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Calculus Kinematics

Van is travelling constant speed of 13ms (46.8kmh)

It passes a sign changing the speed limit to 22.22ms (80kmh)

The acceleration a ms of car t seconds after passing the sign is given a=t/8 + 3/4

Find the distance the van has travelled since passing the sign when it first reaches a speed of 80kmh

Thanks :)

http://askmemathsquestions.tumblr.com/post/33567071382/sorry-for-late-reply-was-out-of-town

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so it goes: can someone please explain to me how you find a definite integral. i...

janedoezxzxzxzx:

can someone please explain to me how you find a definite integral. i can solve most integrals. but when you ask me to find the limit from 0 to x i get so confused.

so i solved the integral of cos^3(1/2x) * sin^5(1/2x) dx and got

1/3 sin ^6(1/2x) - 1/4 sin^8(1/2x) + C

(i thought it should be…

I will give you a general rule

suppose you have to integrate a definite integral over limits from 0 to x

the function that needs to be integrated is f(x) suppose, and when you integrate it, you get g(x) , not g(x) + c because it is a definite integral and when you put in the limits, the constant c is taken care of

so when you get g(x) as your answer,

just evaluate g(x) - g(0)

this is your final answer for the definite integral evaluation of f(x) from 0 to x

if you still have problem, message me, i will sort it out

(Source: janedoez)

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1 note &

Calculus!

haboltingin:

We had this topic on our calculus subject, and we will hate it so much! The topic is Implicit Differentiation. It’s me the ‘headache’.

Here are some of the examples that our teacher gave us few hours ago.

1. If x^2-8xy+16y^2=0, find dy/dx

2. x^2/3+y^2/3=7

3. (square root) xy^2+yx^2=0

We really don’t know what to do! GAH!!

it is really very easy, all you need to know is how to differentiate, you will have no problems in doing implicit differentiation

if you run into some problems, ask me, I will help you out

(Source: jmnendico)

Filed under math implicit differentiation differentiation calculus maths help math help


3 notes &

Gah, so confused

alphamisery:

The problem says “show why the function does not have a derivative at x=1”, but I just did it

AND IT DOES HAVE A DERIVATIVE AT X=1.

lim x → 1  (x^3 - (1)^3)/(x - 1)

lim x → 1 (x - 1)(x^2 + x + 1)/(x - 1)

             (1)^2 + (1) + 1 = 3

and then

lim x → 1 (3x - 3(1))/(x-1)

lim x → 1 3(x - 1)/(x - 1)

lim x → 1 = 3

MY LIFE IS SO HARD.

And the worst part is they’re both super simple, too. I know I have the math correct. Maybe the book is just trying to fuck with me.

nopes, you did not noticed that (x-1) is in denominator, so basically, it is an example of a function which is asymptotic at x = 1, and when ever the function has an asymptote at point x = p, the function is not differentiable at x = p

you cancelled out (x-1) from numerator and denominator so you got both limits existent at x = 1

(Source: suddenlyslug)

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3 notes &

Who is just so fantastic at Calc that they want to help me?

burning-atmosphere:

I missed two days lessons and I’m behind. I’m sure this is a crazy easy problem, but every Google search I use keeps using L’Hospital’s rule and we haven’t learned that so I can’t very well show up with my paper using that. So help!

lim (x->0) = (sin2x)/x 

HELP ME PLEASE.

if you don’t want to use L’ Hospital’s rule

there is an easy way out too

in fact there are two

1st

sin(2x)/x = 2. sin(2x)/2x.just multiplying and dividing by 2 

let 2x = u, since x ->0, 2x ->0  so u-> 0

so your question becomes, limit u -> 0 (2 * sinu/u)

and it  limit u -> 0 ( sinu/u)is a standard result  = 1

so your answer  = 2

2nd method

just expand sin2x 

http://mathworld.wolfram.com/SeriesExpansion.html

the formula for expansion of sinx is given here( 24th formula), put 2x in place of x in the series expansion , and then divide by x and put x = 0, you shall have your answer as 2

(via burning-atmosphere-deactivated2)

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1 note &

y = -x

Does this formula describe y as a function of x?

yes, absolutely.

x is the independent variable and y is the dependent variable.

changes in x makes changes in y, so it is a function of x here

I will give you a tip

If you want to know whether a given expression in x and y is a function of x or not

step 1:


write the function in form of function y = f(x)

step 2


if possible, draw/visualize  the graph of f(x)

step 3


if any straight line parallel to y axis cuts the graph more than once, then it is not a function, else it is a function

why is it so?


a one to many relation is a function, i.e. a line parallel to x axis can cut the graph at multiple points, in fact, you know it also, in a quadratic equation, y= ax^2 + bx + c

the x axis cuts this graph twice, and the point of intersection of x axis with this graph is what we call are the roots of quadratic equations

so y = ax^2 + bx + c is a function

but, 

x = ay^2 + by + c is not a function

because it is a horizontal parabola, and a line parallel to y axis cuts its graph more than once( heck, even the y axis cuts this graph more than once if this has two real roots)

simply because, a many to one relation is not a function.

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1 note &

girlbee:

IS IT A NOT A FUNCTION IF THE DOMAIN REPEATS OR IF THE RANGE REPEATS PLEASE HELP

not exactly sure what you man by domain repeats itself or range repeats itself

if you are referring to x and y values,

two or more values of x can have a single value of y

but if you are getting multiple values of y for single value of x , it is not a function

in simple words,

a many to one relation is a function

but, a one to many relation is not

(Source: butter8)

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327 notes &

leelubell:

I was looking something up for my math homework, and this gif came up and I have no clue why this makes me so happy. It just does. I mean look at that little line it just looks so happy like it’s on a little f’(x) roller coaster.

love it

leelubell:

I was looking something up for my math homework, and this gif came up and I have no clue why this makes me so happy. It just does. I mean look at that little line it just looks so happy like it’s on a little f’(x) roller coaster.

love it

Filed under math calculus calc


1 note &

Lim x -> 0 sinx(1-cosx)/2x^2

ohmyfrkncharlie:

Okay, so like…

I know sinx/x = 1 and 1-cosx/x = 0

But I have no clue how to manipulate this problem to where it equals 0.

Halp.

just a quick heads up

it is of the form 0/0

and you don’t want to do it with L hospital rule

so here is a long and interesting way to manipulate this to get what you want

 taylors series expansion for sinx you can use that 
sin(x) = x - x^3/3! + x^5/5! - x^7/7! +…

same way cos x = 1- x^2/2!+…….

substitute and then divide by 2x^2

the expression you will obtain will be like x(1 +all the other garbage)

put x = 0

you shall have your answer


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