1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
tatiyna
3 years ago
14

Trouble finding arclength calc 2

Mathematics
1 answer:
kiruha [24]3 years ago
7 0

Answer:

S\approx1.1953

Step-by-step explanation:

So we have the function:

y=3-x^2

And we want to find the arc-length from:

0\leq x\leq \sqrt3/2

By differentiating and substituting into the arc-length formula, we will acquire:

\displaystyle S=\int\limits^\sqrt3/2}_0 {\sqrt{1+4x^2} \, dx

To evaluate, we can use trigonometric substitution. First, notice that:

\displaystyle S=\int\limits^\sqrt3/2}_0 {\sqrt{1+(2x)^2} \, dx

Let's let y=2x. So:

y=2x\\dy=2\,dx\\\frac{1}{2}\,dy=dx

We also need to rewrite our bounds. So:

y=2(\sqrt3/2)=\sqrt3\\y=2(0)=0

So, substitute. Our integral is now:

\displaystyle S=\frac{1}{2}\int\limits^\sqrt3}_0 {\sqrt{1+y^2} \, dy

Let's multiply both sides by 2. So, our length S is:

\displaystyle 2S=\int\limits^\sqrt3}_0 {\sqrt{1+y^2} \, dy

Now, we can use trigonometric substitution.

Note that this is in the form a²+x². So, we will let:

y=a\tan(\theta)

Substitute 1 for a. So:

y=\tan(\theta)

Differentiate:

y=\sec^2(\theta)\, d\theta

Of course, we also need to change our bounds. So:

\sqrt3=\tan(\theta), \theta=\pi/3\\0=\tan(\theta), \theta=0

Substitute:

\displaystyle 2S= \int\limits^{\pi/3}_0 {\sqrt{1+\tan^2(\theta)}\sec^2(\theta) \, d\theta

The expression within the square root is equivalent to (Pythagorean Identity):

\displaystyle 2S= \int\limits^{\pi/3}_0 {\sqrt{\sec^2(\theta)}\sec^2(\theta) \, d\theta

Simplify:

\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta

Now, we have to evaluate this integral. To do this, we can use integration by parts. So, let's let u=sec(θ) and dv=sec²(θ). Therefore:

u=\sec(\theta)\\du=\sec(\theta)\tan(\theta)\, d\theta

And:

dv=\sec^2(\theta)\, d\theta\\v=\tan(\theta)

Integration by parts:

\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)-(\int\limits^{\pi/3}_0 {\tan^2(\theta)\sec(\theta)} \, d\theta)

Again, let's using the Pythagorean Identity, we can rewrite tan²(θ) as:

\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)-(\int\limits^{\pi/3}_0 {(\sec^2(\theta)-1)\sec(\theta)} \, d\theta)

Distribute:

\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)-(\int\limits^{\pi/3}_0 {(\sec^3(\theta)-\sec(\theta)} \, d\theta)

Now, let's make the single integral into two integrals. So:

\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)-(\int\limits^{\pi/3}_0 {\sec^3(\theta)\, d\theta-\int\limits^{\pi/3}_0 {\sec(\theta)}\, d\theta)

Distribute the negative:

\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)-\int\limits^{\pi/3}_0 {\sec^3(\theta)\, d\theta+\int\limits^{\pi/3}_0 {\sec(\theta)}\, d\theta

Notice that the integral in the first equation and the second integral in the second equation is the same. In other words, we can add the second integral in the second equation to the integral in the first equation. So:

\displaystyle 2S= 2\int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\sec(\theta)\tan(\theta)+\int\limits^{\pi/3}_0 {\sec(\theta)}\, d\theta

Divide the second and third equation by 2. So: \displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\frac{1}{2}(\sec(\theta)\tan(\theta)+\int\limits^{\pi/3}_0 {\sec(\theta)}\, d\theta)

Now, evaluate the integral in the second equation. This is a common integral, so I won't integrate it here. Namely, it is:

\displaystyle 2S= \int\limits^{\pi/3}_0 (\sec(\theta))\sec^2(\theta) \, d\theta=\frac{1}{2}(\sec(\theta)\tan(\theta)+\ln(\tan(\theta)+\sec(\theta))

Therefore, our arc length will be equivalent to:

\displaystyle 2S=\frac{1}{2}(\sec(\theta)\tan(\theta)+\ln(\tan(\theta)+\sec(\theta)|_{0}^{\pi/3}

Divide both sides by 2:

\displaystyle S=\frac{1}{4}(\sec(\theta)\tan(\theta)+\ln(\tan(\theta)+\sec(\theta)|_{0}^{\pi/3}

Evaluate:

S=\frac{1}{4}((\sec(\pi/3)\tan(\pi/3)+\ln(\tan(\pi/3)+\sec(\pi/3))-(\sec(0)\tan(0)+\ln(\tan(0)+\sec(0))

Evaluate:

S=\frac{1}{4}((2\sqrt3+\ln(\sqrt3+2))-((1)(0)+\ln(0+1))

Simplify:

S=\frac{1}{4}(2\sqrt 3+\ln(\sqrt3+2)}

Use a calculator:

S\approx1.1953

And we're done!

You might be interested in
A triangle has sides with lengths of 6 inches, 13 inches, and 17 inches. Is it a right triangle?
Tems11 [23]

Answer:

Nope

Step-by-step explanation:

To now if a triangle is a right angle, use Pythagoras Theorem,

a^2 + b^2 = c^2

C usually be the longest side of the triangle, C = 17

Then also let a and b be 6 and 13 respectively.

6^2 + 13^2 =205

C^2 = 205

C = sqrt(205) = 14.32

therefore no this is not a right angle triangle

4 0
2 years ago
PLEASE HELP I NEED A GOOD GRADE ON THIS
VladimirAG [237]

Answer:

B = 1/243

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Find the perimeter of the figure.
Alika [10]

1+1=2

2+3=5

5+3=8

8+5=13 inches

3 0
2 years ago
About 1.15 million people live in a circular region with a population density of about 18,075 people per square kilometer. Find
Tanzania [10]

Step 1

<u>Find the total area</u>

Total area=(1,150,000/18,075)=63.62 square kilometer

The area of a circle is equal to

A=\pi * r^{2} \\ 63.62= \pi * r^{2}\\r=\sqrt{(63.62/\pi)} \\r=4.5\ km

therefore

the answer is

4.5\ km

7 0
3 years ago
The cost of coffee is determined by the type of coffee beans that are mixed together. The cost of a local mix of Arabica and Rob
guapka [62]
700a+1200r=1000000       (then plug in the a)
700(1000) +1200r=1000000        (simplify)
700000+1200r=1000000        (subtract 700000 from both sides)
1200r=300000             (divide by 1200 on both sides)
r=250
250 pounds
5 0
3 years ago
Other questions:
  • In the diagram of circle A, what is the measure of ∠XYZ?<br><br> 35°<br> 70°<br> 75°<br> 140°
    8·2 answers
  • A particular plant root grows 2.5 inches per month. How many centimeters is the plant root growing per month? (1 inch = 2.54 cen
    14·2 answers
  • A two-runway (one runway for landing, one runway for taking off) airport is being designed for propeller-driven aircraft. The ti
    13·1 answer
  • C is a point of tangency. Segments PC and PB have a common point at P. Find the length PC.
    13·2 answers
  • I don't know what to do.
    6·1 answer
  • Given the figure below, find the values of x and z.
    8·1 answer
  • Which numbers are a distance of 4 units from −6 on a number line? Select the location on the number line to plot each point.
    9·1 answer
  • vJoey has 2/3 of a gallon of ice cream. He divides the ice cream for his friends so that 1/12 of a gallon is in each bowl. How m
    10·2 answers
  • What does y=kx mean​
    10·2 answers
  • If a liquid has a volume of 7,000 milliliters, what is the volume in liters?
    11·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!