You CAN Ace Calculus

### Topics You Need To Understand For This Page

 derivatives integrals related rates

### 17Calculus Subjects Listed Alphabetically

Single Variable Calculus

 Absolute Convergence Alternating Series Arc Length Area Under Curves Chain Rule Concavity Conics Conics in Polar Form Conditional Convergence Continuity & Discontinuities Convolution, Laplace Transforms Cosine/Sine Integration Critical Points Cylinder-Shell Method - Volume Integrals Definite Integrals Derivatives Differentials Direct Comparison Test Divergence (nth-Term) Test
 Ellipses (Rectangular Conics) Epsilon-Delta Limit Definition Exponential Derivatives Exponential Growth/Decay Finite Limits First Derivative First Derivative Test Formal Limit Definition Fourier Series Geometric Series Graphing Higher Order Derivatives Hyperbolas (Rectangular Conics) Hyperbolic Derivatives
 Implicit Differentiation Improper Integrals Indeterminate Forms Infinite Limits Infinite Series Infinite Series Table Infinite Series Study Techniques Infinite Series, Choosing a Test Infinite Series Exam Preparation Infinite Series Exam A Inflection Points Initial Value Problems, Laplace Transforms Integral Test Integrals Integration by Partial Fractions Integration By Parts Integration By Substitution Intermediate Value Theorem Interval of Convergence Inverse Function Derivatives Inverse Hyperbolic Derivatives Inverse Trig Derivatives
 Laplace Transforms L'Hôpital's Rule Limit Comparison Test Limits Linear Motion Logarithm Derivatives Logarithmic Differentiation Moments, Center of Mass Mean Value Theorem Normal Lines One-Sided Limits Optimization
 p-Series Parabolas (Rectangular Conics) Parabolas (Polar Conics) Parametric Equations Parametric Curves Parametric Surfaces Pinching Theorem Polar Coordinates Plane Regions, Describing Power Rule Power Series Product Rule
 Quotient Rule Radius of Convergence Ratio Test Related Rates Related Rates Areas Related Rates Distances Related Rates Volumes Remainder & Error Bounds Root Test Secant/Tangent Integration Second Derivative Second Derivative Test Shifting Theorems Sine/Cosine Integration Slope and Tangent Lines Square Wave Surface Area
 Tangent/Secant Integration Taylor/Maclaurin Series Telescoping Series Trig Derivatives Trig Integration Trig Limits Trig Substitution Unit Step Function Unit Impulse Function Volume Integrals Washer-Disc Method - Volume Integrals Work

Multi-Variable Calculus

 Acceleration Vector Arc Length (Vector Functions) Arc Length Function Arc Length Parameter Conservative Vector Fields Cross Product Curl Curvature Cylindrical Coordinates
 Directional Derivatives Divergence (Vector Fields) Divergence Theorem Dot Product Double Integrals - Area & Volume Double Integrals - Polar Coordinates Double Integrals - Rectangular Gradients Green's Theorem
 Lagrange Multipliers Line Integrals Partial Derivatives Partial Integrals Path Integrals Potential Functions Principal Unit Normal Vector
 Spherical Coordinates Stokes' Theorem Surface Integrals Tangent Planes Triple Integrals - Cylindrical Triple Integrals - Rectangular Triple Integrals - Spherical
 Unit Tangent Vector Unit Vectors Vector Fields Vectors Vector Functions Vector Functions Equations

Differential Equations

 Boundary Value Problems Bernoulli Equation Cauchy-Euler Equation Chebyshev's Equation Chemical Concentration Classify Differential Equations Differential Equations Euler's Method Exact Equations Existence and Uniqueness Exponential Growth/Decay
 First Order, Linear Fluids, Mixing Fourier Series Inhomogeneous ODE's Integrating Factors, Exact Integrating Factors, Linear Laplace Transforms, Solve Initial Value Problems Linear, First Order Linear, Second Order Linear Systems
 Partial Differential Equations Polynomial Coefficients Population Dynamics Projectile Motion Reduction of Order Resonance
 Second Order, Linear Separation of Variables Slope Fields Stability Substitution Undetermined Coefficients Variation of Parameters Vibration Wronskian

### Search Practice Problems

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

### Practice Exam Tips

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.

Exam Details

Time

2.5 hours

Questions

5

Total Points

70

Tools

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

[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

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

Problem Statement

[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?

$$\displaystyle{ \frac{1}{3\pi} }$$ meters/min

Problem Statement

[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

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

Solution

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 t. $$\displaystyle{ \frac{dV}{dt} = -\frac{\pi}{12}(3)(10-w)^2 (-1)\frac{dw}{dt} }$$ $$\displaystyle{ \frac{dV}{dt} = \frac{\pi}{4}(10-w)^2 \frac{dw}{dt} }$$ 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

$$\displaystyle{ \frac{1}{3\pi} }$$ meters/min

[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

[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?

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

Problem Statement

[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?

Solution

(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.

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

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

Problem Statement

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

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

Problem Statement

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

Solution

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

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

[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

[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.

$$\displaystyle{ y = \frac{-4x}{5}+ \frac{24}{5} }$$

Problem Statement

[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.

Solution

 $$y+\sqrt{xy} = 6$$ $$\displaystyle{ \frac{dy}{dx} + \frac{d}{dx}[(xy)^{1/2}] = \frac{d}{dx}[6] }$$ $$\displaystyle{ \frac{dy}{dx} + \frac{1}{2}(xy)^{-1/2}\frac{d}{dx}[xy] = 0 }$$ $$\displaystyle{ \frac{dy}{dx} + \frac{1}{2\sqrt{xy}}\left[ x\frac{dy}{dx}+ y(1) \right] = 0 }$$ $$\displaystyle{ \frac{dy}{dx} + \frac{x}{2\sqrt{xy}} \frac{dy}{dx} + \frac{y}{2\sqrt{xy}} = 0 }$$ $$\displaystyle{ \frac{dy}{dx} \left[ 1 + \frac{x}{2\sqrt{xy}} \right] = \frac{-y}{2\sqrt{xy}} }$$ 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 $$\displaystyle{ y = \frac{-4x}{5}+ \frac{24}{5} }$$

$$\displaystyle{ y = \frac{-4x}{5}+ \frac{24}{5} }$$

[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

[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

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