The Official Math/Physics/Whatever Homework questions thread - Page 278

I think i see the problem. If it was y = arctan((8x +1)^2) I would have another step in the chain rule, but because the "x^2" in y = arctan(8x +1) (the x^2 from the formula for d/dx(arctan) ) is already there, so you only continue doing the chainrule on whatever gets pasted there as the x.

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03-16-2014 , 10:16 PM

#6928

BruceZ

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Yeah, arctan((8x+1)^2) has 3 links in the chain. The +1 doesn't require an extra link.

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03-16-2014 , 11:45 PM

#6929

Ryanb9

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Quote:

Originally Posted by BruceZ

Yeah, arctan((8x+1)^2) has 3 links in the chain. The +1 doesn't require an extra link.

Could you explain it real quick? I dont see why sometimes you "bring down the n and make it n-1" and sometimes you dont. When do you and when dont you?

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03-17-2014 , 12:05 AM

#6930

BruceZ

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y = arctan((8x+1)^2)

u = (8x+1)^2

v = 8x+1

y = arctan(u)

u = v^2

dy/dx = dy/du * du/dv * dv/dx

= 1 / (1+u^2) * 2v * 8

= 1 / [1 + (8x+1)^4] * 2*(8x+1) * 8

= 16(8x+1) / [1 + (8x+1)]^4.

If you get confused, you can spell it out like that, but you just immediately do

1 / [1 + (8x+1)^4] * 2*(8x+1) * 8

= 16(8x+1) / [1 + (8x+1)^4].

You can always reduce the exponent by 1, but if it's 1 to begin with, it just becomes 0 and the term goes away.

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03-17-2014 , 02:12 PM

#6931

non-self-weighter

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In general, is the inverse of a CDF function of a particular distribution equal to the CDF of the inverse distribution? It's seems intuitive, but I can't be sure. I'm particularly interested in the normal and inverse normal CDFs, which of course are analytically undefined.

This is not a homework question or anything. Just wondering because of an article I'm reading which uses both of these functions.

Thanks!

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03-17-2014 , 03:33 PM

#6932

BruceZ

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The inverse of the CDF is the quantile function which is defined on [0,1], and for the normal distribution takes on values from - infinity to infinity. That can't be the CDF of anything since a CDF takes on values from 0 to 1. The normal pdf doesn't even have an inverse as there are 2 values that can produce every value for the pdf except for the peak. It's

f(x) = +/- sqrt(-2*ln(x)) for x in (0,1/sqrt(2*pi)]

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03-17-2014 , 04:16 PM

#6933

Ryanb9

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Quote:

Originally Posted by BruceZ

There's 1 thing you left out from this post that I really wanted you to try to explain to me but it is a hard question to ask (or for you to understand what it is that im asking). I guess if i do it Socratically it will be easier to get my point across so bare with me here...

assume u = (8x^2 + x)^2

1) what is d/dx[u]

2) what is d/dx [arctan(u)]

3) if I know that arctan(x) = [ 1 / (1 + x^2) ], why am I not doing the same thing with this bolded power-of-two here that I did with the other bolded power-of-two in #1?

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03-17-2014 , 05:32 PM

#6934

BruceZ

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Quote:

Originally Posted by Ryanb9

2(8x^2 + x) * (16x + 1)

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2) what is d/dx [arctan(u)]

1 / (1 + u^2) * du/dx

Substitute for u and du/dx from 1.

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3) if I know that arctan(x) = [ 1 / (1 + x^2) ]

That's the derivative of arctan(x).

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why am I not doing the same thing with this bolded power-of-two here that I did with the other bolded power-of-two in #1?

Because one has nothing to do with the other. In 1 we differentiated u. We're not differentiating arctan(x).

Are you going to take integral calculus?

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03-17-2014 , 05:52 PM

#6935

Ryanb9

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Quote:

Originally Posted by BruceZ

2(8x^2 + x) * (16x + 1)

1 / (1 + u^2) * du/dx

Substitute for u and du/dx from 1.

That's the derivative of arctan(x).

Because one has nothing to do with the other. In 1 we differentiated u. We're not differentiating arctan(x).

Are you going to take integral calculus?

for 2 i meant to say whats d/dx of arctan(u) i think. And idk whats the next math i have to take, probably that yeah.

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03-17-2014 , 05:56 PM

#6936

BruceZ

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Quote:

Originally Posted by Ryanb9

for 2 i meant to say whats d/dx of arctan(u) i think. And idk whats the next math i have to take, probably that yeah.

I answered d/dx of arctan(u). That was d/du of arctan(u) times du/dx.

Integral calculus usually follows differential calculus. This is a joke compared to that.

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03-17-2014 , 06:26 PM

#6937

Ryanb9

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Quote:

Originally Posted by BruceZ

I answered d/dx of arctan(u). That was d/du of arctan(u) times du/dx.

Integral calculus usually follows differential calculus. This is a joke compared to that.

In the first one, we used the power rule b/c we had (x)^2
In the second one, when u plug in u on the bottom for x, u get (x)^2, and u dont use the power rule.
In both im assuming you have to use the chain rule, do you not have to for d/dx of arctan or arcsin etc?

About integral, yeah **** this **** next math class im hiring a personal tutor. I can work for 1 hr, ship that money to a tutor, save 4 hours of not understanding wtfs going on, and profit with 3hrs of ANYTHING but math.

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03-17-2014 , 06:43 PM

#6938

Ryanb9

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When I do the quotient rule, do I still have to do the chain rule after that or does the quotient rule take that into consideration lol. If I had to guess I'd say I have to do the chain rule in this problem after the quotient rule and just keep multiplying

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03-17-2014 , 06:48 PM

#6939

BruceZ

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Quote:

Originally Posted by Ryanb9

You don't get x^2, you get u^2. You use the chain rule to compute

d[arctan(u)]/du * du/dx

= 1/ (1 + u^2) * du/dx

There's no power rule involved with the u^2.

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03-17-2014 , 08:36 PM

#6940

coon74

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Quote:

Originally Posted by Ryanb9

Generally, you'd be still supposed to apply both the chain rule (twice here - for the logarithm and the root) and the quotient rule after it (to find the derivative of the function under the root).

But here the expression simplifies into 0.5 ln(x^2-7) - 0.5 ln (x^2+7), so the quotient rule is not needed.

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03-18-2014 , 01:50 AM

#6941

Ryanb9

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I was given u(5) = 6, u'(5) = -3, v(5) = 9, v'(5) = 6
and asked to find d/dx(8uv)

If the 8 wasnt there i'd be fine and just do the product rule, but what do I do with this 8 here?

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03-18-2014 , 01:57 AM

#6942

Cangurino

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8 is just a constant... which you can pull out of the derivative.

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03-18-2014 , 08:28 AM

#6943

Kvitlekh

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Quote:

Originally Posted by Kvitlekh

Coughing forces the trachea (windpipe) to contract, which affects the velocity v of the air through the trachea. The velocity of the air during coughing is
v = k(R − r)r², 0 ≤ r < R
where k is a constant, R is the normal radius of the trachea, and r is the radius during coughing. What radius r will produce the maximum air velocity?

I imagine that they want to know for what value of r the derivative is 0, but how to find the derivative with more than one variable in the equation for V?

I need a hint. I've been thinking about this for too long and I'm not seeing it. Should r be in terms of the other constants or is there an absolute number?

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03-18-2014 , 08:35 AM

#6944

Mr.mmmKay

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The solution will be expressed in terms of the constants

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03-18-2014 , 09:11 AM

#6945

BruceZ

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It will actually be in terms of just 1 of the other constants. The other cancels out.

That assumes k > 0. If k < 0, the answer is r=0. If k=0, there is no maximum.

Last edited by BruceZ; 03-18-2014 at 09:23 AM.

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03-18-2014 , 10:44 AM

#6946

Kvitlekh

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Quote:

Originally Posted by BruceZ

It will actually be in terms of just 1 of the other constants. The other cancels out.

That assumes k > 0. If k < 0, the answer is r=0. If k=0, there is no maximum.

I must be blind. Here's what I get:

v' = k[2r(R-r)-r²]

= k(2Rr-3r²)

I can't see how to get rid of one of the constants.

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03-18-2014 , 10:50 AM

#6947

BruceZ

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Set that equal to zero and solve for r.

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03-18-2014 , 11:06 AM

#6948

Kvitlekh

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r=2/3R is what I get.

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03-18-2014 , 11:14 AM

#6949

BruceZ

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Right, or r = 0. When k > 0, 2/3 * R is a maximum, and r=0 is a minimum. When k < 0, the maximum and minimum reverse. You can show that by taking the second derivative. When the second derivative is negative, you have a maximum, and when it's positive, you have a minimum. When k=0, v is always 0, so there is no maximum, or technically, all points are a maximum.