Recent content by ForceBoy

I will keep this in mind as I learn more. While it may not be immediately relevant, it's bound to be in the future.

Okay. Then my distance to the center of the planet will be some ##q## and my ##g## should be ##g = -\frac{Gm_{planet}}{(h+q)^{2}} ## Then, would my final equation come from the following? ## h(t) =\int\int\frac{Gm_{planet}}{(h(t)+q)^{2}} dt^{2} ## And if this is so, how could I solve...

The typical equation for the height of a projectile on Earth after ##t ## seconds is ## h = -4.9t^{2}+vt+c## where ##v## is the velocity of the projectile and ##c## the initial height. This is nice and all but what happens if the height is very large? The leading coefficient of the equation...

Oh, I hadn't caught that! Thanks alot. Here is the correct version: ## Q(t) = \frac{r-Ae^{-kt}}{k} ## This is a great tip. Thanks a lot this will save me work. ## \displaystyle\frac{-1}{k}\int_{Q(0)}^{Q(t)} \frac{-kdQ}{r- kQ} = \int_{0}^{t} dt' ## ## \displaystyle\ln(r-kQ)_{Q(0)}^{Q(t)}...

Thank you. I put my equation in the form you gave and solved just fine: ## \frac{dQ}{dt} = Q'(t)## ## r - kQ(t) = g(Q(t)) ## ________________________ ##Q'(t) = g(Q(t)) ## ## \frac{Q'(t)}{g(Q(t))} =1 ## ## Q'(t) f(Q(t)) = 1## ##\frac{dQ}{dt} \frac{1}{r-kQ} = 1 ## ##\frac{dQ}{r-kQ} = dt ##...

The rate at which glucose enters the bloodstream is ##r## units per minute so: ## \frac{dI}{dt} = r ## The rate at which it leaves the body is: ##\frac {dE}{dt} = -k Q(t) ## Then the rate at which the glucose in the body changes is: A) ## Q'(t) = \frac{dI}{dt} + \frac {dE}{dt} = r - k...

The equation I was solving was meant to be analogous to solving an indeterminate integral. For this reason I never really established my limits. Since you brought this to my attention I decided to evaluate the sum from 0 to n. I then used the equation I had gotten and the fundamental theorem...

Hello all. I've come across some math which consists of just applying the basic ideas of calculus (derivatives and integrals) onto discrete functions. (The link: http://homepages.math.uic.edu/~kauffman/DCalc.pdf ) The discrete derivative with respect to n is defined as ## \Delta_n f(n) = f(n+1)...

In line seven, I meant ## b## lies in the range ##(0,1)##

I'm starting to see what you mean. I'll try the polar approach then look at this algebraic approach better. Thanks.

I see what you're saying. I just can't understand why ##t^2 +1## should be used. I know it evaluates to zero if ##t=i## but I don't understand why ##t^2 +1 ## should be used. I would also like to thank everyone at this point for your insight and help.

I'm sorry, I don't understand what the ##R[t]/(t^{2}+1)## means.

D'oh! I just saw that! I can't believe I didn't see this before! Now, I plugged the correct formula into wolfram alpha and got ## (1+i)^1*\frac{\sqrt{2}^{i}}{e^{\frac{\pi}{4}}} \approx 0.274 +i0.584 ## https://www.wolframalpha.com/input/?i=(1+i)^1+((sqrt(2)^(i))/(e^(pi/4))) This answer is now...

The theorem only works for real exponents, right?

Thanks. I understand better why my way doesn't work. Just to be clear, ##z^{Re(w)} * e^{(\ln|z| +\theta i)i Im(w)} ## ##z^{Re(w)} * e^{iIm(w)\ln(w)-\theta Im(w)} ## Is this where I messed up? I think that here I used the rule ## (x^{a})^{b} = x^{ab} ##, where I multiplied the constants. The...