/sci/ - Science & Math

[Anonymous No.5332017]

Currently going through old tests for calculus final monday. Desperately need an A on the exam. I'm stuck on question 4 from test 3. If anyone could be a bro and explain it to me I would be eternally in your debt.

I put all the tests in my dropbox's public folder: https://www.dropbox.com/sh/bi12fhvttfiehoy/1ipF60PJoK

[Blackman No.5332029]

Thanks for the virus, faggot

[Anonymous No.5332035]

>>5332017

Take a screenshot of the question and post it

[Anonymous No.5332038]

test

[Anonymous No.5332043]

>>5332035
Here's the cap.

>>5332029
There's no virus asshole.

[Anonymous No.5332050]

I'll give you a hint

find the formula for the volume of a circular cone, next take the derivative of that formula being sure to use implicit differentiation

that should help get you started

[Anonymous No.5332062]

>>5332043
>imagine an inverted circular cone
>inverted circular cone

implying this is a possible object.
2/10

[Anonymous No.5332073]

>>5332043

Aint got my shit here, let's see if I can do it with paint while watching starcraft

[Anonymous No.5332079]

>>5332062
Never seen a funnel or upside-down traffic cone?

[Anonymous No.5332089]

Trying imgur since 4chan doesnt want to cooperate

http://imgur.com/CZIhG

[Anonymous No.5332093]

>>5332089

oups, I made an error at the integral, anyways use (pi)(f(x))^2

[Anonymous No.5332096]

>>5332093
>>5332089
Thanks brohan

[Anonymous No.5332099]

>>5332096

I'm not 100% sure it's right since paint aint exactly the best program to solve math...

[Anonymous No.5332100]

>>5332096

Just lemme know if you need help on anything else

[Anonymous No.5332105]

I don't feel like solving it for you, but I get the feeling that after you differentiate the volume of a circular cone, you're going to have to use similar cones to be able to substitute values.

That is, the tank is in the shape of an inverted cone. The water inside this cone will also form to the shape of a cone. Even though the water and the tank itself will have different dimensions, they will follow the same ratios, so you can write an expression to substitute undesired values in your equation after you've differentiated.

[Anonymous No.5332106]

<span class="math">r(y)=\frac{y}{2}. V(t)=\int_{0}^{y(t)}r^2(y)\cdot \pi dy \rightarrow \frac{dV}{dt}=\frac{\pi\cdot y^2}{4} \frac{dy}{dt}[/spoiler] Use the chain rule.

[Anonymous No.5332114]

>>5332100
This one is being tricky too.

[Anonymous No.5332125]

>>5332105

see >>5332089

[Anonymous No.5332165]

>>5332114
OP here.

I put this on the actual test but I got -2 out of 5 for some reason:

C = 2pi(r) SA = 4pi(r^2)
50 = 2pi(r) (SA)' = 8pi(r)
solve for r
=> r = 25/pi

Max.Error = SA'(r) x (dr)
=> 8pi(25/pi) x (0.4)
=> 8(200)(0.4)
=> 200(0.4)
=> 80

Can anyone point out where I went wrong?

My buddy says that I shouldn't have taken the derivative of the SA. I think he may be correct but I'd like 2nd opinion

[Anonymous No.5332171]

>>5332165
Pic of my actual test answer

[Anonymous No.5332177]

Find the equation of the line through the point (3,5) that cuts off the least area from the first quadrant.

What do?

Using the origin the line comes out as 5/3(x). Do I try to make the line as horizontal as possible?

[Anonymous No.5332198]

This now the official /sci/ finals study thread. Post all related questions.

GET ON IT /SCI/

[Anonymous No.5332201]

>>5332198
Taking Math 205 (Discrete). Can someone explain all this path/cycle shit?

[Anonymous No.5332204]

>>5332177
Please help. Any way I see the line it cuts the 1st quadrant. How do I minimize the cut?

[Anonymous No.5332220]

Answer check please.

anti-derivative of sqt(x-1) is (2/3)(x-1)^3/2?

[Anonymous No.5332265]

>>5332220
You are correct.

The power rule for antiderivatives is basically backward. Add one to the power, then divide by the resulting power.

Or you can look at it this way, which is how I like to think about it: what number or function, if derived, would equal the number you are trying to take the antiderivative of?

[Anonymous No.5332272]

>>5332204
describe the line depending on one parameter (x_0, y_0 or its derivative)
Compute Area(parameter)
minimize (set derivative=0 to get min_parameter, compute Area(min_parameter)

[Anonymous No.5332283]

>>5332165
>>5332171

Can anyone look over the above and tell me why I was deducted two points?

[Anonymous No.5332616]

bumpity bump bump

[Anonymous No.5332672]

>>5332201
A path is just a sequence of vertices that you visit. It becomes a cycle if the start and end vertex are the same.

A simple path means you don't repeat any edges or vertices. A simple cycle means you don't repeat any edges or vertices, and start and end at the same place.

[Anonymous No.5332781]

>>5332272
could you draw lines any worse holy shit

[Anonymous No.5333131]

You were deducted points because you wrote the equation for a line, not for the curve that you drew.

[Anonymous No.5333145]

>>5332283
Maybe because you didn't do the last step, 200 x 0.4 = 80 duh

[Anonymous No.5333148]

I am breaking my balls over this problem....

Show that int tanx dx is ln |secx| + C

No matter what I do with int tanx dx I get -ln|secx|+C

[Anonymous No.5333198]

>>5333148
The anti derivative of tanx is -ln(cosx) + C

[Anonymous No.5333212]

>>5333198
The problem is to show that int of tanx dx is ln|secx|+C

[Anonymous No.5333248]

>>5333148
<span class="math"> \int\tan xdx=\int\frac{\sin x}{\cos x}dx [/spoiler]
Now let <span class="math"> u=\cos x [/spoiler]
Then <span class="math">du=-\sin xdx [/spoiler]
Thus <span class="math"> \int\tan xdx=\int\frac{\sin x}{\cos x}dx = -\int\frac{du}{u} = \log|\cos(x)| + C [/spoiler]
That should be your answer

[Anonymous No.5333254]

>>5333248
sorry, that's <span class="math"> -log|\cos(x)| +C [/spoiler]
I dropped a negative there

[Anonymous No.5333353]

>>5333248
I am sorry for being retarded but I don't see how you get ln|cosx| from sinxdx/cosx

or how is -ln|cosx|+C = ln|secx|+C

[Anonymous No.5333364]

>>5333353
nvm I figured out sinxdx/cosx

But still dont see how -log|cosx|+C is suppose to be the same as ln|secx|+C

[Anonymous No.5333372]

>>5333353
>>5333364
Using substitution:

from >>5333248

1/u du , which is the anti-deriavtive of 1/u which is ln|u| so its 1/cosx which is same as secx.

[Anonymous No.5333378]

>>5333364
Review your logarithmic properties.

[Anonymous No.5333394]

>>5333364
refer to this >>5333378
But it's because <span class="math"> -\log(a)=\log(\frac{1}{a}) [/spoiler]
So then <span class="math"> -\log|\cos(x)|=\log|\frac{1}{\cos(x)}|=\log|\sec(x)| [\math][/spoiler]

[Anonymous No.5333408]

>>5333394
sorry. Fucked up the TeX
In this equation you have<span class="math"> -\log|\cos(x)|=\log|\frac{1}{\cos(x)}|=\log|\sec(x)| [/spoiler]

[Anonymous No.5333410]

>>5333378
ffffffffffffffffffffffffffffffff-

this is why I am so bad at math....I always forget shit like this and don't recognize them....