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4 years ago

Find the following product, and write the product in rectangular form, using exact values.

Mathematics

1 answer:

TEA [102]4 years ago

3 0

Writing each complex number in exponential form makes this very easy. Recall Euler's formula:

$e^{i\theta}=\cos\theta+i\sin\theta$

Then

$8(\cos45^\circ+i\sin45^\circ)=8e^{i\pi/4}$

(since 45º = π/4 rad)

$7(\cos165^\circ+i\sin165^\circ)=7e^{i(11\pi/12)}$

(since 165º = 11π/12 rad)

The product is

$56e^{i(\pi/4+11\pi/12)}=56e^{i(7\pi/6)}$

and in Cartesian form this is

$56(\cos210^\circ+i\sin210^\circ)=56\left(-\dfrac{\sqrt3}2-i\dfrac12\right)=\boxed{-28\sqrt3-28i}$

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(d). Use an appropriate technique to find the derivative of the following functions:

natima [27]

(i) I would first suggest writing this function as a product of the functions,

$\displaystyle y = fgh = (4+3x^2)^{1/2} (x^2+1)^{-1/3} \pi^x$

then apply the product rule. Hopefully it's clear which function each of f, g, and h refer to.

We then have, using the power and chain rules,

$\displaystyle \frac{df}{dx} = \frac12 (4+3x^2)^{-1/2} \cdot 6x = \frac{3x}{(4+3x^2)^{1/2}}$

$\displaystyle \frac{dg}{dx} = -\frac13 (x^2+1)^{-4/3} \cdot 2x = -\frac{2x}{3(x^2+1)^{4/3}}$

For the third function, we first rewrite in terms of the logarithmic and the exponential functions,

$h = \pi^x = e^{\ln(\pi^x)} = e^{\ln(\pi)x}$

Then by the chain rule,

$\displaystyle \frac{dh}{dx} = e^{\ln(\pi)x} \cdot \ln(\pi) = \ln(\pi) \pi^x$

By the product rule, we have

$\displaystyle \frac{dy}{dx} = \frac{df}{dx}gh + f\frac{dg}{dx}h + fg\frac{dh}{dx}$

$\displaystyle \frac{dy}{dx} = \frac{3x}{(4+3x^2)^{1/2}} (x^2+1)^{-1/3} \pi^x - (4+3x^2)^{1/2} \frac{2x}{3(x^2+1)^{4/3}} \pi^x + (4+3x^2)^{1/2} (x^2+1)^{-1/3} \ln(\pi) \pi^x$

$\displaystyle \frac{dy}{dx} = \frac{3x}{(4+3x^2)^{1/2}} \frac{1}{(x^2+1)^{1/3}} \pi^x - (4+3x^2)^{1/2} \frac{2x}{3(x^2+1)^{4/3}} \pi^x + (4+3x^2)^{1/2} \frac{1}{ (x^2+1)^{1/3}} \ln(\pi) \pi^x$

$\displaystyle \frac{dy}{dx} = \boxed{\frac{\pi^x}{(4+3x^2)^{1/2} (x^2+1)^{1/3}} \left( 3x - \frac{2x(4+3x^2)}{3(x^2+1)} + (4+3x^2)\ln(\pi)\right)}$

You could simplify this further if you like.

In Mathematica, you can confirm this by running

D[(4+3x^2)^(1/2) (x^2+1)^(-1/3) Pi^x, x]

The immediate result likely won't match up with what we found earlier, so you could try getting a result that more closely resembles it by following up with Simplify or FullSimplify, as in

FullSimplify[%]

(% refers to the last output)

If it still doesn't match, you can try running

Reduce[<our result> == %, {}]

and if our answer is indeed correct, this will return True. (I don't have access to M at the moment, so I can't check for myself.)

(ii) Given

$\displaystyle \frac{xy^3}{1+\sec(y)} = e^{xy}$

differentiating both sides with respect to x by the quotient and chain rules, taking y = y(x), gives

$\displaystyle \frac{(1+\sec(y))\left(y^3+3xy^2 \frac{dy}{dx}\right) - xy^3\sec(y)\tan(y) \frac{dy}{dx}}{(1+\sec(y))^2} = e^{xy} \left(y + x\frac{dy}{dx}\right)$

$\displaystyle \frac{y^3(1+\sec(y)) + 3xy^2(1+\sec(y)) \frac{dy}{dx} - xy^3\sec(y)\tan(y) \frac{dy}{dx}}{(1+\sec(y))^2} = ye^{xy} + xe^{xy}\frac{dy}{dx}$

$\displaystyle \frac{y^3}{1+\sec(y)} + \frac{3xy^2}{1+\sec(y)} \frac{dy}{dx} - \frac{xy^3\sec(y)\tan(y)}{(1+\sec(y))^2} \frac{dy}{dx} = ye^{xy} + xe^{xy}\frac{dy}{dx}$

$\displaystyle \left(\frac{3xy^2}{1+\sec(y)} - \frac{xy^3\sec(y)\tan(y)}{(1+\sec(y))^2} - xe^{xy}\right) \frac{dy}{dx}= ye^{xy} - \frac{y^3}{1+\sec(y)}$

$\displaystyle \frac{dy}{dx}= \frac{ye^{xy} - \frac{y^3}{1+\sec(y)}}{\frac{3xy^2}{1+\sec(y)} - \frac{xy^3\sec(y)\tan(y)}{(1+\sec(y))^2} - xe^{xy}}$

which could be simplified further if you wish.

In M, off the top of my head I would suggest verifying this solution by

Solve[D[x*y[x]^3/(1 + Sec[y[x]]) == E^(x*y[x]), x], y'[x]]

but I'm not entirely sure that will work. If you're using version 12 or older (you can check by running $Version), you can use a ResourceFunction,

ResourceFunction["ImplicitD"][<our equation>, x]

but I'm not completely confident that I have the right syntax, so you might want to consult the documentation.

3 0

3 years ago

This is confusinggggg

MakcuM [25]

In order to find the area of a circle. You need to find the radius of the circle, which is the diameter of the circle, halved. In your question all you do is half 62 which is 31, then do Area=pie x radius squared which is 3019.07 but then to give in 1DP the answer is
3019.1 hope this helps!! Mark me the brainiest!!

ANSWER- 3019.07

8 0

3 years ago

Flora made 7 withdrawals of $75 each from her bank account. What was the overall change in her account ?

qaws [65]

525 dollars 75x5= 525

6 0

3 years ago

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Can someone plz help me quickly !!!!!!!!!!!

Andrews [41]

Answer:

a. 80

b. 34

Step-by-step explanation:

45 percent of 80 is 36.

85 percent of 40 is 34.

You can find these online, through a Percentages Calculator.

<u><em>Have a good day and a good rest of 2021. :) Please feel free to give me Brainliest if you feel this helped. It's up to you though :)</em></u>

6 0

3 years ago

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In the system of equations below, which variable would it be easiest to solve for?

Maru [420]

Answer:

<h3> The first: The easiest to solve for is x in the first equation.</h3>

Step-by-step explanation:

To get x from first equation you just need to subtract 4y from both sides.

The rest of variables needs two operations to get variable - subtracting and then dividing.

7 0

3 years ago

Read 2 more answers