Calc Solutions: January 2012

Thursday, January 26, 2012

Day 7: Double Conjugate Limit Problem

Find the limit:

$\lim_{x\rightarrow2}{\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}}$

Solution: Limits with square roots in the top and bottom are daunting. They require a bit of work as well. Had there been only one square root, the task would be to multiply by the conjugate on the top and bottom. The task at hand is not much different, it requires multiply the top and bottom by both conjugates as follows:

$\lim_{x\rightarrow2}{\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}}=\lim_{x\rightarrow2}{\frac{(\sqrt{6-x}-2)(\sqrt{6-x}+2)(\sqrt{3-x}+1)}{(\sqrt{3-x}-1)(\sqrt{6-x}+2)(\sqrt{3-x}+1)}}$

Notice that the operation applied above does not change anything because the top and bottom was multiplied by the same thing, hence, we multiplied the limit by 1. However, the limit simplifies nicely from here:

$=\lim_{x\rightarrow2}{\frac{(6-x-4)(\sqrt{3-x}+1)}{(\sqrt{6-x}+2)(3-x-1)}}=\lim_{x\rightarrow2}{\frac{(\sqrt{3-x}+1)}{(\sqrt{6-x}+2)}}=\frac{(\sqrt{3-2}+1)}{(\sqrt{6-2}+2)}=\frac{1}{2}$

And now these problems are cake!

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Day 6: Approximating Roots for e^x = x + 2

Show that the equation e^x = x + 2 has a root in the interval [1, 2]. Then determine an interval with a width of at least 0.25 that contains this root.

Solution:
$\text{if }e^x=x+2, \text{ then } e^x-x-2=0.$
This reduces the problem to showing that the following function has a zero on the interval [1,2]: $f(x)=e^x-x-2$
To solve this, we can use the intermediate value theorem (if that's not a familiar theorem, the idea should still make sense). Test f(1) and f(2).
$f(1)=e-1-2=\text{negative, } f(2)=e^2-2-2=\text{positive}$
Since f is negative on one end of the interval and positive on the other, it's clear that, at some point, it must have been zero. This shows that the equation has a root on the interval [1,2].

Next, choose values within the interval to narrow down the error. First I'll try 1.1:

$f(1.1)\approx-0.0958$
Since this value is BARELY negative and our goal is a range of .25, I'll try 1.35 and hope that it's positive (if it is, we've show that the root lies on the interval [1.1,1.35], whose width is .25 as required):
$f(1.35)\approx0.507$

As noted above, this completes the problem. An interval of size .25 containing a root of the equation is [1.1,1.35]. Notice that there are many different (but correct) solutions to this problem.

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Day 5: Discontinuity Problem

1. Determine all points, if any, at which each of the following functions is discontinuous, and state the type of discontinuity. Justify your answer with an appropriate analysis:
(a)
$f(x)=\frac{x^2-3x+2}{|x-2|}$
(b)

$g(x) = \begin{cases} \frac{4-\sqrt{x}}{16-x}, & \text{if }x\neq16 \\ -8, & \text{if }x=16 \end{cases}$

Solution:
(a) To determine where f(x) is discontinuous, it's necessary to know what could make it discontinuous. Dividing by 0, taking the square root of a negative number are ways a graph can be discontinuous. The only one that applies to f(x) is dividing by zero. Suppose x=2, then 0/0 is obtained:

$f(2)=\frac{0}{0}$
To identify which type of discontinuity this is, you must know how each type of discontinuity occurs. When dividing by zero, there are two possible discontinuities. If, at the number that causes the denominator to be zero, the numerator is also zero, there is a removable discontinuity called a "hole". If the numerator is nonzero, but the denominator is zero, then the function has an "asymptote". We found that f is discontinuous at x=2 and that f(2)=0/0, so f has a hole at 2.

(b) Now g(x) has a square root and is a fraction, so there are two ways for g to be discontinuous. First of all, let's determine the following limit:

$\lim_{x\rightarrow16}{\frac{4-\sqrt{x}}{x-16}}$
To do so, multiply the top and bottom by the conjugate of the numerator (which is the same as multiplying by one, so it doesn't change the value of the limit!). The conjugate is simply the numerator, but change the minus to a plus:

$\lim_{x\rightarrow16}{\frac{4-\sqrt{x}}{x-16}}=\lim_{x\rightarrow16}{\frac{(4-\sqrt{x})(4+\sqrt{x})}{(x-16)(4+\sqrt{x})}}=\lim_{x\rightarrow16}{\frac{16-x}{(x-16)(4+\sqrt{x})}}=\lim_{x\rightarrow16}{\frac{1}{(4+\sqrt{x})}}=\frac{1}{8}$

Hence, the following is true:

$\lim_{x\rightarrow16}{f(x)}\neq f(16)$
i.e., there is a hole at x=16.

Finally, if x is negative (x<0), then the square root of x is undefined. Thus, x<0 is not in the domain of g.

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Monday, January 23, 2012

Day 4: L'Hospital/Special Limit Challenge Problem

Citation: "Calculus Early Transcendentals, 7th Edition", by James Stewart Section 3.4 The Chain Rule Problem 51: The figure shows a circular arc of length s and a choard of length d, both suspended by a central angle theta. Find the limit as theta approach 0 from the right of s/d:

Solution: The first task will be identifying s and d in terms of theta, since the limit is as theta approaches 0. Recall the formula for arc-length:

$Arclength=s=r\theta$

In order to rename d in terms of theta and r, we must employ trigonometry.

Notice that we can now relate d with theta and r via the sine function:

$sin(\frac{\theta}{2})=\frac{d}{2}/r$
Ergo,
$d=2rsin(\frac{\theta}{2})$
And this finally gives:

$\lim_{\theta\rightarrow 0^+}{\frac{r\theta}{2rsin{(\frac{\theta}{2})}}}=\lim_{\theta\rightarrow 0^+}\frac{\theta/2}{sin{(\frac{\theta}{2})}}$
There are two ways to complete this problem. One involves recalling a special limit, and the other involves L'Hospital's Rule. I will show each respectively:

Special Limit:
This route using the special sine limit:

$\lim_{x\rightarrow 0^+}{\frac{sin(x)}{x}}=1$
We can manipulate our limit to put it in this special form as follows:

$\lim_{\theta\rightarrow 0^+}{\frac{\theta/2}{sin(\theta/2)}}=\frac{1}{\lim_{\theta\rightarrow 0^+}{\frac{sin(\theta/2)}{\theta/2}}}$
By substituting x=theta/2 we can put this limit into a know form, particularly the special form sin(x)/x:

$\frac{1}{\lim_{\theta\rightarrow 0^+}{\frac{sin(\theta/2)}{\theta/2}}}=\frac{1}{\lim_{x\rightarrow 0^+}{\frac{sin(x)}{x}}}=\frac{1}{1}=1$
This completes the problem using the first method, which is what the section was asking for. However, for the more advanced students, and myself, I would initially approach the limit with the following method.

L'Hospital's Rule:

$\lim_{\theta\rightarrow 0^+}\frac{\theta/2}{sin{(\frac{\theta}{2})}}$
This limit has the misfortune of being 0/0 when direct substitution is applied. As you should remember (or note) this indicates a removable discontinuity. In this situation, we can also use L'Hospital's Rule which states:

Theorem: L'Hospital's Rule:
If

$\lim_{x \to c}f(x)=\lim_{x \to c}g(x)=0 \text{ or } \pm\infty$ , and

$\lim_{x\to c}\frac{f'(x)}{g'(x)}$ exists, and

$g'(x)\neq 0$ on an open interval containing c,

then

$\lim_{x\to c}\frac{f(x)}{g(x)} = \lim_{x\to c}\frac{f'(x)}{g'(x)}$ .

In our case, we have the following:
$f'(\theta)=1/2 \quad g'(\theta)=\frac{1}{2}cos(\theta)$
The conditions of L'Hospital's Rule (red) are certainly fulfilled, so we can apply the implication of the rule, as follows:

$\lim_{\theta\rightarrow\0^+}{\frac{\theta/2}{sin(\theta/2)}}=\lim_{\theta\rightarrow\0^+}{\frac{1/2}{1/2\,cos(\theta/2)}}=\lim_{\theta\rightarrow\0^+}{\frac{1}{cos(\theta/2)}}=\frac{1}{1}=1$

It's clear that both methods are valid approaches to the problem, yielding the same answer. From a personal standpoint, it's easier to remember the rule (L'Hospital's) that can be applied to numerous problems than to systematically manipulate equations so that the special (and important!) limit can be used

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Sunday, January 22, 2012

Day 3 Trig Substitution

Citation: "Calculus Early Transcendentals, 6th Edition", by James Stewart Section 7.3 Trigonometric Substitution Integrals Problem 4:

$\int \frac{dx}{\sqrt{16-x^2}}$
Solution: This problem will be a very basic and instructive example of integrating with trig substitutions. The key to doing these problems is the following identity:

$Identity: sin^2(\theta)+cos^2(\theta)=1$
Suppose we make a substitution for x:
$$x=4sin(\theta)$
Then,
$\int \frac{dx}{\sqrt{16-x^2}}=\int \frac{4cos(\theta)d\theta}{\sqrt16-16sin^2(\theta)}=\int \frac{cos(\theta)d\theta}{\sqrt{1-sin^2(\theta)}}=\int \frac{cos(\theta)}{cos(\theta)}= \theta$
Noting that the following comes from the Pythagorean Identity written above:
$$cos^2(\theta)=1-sin^2(\theta)$
Finally, we rewrite theta in terms of x.
$\theta=sin^{-1}(x)$

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Day 2: Optimization Fence

Citation: "Calculus Early Transcendentals, 6th Edition", by James Stewart Section 4.7 Optimization Problem 11: A farmer wants to fence an area of 1.5 million square feet in a rectangular field and then divide it in half with a fence parallel to one of the sides of the rectangle. How can he do this so as to minimize the cost of the fence.

Solution: So first of all, we will need a visualization of the fence we're optimizing. The description of the fence can be drawn with three side by side identical vertical lines and two horizontal lines connecting the tops of the vertical lines. Since all the vertical lines are equal distance, call their length x and call the length of the horizontal lines y. The total area of the fence will be:
$A=xy=1.5\,\,million$

And hence

$y=\frac{1.5\,\,million}{x}$
The total perimeter, or length of fencing, is given by:

$P=x+x+x+y+y=3x+2y=3x+\frac{1,500,000}{x}$

Keep in mind the problem is asking to minimize the cost of the fence, which, we assume, is analogous to minimizing the length of the fence, identified as "P". We can use the Calculus to minimize the function P(x). Finding P'(x):

$P'(x)=3-\frac{1,500,000}{x^2 }$
$3-\frac{1,500,000}{x^2}=0$
$x=\sqrt{500,000}\approx{707.1}$
This, so far, has provided the optimal x value. We can use this to find the optimal y value:

$y=\frac{1,500,000}{1,000}=1,500$
...and voila, we found the optimal dimensions of the fence with given area.

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Day 1: Related Rates Boat Problem

Citation: "Calculus Early Transcendentals, 6th Edition", by James Stewart Section 3.9 Related Rates Problem 20: A boat is pulled into a dock by a rope attached to the bow of the boat and passing through a pulley on the dock that is 1 m higher than the bow of the boat. If the rope is pulled in at a rate of 1 m/s, how fast is the boat approaching the dock when it is 8 m from the dock?
Solution: Label the distance along the water from the boat to the dock as x and the height of the dock y and the length of the rope z. Then, identify the rates of change in terms of time. By this, we wish to identify dx/dt, dy/dt, and dz/dt which are, respectively, the change in the distance from the boat to the dock along the water in terms of time t (which is what the problem is asking us), the change in the height of the dock in terms of time t (which doesn't change!), and the rate of change of the length of the rope from the boat to the dock (which is given as 1 m/s in the problem). Now that we have the "rate" from "related rates' we need to find the "related" part, or the relationship. From the Pythagorean Theorem:
$x^2 + y^2 = z^2$
Now, IMPLICITLY differentiate with respect to time t. Related rates problems will always be differentiated with respect to time because the way things change is some measurement (distance, volume, etc..) over time t (seconds, minutes, etc.). This yields the following:
$\inline 2x\frac{dx}{dt}+2y\frac{dy}{dt}=2z\frac{dz}{dt}$
Next, gather the knowns in the equation: The problem tells us at time we are considering, the boat is 8m from the dock--the value of x. Also, the height of the dock is permanently 1m -- y (change in tides, etc are not part of the problem). By applying the Pythagorean Theorm, we find z:
$\inline $1^2+8^2=65=z^2 \\Hence, z=\sqrt{65}\approx8.06$
We also know that dz/dt is 1m/s dy/dt is 0m/s because the height of the dock is not changing, and the problem is asking us to identify dx/dt. By substituting the known values, we get:
$\inline $8\times\frac{dx}{dt}+1\times0m/s=8.06\times1 m/s$
$\inline \frac{dx}{dt}=1.0075$

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Friday, January 20, 2012

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