Bay Area Tutoring

Dec 26

Related Rates: Defining Variables

By Tutor GuyNo Comments

Related rate problems will no longer seem so complicated once you master the art of defining rates as variables. And not just any variables, but variables in the form of derivatives! In other words, express the known and unknown rates as derivatives such as $dV/dt$ or $dA/dt$ . Remember that $dA/dt$ means “the rate of change of the Area”. Your task is to find expressions for rates in the problem and “translate” them from English into Calculus. Here are some examples:

“The radius is increasing at a rate of 3.0 cm/s.”

$\dfrac{dr}{dt}=3.0$

“If the volume is decreasing at a rate of 1.5 liters per minute…”

$\dfrac{dV}{dt}=-1.5$

“How fast is the ladder falling down the side of the building?”

$\dfrac{dy}{dt}= \, ?$

Remember that speed at which something moves is the rate of change of its position. Also, this derivative expresses the unknown we are going to find in the problem.

“The temperature is dropping by 2.3° per hour.”

$\dfrac{dT}{dt}=-2.3$

Note that when a value is falling or decreasing, the value of the derivative is negative.

Once you have defined all your rates as variables, your next step is to write the equation that expresses one variable in terms of the others. However, the equation has to be in terms of the quantities in the numerators, not in terms of the derivatives. Yes, this seems all wrong, but you will get the derivatives when you differentiate each side of the equation. For example, if your two rates are dV/dt and dr/dt, then you need to write an equation that expresses V in terms of r. When you differentiate this equation, you get the related rates equation and can plug in all the values you have.

Example: A hot air balloon in the shape of a sphere is being filled with gas at the rate of 20 cubic meters per minute. How fast is the radius changing when the volume is 288π cubic meters?

Solution: First, find all the rates mentioned in the problem:

$\dfrac{dV}{dt}=20; \; \dfrac{dr}{dt}= \, ?$

Now, you need an equation that relates volume (V) to radius (r). Because the balloon is a sphere, use the formula for the volume of a sphere:

$V= \dfrac{4}{3} \pi r^3$

Next, differentiate with respect to time:

$\dfrac{dV}{dt}=4 \pi r^2 \dfrac{dr}{dt}$

This equation includes r. You need to plug in the value of V in the original equation to solve for r:

$V= \dfrac{4}{3} \pi r^3 \rightarrow 288 \pi= \dfrac{4}{3} \pi r^3 \rightarrow 216=r^3 \rightarrow r=6 \, m$

Now plug in all the values to solve for the unknown rate:

$\dfrac{dV}{dt}=4 \pi r^2 \dfrac{dr}{dt} \rightarrow 20=4 \pi (6)^2 \dfrac{dr}{dt} \rightarrow \dfrac{dr}{dt}= \dfrac{20}{4 \pi (36)} \approx 0.044 \, m/min$

(Look for other tips on solving related rate problems on this website if you need help with other aspects of related rate problems.)

Calculus

calculus, math help, math homework, related rates

Dec 19

Translating functions

By Tutor GuyNo Comments

Any function f(x) is translated h units to the right when you replace x with x-h. Any function is translated k units up when you replace y with y-k. This is the basis of the vertex form of parabolas and standard form of the other conic sections, but it will help you graph almost any function you encounter without having to write out a “tree” first. If you know what the “parent” functions look like (and by the time you get to Algebra 2, you should memorize the parent functions), then you can graph translations without too much trouble.

Example 1: Graph

f(x) = ln (x-3).

Here the blue dotted graph is the parent function f(x) = ln (x). The red graph shows the solution, by translating the graph 3 units to the right.

In the graphs below, the parent function has been translated 3 units to the left and 2 units up.

Algebra 2

algebra 2, functions, math help, math homework, translations

Dec 15

Pulley problems

By Tutor GuyNo Comments

In a pulley problem, if you have a chain of boxes all connected by ropes, first find the tension in the rope that goes over the pulley (treat all of the boxes on the table as a single box and all of the boxes hanging down as a single box). Then work your way to each end of the chain one box at a time to find the tension in each rope along the chain.

Example: Find the tensions in each cable and the acceleration of the blocks in the diagram below. Assume that the pulley is massless and the table and pulley are frictionless.

Solution: First, treat the two blocks on the table as a single unit of 5 kg, and treat the two hanging blocks as a single unit of 5 kg:

The net force equation for the “block” on the table is:

T₂ = ma = 5a (1)

The net force equation for the hanging “block” is:

mg – T₂ = ma → 5g – T₂ = 5a (2)

Plugging equation (1) into equation (2) gives

a = g/2 = 4.9 m/s² and T₂ = 5g/2 = 24.5 N.

Now you can find the other tensions. Find T₃ by doing an FBD on the 3 kg block:

3g – T₃ = 3a = 3g/2

Solve to get T₃ = 3g/2 = 14.7 N. Find T₁ by doing an FBD on the 4 kg block:

T₁ = 4a = 2g, so T₁ = 19.6 N.

Physics

FBD, physics, pulley problem

Dec 12

Solving uniform circular motion problems

By Tutor GuyNo Comments

In order for an object to move in a uniform circular path, the net force on the object must be mv²/r. Draw the FBD, and any forces not pointed along a radius of the circle must cancel out. All the remaining forces must add up to mv²/r.

Example 1: You attach a ball to the end of a rope and swing it around vertically in a circle. If the rope is 0.50 m long, the ball has a mass of 0.50 kg, and the ball is traveling at a constant speed of 6.0 m/s, what is the tension in the rope at the bottom of the circle? At the top of the circle?

Solution: The FBD is shown for the ball at the bottom of the circle and at the top (I’ve shown the ball in its orbit, but left out the rope to make the picture clearer). Let’s solve the bottom case first.

The net force is T– F_g. Since the ball is traveling in a circle, the net force must be the centripetal force that keeps the ball in a circle, so

$T-F_g= \dfrac{mv^2}{r} \rightarrow T= \dfrac{mv^2}{r}+mg \rightarrow T= \dfrac{(0.5)(6.0)}{(0.5)}+(0.5)(9.8)=10.9 \, N$

Finding the solution at the top of the circle is just as simple:

$T+F_g= \dfrac{mv^2}{r} \rightarrow T= \dfrac{mv^2}{r}-mg \rightarrow T= \dfrac{(0.5)(6.0)}{(0.5)}-(0.5)(9.8)=1.1 \, N$

Note that in problems like this one, the tension is greatest at the bottom and a minimum at the top.

Example 2 : One of the obstacles at your favorite miniature golf course is a loop-the-loop as seen in the picture below. What is the minimum velocity necessary for the golf ball to stay in contact with the loop-the-loop at the top of the loop? The loop has a radius of 1.0 m.

Solution: Draw an FBD of the situation.

The two forces on the golf ball are the force due to gravity and the normal force of the loop against the golf ball. As in the last example, we set the net force equal to the centripetal force.

$F_N+F_g= \dfrac{mv^2}{r}$

You can see from this equation that minimizing the velocity means minimizing the two forces. If the normal force is zero (i.e., if the ball is just barely making contact with the loop, this satisfies our requirement:

$0+mg= \dfrac{mv^2}{r} \rightarrow v= \sqrt{rg} \rightarrow v= \sqrt{(1)(9.8)}=3.1 \, m/s$

Note that the answer does not depend on the mass of the golf ball!

Physics

centripetal force, physics, uniform circular motion

Dec 08

Using the test statistic

By Tutor GuyNo Comments

The test statistic is a standardized measure that allows us to calculate the probability of an event. Although the formulas appear to change in every section, the test statistic is almost always the number of standard deviations the statistic is from the mean. Here are four different examples that you probably learned in class.

1-sample means test (Z-test)

1-proportion means test (Z-test)

2-sample means test (Z-test)

2-sample means test (dependent samples)

Although the four equations look very different, they all have exactly the same structure:

The reason the formulas all look different is because we determine the population mean and standard deviation differently for different sample tests. But if you remember that all the test statistics have the same basic structure, you will find your calculations are a lot easier to figure out.

Statistics

statistics, t score, test statistic, z score

Dec 05

What do probability distributions tell us?

By Tutor GuyNo Comments

A probability distribution can be almost completely described if we know its shape, its center and its variation. Most of Descriptive Statistics (typically the first semester) is about describing the shape, center and variation of our data set.

There are a lot of formulas to learn in your Stats class. Keep in mind that almost all of them in the first semester are used to help us characterize each type of distribution we learn.

Here are some examples:

Binomial Distribution—its shape, center and distribution are defined by:

$p(x)= \displaystyle \binom{n}{x} p^x(1-p)^{n-x}; \; \mu = np; \; \sigma= \sqrt{np(1-p)}$

Geometric Distribution—its shape, center and distribution are defined by:

$p(x)=p(1-p)^{x-1}; \; \mu = \dfrac{1}{p}; \; \sigma = \dfrac{\sqrt{1-p}}{p}$

Exponential Distribution—its shape, center and distribution are defined by:

$p(x)= \dfrac{1}{\beta}e^{-x/\beta}; \; \mu = \beta; \; \sigma = \beta$

Where do these formulas come from? Well, we calculate the mean and standard deviation for a distribution just the way you do for a sample. But the algebra can get a little messy, and unless you are a statistics major, you probably don’t care. Just identify the type of distribution you have, and use the formulas to determine the mean and standard deviation.

Statistics

distributions, pdf, statistics

Dec 01

Simplifying calculus by simplifying equations at each step

By Tutor GuyNo Comments

Get in the habit of simplifying your equations as you go. If you need to find a second (or higher) derivative, simplify f ′(x) before you take the derivative again. If you are finding the volume of revolution, simplify your integral before you evaluate it. If you are applying l’Hôpital’s rule, simplify your new expression before you plug in the value of x again. If you don’t simplify, you usually make your task much harder.

Example: Find f ″(x).

$f(x)= \dfrac{2x+1}{x+1}$

Solution: The first derivative is straightforward:

$f'(x)= \dfrac{(x+1)2-(2x+1)1}{(x+1)^2}$

Taking the derivative of this is somewhat messy, unless you simplify first, as follows:

$f'(x)= \dfrac{(x+1)2-(2x+1)1}{(x+1)^2}= \dfrac{(2x+2)-(2x+1)}{(x+1)^2}= \dfrac{1}{(x+1)^2}$

Now the second derivative is much easier to evaluate:

$f''(x)= \dfrac{-2}{(x+1)^3}$

Calculus

calculus, math help, math homework

Nov 28

Optimization problems

By Tutor GuyNo Comments

Optimization problems can be tamed if you remember one simple idea. The purpose of an optimization problem is to optimize the situation. That is, you want to find where the results are the biggest, or the smallest, or the fastest, or the highest. In other words, if you can find an equation that describes the situation, you are being asked to find the relative minimum or relative maximum of that function. So break these problems down into three steps:

Determine what parameter you are optimizing.
Write a formula that gives that parameter in terms of an independent variable.
Take the derivative to find the relative min and/or max.

Example: What are the dimensions of the rectangle with the largest area that can be inscribed above the x-axis and under the function f(x) = 9 – x²?

Solution: (Step 1) In this case, you are trying to maximize area. So you need (Step 2) a formula that gives the area in terms of the x-variable. Let’s draw a picture:

We can see from the sketch that if the upper right corner of the rectangle is on the point (x,y), the base of the rectangle is 2x and the height is y. So our first effort at an equation for the area is

A = 2xy.

To finish the problem, we need to express y in terms of x. Since the point is on the parabola, y = 9 – x². Therefore, the equation we seek is

A = 2x(9 – x²) = 18x – 2x³.

To find the dimensions of the largest possible area, we find the relative maximum of the equation (Step 3).

A’ = 18 – 6x² = 0 → x² = 3

Plugging back into the equation for y gives y = 6.

Calculus

calculus, math help, math homework, optimization problem

Tutor Guy Homework Tips