frac{10}{x^2+1}(2x) }\) |

\(\displaystyle{ \frac{dy}{dx} = \left[ \frac{1}{2x} + 2x + \frac{20x}{x^2+1} \right] \sqrt{x}e^{x^2}(x^2+1)^{10} }\) |

Final Answer |
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\(\displaystyle{ \frac{dy}{dx} = \left[ \frac{1}{2x} + 2x + \frac{20x}{x^2+1} \right] \sqrt{x}e^{x^2}(x^2+1)^{10} }\) |

close solution |

[10 points] Calculate the equation of the tangent line, in slope-intercept form, to the graph of the equation \(y+\sqrt{xy} = 6\) at \(y=4\). Accurately plot your tangent line on the graph. Note: Your line on the graph must match your calculations.

Problem Statement |
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[10 points] Calculate the equation of the tangent line, in slope-intercept form, to the graph of the equation \(y+\sqrt{xy} = 6\) at \(y=4\). Accurately plot your tangent line on the graph. Note: Your line on the graph must match your calculations.

Final Answer |
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\(\displaystyle{ y = \frac{-4x}{5}+ \frac{24}{5} }\) |

Problem Statement |
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Solution |
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\( y+\sqrt{xy} = 6 \) |

\(\displaystyle{ \frac{dy}{dx} + \frac{d}{dx}[(xy)^{1/2}] = \frac{d}{dx}[6] }\) |

\(\displaystyle{ \frac{dy}{dx} + \frac{1}{2}(xy)^{-1/2}\frac{d}{dx}[xy] = 0 }\) |

\(\displaystyle{ \frac{dy}{dx} + \frac{1}{2\sqrt{xy}}\left[ x\frac{dy}{dx}+ y(1) \right] = 0 }\) |

\(\displaystyle{ \frac{dy}{dx} + \frac{x}{2\sqrt{xy}} \frac{dy}{dx} + \frac{y}{2\sqrt{xy}} = 0 }\) |

\(\displaystyle{ \frac{dy}{dx} \left[ 1 + \frac{x}{2\sqrt{xy}} \right] = \frac{-y}{2\sqrt{xy}} }\) |

Now substitute \(y=4\). |

\(\displaystyle{ \frac{dy}{dx}\left[ 1 + \frac{1}{4} \right] = \frac{-4}{2(2)} }\) |

\(\displaystyle{ \frac{dy}{dx} (5/4) = -1 }\) |

\(\displaystyle{ \frac{dy}{dx} = -4/5 }\) |

So the slope is \(m=-4/5\). To find the point, we plug in \(y=4\) in the original equation to get \(x=1\).

\(y-y_1 = m(x-x_1) \)

\(y-4 = (-4/5)(x-1) \)

Final Answer |
\(\displaystyle{ y = \frac{-4x}{5}+ \frac{24}{5} }\) |

Final Answer |
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\(\displaystyle{ y = \frac{-4x}{5}+ \frac{24}{5} }\) |

close solution |

[15 points] Calculate the derivative of \(f(t) = \tan(\arccos(3t)) \) two ways. First, directly and secondly, by converting \(f(t)\) to algebraic form then taking the derivative. You do not need to show that your answers are equal.

Problem Statement |
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[15 points] Calculate the derivative of \(f(t) = \tan(\arccos(3t)) \) two ways. First, directly and secondly, by converting \(f(t)\) to algebraic form then taking the derivative. You do not need to show that your answers are equal.

Solution |
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First, take the derivative directly.

\(f(t) = \tan(\arccos(3t)) \) |

\(\displaystyle{ f'(t) = \sec^2(\arccos(3t))\frac{d}{dt}[\arccos(3t)] }\) |

\(\displaystyle{ f'(t) = \sec^2(\arccos(3t))\frac{-3}{\sqrt{1-9t^2}} }\) |

Final Answer - First Part |
\(\displaystyle{ f'(t) = \frac{-3\sec^2(\arccos(3t))}{\sqrt{1-9t^2}} }\) |

Secondly, converting to algebraic terms then taking the derivative.

Starting with the inside function, \(\theta = \arccos(3t) \to \cos\theta = 3t\).

Now draw a triangle matching \( \cos\theta = 3t \). From this triangle, the expression for \(\tan\theta\) is
\(\displaystyle{ f(t) = \tan\theta = \frac{\sqrt{1-9t^2}}{3t} }\). Now take the derivative.

\(\displaystyle{ f(t) = \frac{\sqrt{1-9t^2}}{3t} }\) |

Use the quotient rule. |

\(\displaystyle{ f'(t) = \frac{3t/2(1-9t^2)^{1/2}(-18t) - (1-9t^2)^{1/2}(3)}{9t^2} }\) |

Simplifying, we get the final answer.

Final Answer - Second Part |
\(\displaystyle{ f'(t) = \frac{-1}{3t^2(1-9t^2)^{1/2}} }\) |

close solution |