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3 years ago

10

PLEASE HELP MEEEE THIS IS DUE IN TEN MINUTES ALSO HOW THE HECK DO YOU CHANGE THE THING THAT SAYS COLLEGE BY UR NAME? IM IN FLIPP

ING EIGTH GRADE NOT COLLEGE!!!

2 answers:

irina1246 [14]3 years ago

7 0

This ain’t hard dw I’ll give you the answers :)

Here ya go :)

Y_Kistochka [10]3 years ago

7 0

Answer:

1. 4.05 x 10^-4

2. 49,345,000

3. 2.357897 × 10^9

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4 years ago

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Find the tangent line equation of the curve at the given point. Y=arcsin(7x) at the point where x=sqrt2/14

Mumz [18]

Answer:

$Y-\frac{\pi}{2} =\frac{\pi}{2} (x+\sqrt{\frac{1}{7}})$

Step-by-step explanation:

The equation of the curve is

$Y = sin^{-1}(7x)$

To find the equation of tangent we need to differentiate this equation w.r.t x

So, differentiating we get

$Y'=\frac{7}{\sqrt{1-49x^2} }$

This would give the slope of the tangent line at any given point of which x coordinate is known. In the present case it is $x = \sqrt{\frac{1}{7} }$

Then slope would accordingly be

$Y'=\frac{7}{\sqrt{1-49/49} }$

= ∞

For, $x = \sqrt{\frac{1}{7} }$ , $Y = sin^{-1}(7/7)= \pi/2$

Equation of tangent line, in the point slope form, would be $Y-\frac{\pi}{2} =\frac{\pi}{2} (x+\sqrt{\frac{1}{7}})$

4 0

3 years ago

How do I evaluate this using trigonometric substitution?<br><br>∫dx/(81x^2+4)^2

Daniel [21]

Answer:

$\displaystyle \frac{1}{144}arctan(\frac{9x}{2}) + \frac{x}{8(81x^2 + 4)} + C$

General Formulas and Concepts:

<u>Alg I</u>

Terms/Coefficients
Factor
Exponential Rule [Dividing]: $\displaystyle \frac{b^m}{b^n} = b^{m - n}$

<u>Pre-Calc</u>

[Right Triangle Only] Pythagorean Theorem: a² + b² = c²

a is a leg
b is a leg
c is hypotenuse

Trigonometric Ratio: $\displaystyle sec(\theta) = \frac{1}{cos(\theta)}$

Trigonometric Identity: $\displaystyle tan^2\theta + 1 = sec^2\theta$

TI: $\displaystyle sin(2x) = 2sin(x)cos(x)$

TI: $\displaystyle cos^2(\theta) = \frac{cos(2x) + 1}{2}$

<u>Calc</u>

Integration Rule [Reverse Power Rule]: $\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C$

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

IP [Addition/Subtraction]: $\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx$

U-Substitution

U-Trig Substitution: x² + a² → x = atanθ

Step-by-step explanation:

<u>Step 1: Define</u>

$\displaystyle \int {\frac{dx}{(81x^2 + 4)^2}}$

<u>Step 2: Identify Sub Variables Pt.1</u>

Rewrite integral [factor expression]:

$\displaystyle \int {\frac{dx}{[(9x)^2 + 4]^2}}$

Identify u-trig sub:

$\displaystyle x = atan\theta\\9x = 2tan\theta \rightarrow x = \frac{2}{9}tan\theta\\dx = \frac{2}{9}sec^2\theta d\theta$

Later, back-sub θ (integrate w/ respect to <em>x</em>):

$\displaystyle tan\theta = \frac{9x}{2} \rightarrow \theta = arctan(\frac{9x}{2})$

<u>Step 3: Integrate Pt.1</u>

[Int] Sub u-trig variables: $\displaystyle \int {\frac{\frac{2}{9}sec^2\theta}{[(2tan\theta)^2 + 4]^2}} \ d\theta$
[Int] Rewrite [Int Prop - MC]: $\displaystyle \frac{2}{9} \int {\frac{sec^2\theta}{[(2tan\theta)^2 + 4]^2}} \ d\theta$
[Int] Evaluate exponents: $\displaystyle \frac{2}{9} \int {\frac{sec^2\theta}{[4tan^2\theta + 4]^2}} \ d\theta$
[Int] Factor: $\displaystyle \frac{2}{9} \int {\frac{sec^2\theta}{[4(tan^2\theta + 1)]^2}} \ d\theta$
[Int] Rewrite [TI]: $\displaystyle \frac{2}{9} \int {\frac{sec^2\theta}{[4sec^2\theta]^2}} \ d\theta$
[Int] Evaluate exponents: $\displaystyle \frac{2}{9} \int {\frac{sec^2\theta}{16sec^4\theta} \ d\theta$
[Int] Rewrite [Int Prop - MC]: $\displaystyle \frac{1}{72} \int {\frac{sec^2\theta}{sec^4\theta} \ d\theta$
[Int] Divide [ER - D]: $\displaystyle \frac{1}{72} \int {\frac{1}{sec^2\theta} \ d\theta$
[Int] Rewrite [TR]: $\displaystyle \frac{1}{72} \int {cos^2\theta} \ d\theta$
[Int] Rewrite [TI]: $\displaystyle \frac{1}{72} \int {\frac{cos(2\theta) + 1}{2}} \ d\theta$
[Int] Rewrite [Int Prop - MC]: $\displaystyle \frac{1}{144} \int {cos(2\theta) + 1} \ d\theta$
[Int] Rewrite [Int Prop - A/S]: $\displaystyle \frac{1}{144} [\int {cos(2\theta) \ d\theta + \int {1} \ d\theta]$

<u>Step 4: Identify Sub Variables Pt.2</u>

Determine u-sub for trig int:

u = 2θ

du = 2dθ

<u>Step 5: Integrate Pt.2</u>

[Ints] Rewrite [Int Prop - MC]: $\displaystyle \frac{1}{144} [\frac{1}{2} \int {2cos(2\theta) \ d\theta + \int {1 \theta ^0} \ d\theta]$
[Int] U-Sub: $\displaystyle \frac{1}{144} [\frac{1}{2} \int {cos(u) \ du + \int {1 \theta ^0} \ d\theta]$
[Ints] Integrate [Trig/Int Rule - RPR]: $\displaystyle \frac{1}{144} [\frac{1}{2} sin(u) + \theta + C]$
[Expression] Back Sub: $\displaystyle \frac{1}{144} [\frac{1}{2} sin(2 \theta) + arctan(\frac{9x}{2}) + C]$
[Exp] Rewrite [TI]: $\displaystyle \frac{1}{144} [\frac{1}{2}(2sin(\theta)cos(\theta)) + arctan(\frac{9x}{2}) + C]$
[Exp] Multiply: $\displaystyle \frac{1}{144} [sin(\theta)cos(\theta) + arctan(\frac{9x}{2}) + C]$
[Exp] Back Sub: $\displaystyle \frac{1}{144} [sin(arctan(\frac{9x}{2}))cos(arctan(\frac{9x}{2})) + arctan(\frac{9x}{2}) + C]$

<u>Step 6: Triangle</u>

Find trig values:

$\displaystyle tan\theta = \frac{9x}{2}$

$\displaystyle \theta = arctan(\frac{9x}{2})$

tanθ = opposite / adjacent; solve hypotenuse of right triangle, determine trig ratios:

sinθ = opposite / hypotenuse

cosθ = adjacent / hypotenuse

Leg <em>a</em> = 2

Leg <em>b</em> = 9x

Leg <em>c</em> = ?

Sub variables [PT]: $\displaystyle 2^2 + (9x)^2 = c^2$
Evaluate exponents: $\displaystyle 4 + 81x^2 = c^2$
[Equality Property] Square root both sides: $\displaystyle \sqrt{4 + 81x^2} = c$
Rewrite: $c = \sqrt{81x^2 + 4}$

Substitute into trig ratios:

$\displaystyle sin\theta = \frac{9x}{\sqrt{81x^2 + 4}}$

$\displaystyle cos\theta = \frac{2}{\sqrt{81x^2 + 4}}$

<u>Step 7: Integrate Pt.3</u>

[Exp] Sub variables [TR]: $\displaystyle \frac{1}{144} [\frac{9x}{\sqrt{81x^2 + 4}} \cdot \frac{2}{\sqrt{81x^2 + 4}} + arctan(\frac{9x}{2}) + C]$
[Exp] Multiply: $\displaystyle \frac{1}{144} [\frac{18x}{81x^2 + 4} + arctan(\frac{9x}{2}) + C]$
[Exp] Distribute: $\displaystyle \frac{1}{144}arctan(\frac{9x}{2}) + \frac{x}{8(81x^2 + 4)} + C$

3 0

3 years ago