Can You Solve These Challenging Calculus Problems?

[aetatis]

Hey I have a course project where I have to find/develop 15 slightly difficult calculus problems and solve them. Any suggestions?
(I'm not asking for proofs but if you want to share those as well I really really don't object )

Oh, and this is first/second year calculus not multivariable.

 

[Cyrus]

Hey I have a course project where I have to find/develop 15 slightly difficult calculus problems and solve them. Any suggestions?
(I'm not asking for proofs but if you want to share those as well I really really don't object )

Oh, and this is first/second year calculus not multivariable.

If I were in your shoes I would pick 15 problems that involve the trig identities. Those were the hardest for me, because you have to remember the formulas which you can use as tricks to solve the problems. If your taking calc 2, Id do problems with trig substitution, those are kind of hard and will help you out on tests and stuff.

 

[berkeman]

Here's kind of a fun site from Harvey Mudd College with tutorials and quizzes about calculus. You can go into the tutorials for 1st year calculus subjects, go to the harder subjects, and at the end of some of the tutorial pages, they'll have an "Explore" page where you can play with graphing or other stuff. Check out the Taylor Series subject and its Exploration page -- very cool.

http://www.math.hmc.edu/calculus/

 

[mattmns]

$$\int \sqrt{tan\theta} d\theta$$

I remember this problem from calc 2

 

[whozum]

Prove that

$$\lim_{x\rightarrow 0} \frac{\sin x}{x} = 1$$

 

[The Guru Kid]

Prove that

$$\lim_{x\rightarrow 0} \frac{\sin x}{x} = 1$$

Actually that wouldn't be too hard.

Ignoring the limit and rearranging the equation gives you sinx = x. That's only true when x = 0.

The only reason the limit exists in the original equation is because you cannot divide by zero.

p.s. first post, hope I'm not wrong

 

[amcavoy]

$$\lim_{x\rightarrow 0} \frac{\sin x}{x} = 1$$

To prove this I would set up a unit circle and find the area of the triangle with legs sin(x) and cos(x), area of the sector of angle x, and area of triangle with legs 1 and tan(x). Now rearrange and set up and equation with these which will lead you to the fact that $$1\leq \lim_{x\to 0}\frac{\sin x}{x}\leq 1$$

 

[steven187]

hello there

well i remember this from high school, but i know that some of the people i go to uni with, were not able to do it
$$\int \sec\theta d\theta$$
there would be two ways of doing it, a long and short way, i come to look at this now its pretty simple anyway good luck with solving

by the way it would be a great idea to list some topics that you have went though in your calculus class?

Steven

 

[fourier jr]

$$\int_{0}^{\pi/2} \frac{dx}{1+(tan(x))^{\sqrt2}}$$

(the answer is pi/4, & it doesn't matter what number goes where the sqrt 2 is)

 

[shmoe]

Ignoring the limit and rearranging the equation gives you sinx = x. That's only true when x = 0.

You can't just ignore the limit or declare it to be 1 since sin(x)=x at x=0. Notice sin(x)=x^(1/3) only at x=0 as well, but $$\lim_{x\rightarrow 0} \frac{\sin x}{x^{1/3}} \neq 1$$. You have to treat indeterminant limits like this with care.

 

[The Guru Kid]

You can't just ignore the limit or declare it to be 1 since sin(x)=x at x=0. Notice sin(x)=x^(1/3) only at x=0 as well, but $$\lim_{x\rightarrow 0} \frac{\sin x}{x^{1/3}} \neq 1$$. You have to treat indeterminant limits like this with care.

Well i guess you have to understand why we took the limit in the first place. suppose we just had the equation w/o the limit. Then solving for x you get 0. But in current mathematics, you cannot do that since it gives you division by 0 (in the times befor Sir Wallis, mathematicians like Fermat did just that).

The reason we're saying sin(x)=x and not anything else is because we don't want to change the original equation.

In your equation, the limit would be x^(-2/3) which is solved in exactly the same manner.

 

[wisredz]

The limit involving sinx/x is prooved by drawing a unit circle. What you told is wrong because what you do is intersecting the x=y line and y=sinx curve. sinx/x has a totally different curve.
Using the areas in the unit circle, it can be proved.

Not so difficult but try prooving that as x approaches 0, sinkx/x=k.

 

[The Guru Kid]

I'm sorry i still don't see how my method is wrong. If you want to use unit circles, that's fine, but my method works unless it's one of those math coincidences.

If someone could explain why it's wrong, i would be grateful.

 

[whozum]

I think shmoe did a good job of that. I guess the better problem as wisredz suggested was find

$$\lim_{x\rightarrow 0} \frac{\sin kx}{x}$$

 

[whozum]

Another good one would be using some tabular integration to find a reduction formula for

$$\int \sin ^n x dx$$ for even integers 'n'.

 

[shmoe]

In your equation, the limit would be x^(-2/3) which is solved in exactly the same manner.

Are you saying $$\lim_{x\rightarrow 0} \frac{\sin x}{x^{1/3}}=\frac{1}{x^{2/3}}$$? This wouldn't make any sense at all, could you please clarify?

By your method if we ignore the limit in the 'equation'

$$\lim_{x\rightarrow 0} \frac{\sin x}{x^{1/3}}=1$$

and rearrange we get $$\sin x=x^{1/3}$$ which is true only when x=0 so this limit is correct (of course it's actually wrong).

Please explain to me how this is different from what you've proposed for sin(x)/x.

 

[fourier jr]

here's another one:

maximize $$f(x) = \frac{1}{2^{x}} + \frac{1}{2^{1/x}}$$ for x>0.

 

[aetatis]

wow

wow what a great response thanks so much guys

 

[aetatis]

oh and some of the topics we covered in that class are pretty basic:
improper integrals
parametric eq
taylor mclauren series
length of a curve/along a path
but I've also done multivariable calc.

 

[riddick]

$$\int_{0}^{\pi/2} \frac{dx}{1+(tan(x))^{\sqrt2}}$$

(the answer is pi/4, & it doesn't matter what number goes where the sqrt 2 is)

hello i was trying to solve this problem but i really couldn't could you please show me step by step how to solve it i would really appreciate that, thank you very much..

 

[fourier jr]

hello i was trying to solve this problem but i really couldn't could you please show me step by step how to solve it i would really appreciate that, thank you very much..

- multiply the numerator & denominator of the integrand by cos(x)^sqrt(2)
- set y=(pi/2)-x
- substitute y for x
- but sin((pi/2)-x) = cos(x)
- call the original integral I, and consider I+I, where one has sin in the numerator & the other has cos, & find that the integrand =1
- integrate & get 2I = pi/2 => I=pi/4

 

[LCKurtz]

Here's a tougher problem from Calc III:

Show that an airplane flying a closed course at constant airspeed in the presence of a constant wind (constant velocity and direction) must take longer than if there were no wind. The solution is at math.asu.edu/~kurtz if you have to look.

 

[coki2000]

$$\int{x^x}dx$$ do someone know to solve this integral??

 

[fourier jr]

i think you need to rewrite the integrand as exp(xlogx) but i forget the rest

edit: no that's for differentiating it, never mind

 

[lurflurf]

$$\int{x^x}dx$$ do someone know to solve this integral??

That is an integral not expressible in elementary functions or standard special functions. If the integral on (0,1) in terms of an infinite series is enough we have the sophomore dream.

 

[elemental09]

Can't sin(x)/x -> 1 as x->0 can be shown using L'Hopital's rule once?

 

[lurflurf]

Can't sin(x)/x -> 1 as x->0 can be shown using L'Hopital's rule once?

Yes, but the strength of L'Hopital's rule is not needed since by the definition of the derivative lim sin(x)/x=lim [sin(0+h)-sin(0)]/h=sin'(0). Normally one wants to show lim sin(x)/x=1 before one has L'Hopital's rule available. In either case one must have defined sine in such a way that sin'(0)=1 may be found without the use of lim sin(x)/x to avoid circular reasoning.

 

[l'Hôpital]

Yes, but the strength of L'Hopital's rule is not needed since by the definition of the derivative lim sin(x)/x=lim [sin(0+h)-sin(0)]/h=sin'(0). Normally one wants to show lim sin(x)/x=1 before one has L'Hopital's rule available. In either case one must have defined sine in such a way that sin'(0)=1 may be found without the use of lim sin(x)/x to avoid circular reasoning.

Well, all you really need is some geometry to show that cos x < sin x/x < sec x for 0 < x < pi/2 , then squeeze that limit.

EDIT: Oops, wrong inequality. Should always know to check my work.

 

[fourier jr]

how about this one. find the two lines tangent to y=4x-x^2 that go through (2,5)

 

[weagle2008]

Actually that wouldn't be too hard.

Ignoring the limit and rearranging the equation gives you sinx = x. That's only true when x = 0.

The only reason the limit exists in the original equation is because you cannot divide by zero.

p.s. first post, hope I'm not wrong

Actually all you need is your trig identities:

sinx / x = cosx ; and since cos (0) = 1 ; sin (0) / 0 = 1

 

[l'Hôpital]

Actually all you need is your trig identities:

sinx / x = cosx ; and since cos (0) = 1 ; sin (0) / 0 = 1

What? Take x = pi.

$$\frac{sin \pi}{\pi} = 0$$

But...

$$cos \pi = -1$$

 

[Redbelly98]

Let t = time when this question was asked.

Find t.

 

[weagle2008]

What? Take x = pi.

$$\frac{sin \pi}{\pi} = 0$$

But...

$$cos \pi = -1$$

The question isn't as x approaches Pi it is as x approaches 0. As anyone knows Pi = 180 degrees. 2*Pi = 360 degrees which is = 0 in trig. Thus cos (Pi) = -1, but cos (0) = 1.

 

[Anonymous217]

The question isn't as x approaches Pi it is as x approaches 0. As anyone knows Pi = 180 degrees. 2*Pi = 360 degrees which is = 0 in trig. Thus cos (Pi) = -1, but cos (0) = 1.

But you stated that (sinx)/x = cosx, which means that this is true for all x (which is wrong). This is what he was talking about.

sinx / x = cosx ; and since cos (0) = 1 ; sin (0) / 0 = 1

I would conjecture that you meant $$\lim_{x\rightarrow 0}\frac{sinx}{x} = \lim_{x\rightarrow 0} cosx$$. However, I'm not exactly sure how you got to that without L'H.