Drew Christianson
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 Oct 3 comment Velocity word problem @jordan For me, the answers are: It's a point moving along a line, s(t) shows me where that point is at anytime time t. Knowing the position at every point in time means pretty much everything (velocity, acceleration, location, distance traveled, etc) follows from that. Moreover that function exists inside the logical framework of mathematics, within which I expect everything to be consistent. If I follow the rules anything that begins corresponding to reality should end corresponding to reality. Hence, you shouldn't 'know' ex ante you can't plug in h=25 before you differentiate Oct 3 comment Velocity word problem @jordan Math can be frustrating, it's complicated stuff. Some of us here definitely make it look easy, but that's the product of decades of constant work. If it hasn't been your focus (which isn't bad in the slightest) it's just going to be a bit harder for you. The best advice I can give you is what everyone else has said: don't think about rules, tricks, etc. Instead, try to conceptualize everything you run. Here, the function s(t) = h is good example. What does it look like? What does it tell you about the point in describes? What follows from it? Oct 2 comment Velocity word problem @jordan Ideally, you don't know the correct answer at the outset, only the validity and use of your tools (derivatives, algebra, etc). Then, by careful application either by the predefined 'rules' as you call them (recipes as didier or someone called them) or novel 'recipes' that nevertheless use the tools correctly, you can know completely and undoubtedly that the answer is right. To clarify: there is a way to do math properly, and it is by using the tools you have carefully so that you know every step is valid, and thus the entire result you derive is valid. Oct 2 comment Velocity word problem @jordan remember, tools, not rules. Oct 2 comment Velocity word problem @jordan you can't plug in the h=25 before you take the derivative. As didier, Hennring, and many other members have told you, math is not a set of strict rules. You have to think about the problems and apply the tools you have to solve them, sometimes in new and different ways. If you wanted to know the slope of $f(x)=x^2$ at x=10, you would not plug 10 in then differentiate. You would find $f'(x) = 2x$ and the calculate $f'(10)$. We're doing the same thing here, finding $t_{1}$ such that $s(t_{1}) = 25$ and $s'(t)$ then substituting $t_{1}$ into $s'(t)$ Oct 2 comment What is the best approach when things seem hopeless? Fourier Analysis is not my forte, sorry. Pose it as a question here, and maybe the signal processing SE, someone will have a good reference. Oct 2 comment What is the best approach when things seem hopeless? Ah. That makes things tougher, but I would argue that working backwards still helps. Given that brevity of proofs is that much of a problem, I would also look into getting different & more detailed reference material - I'm sure a reference request here or to your professor would generate useful responses. Oct 2 comment What is the best approach when things seem hopeless? It may seem like everything now, but if the course is well constructed, more recent proofs should build on previous ones. Assume referenced theorems are true and see if the proofs make sense given that truth. Then, when you step back to the earlier theorems later proofs rely on and prove those, you already have a rigorous understanding of later proofs. As I said, the idea is that what seems like everything now is hopefully only a few key concepts - identify those and you're golden. Oct 2 revised What is the formula of the summation of integers from i=0 to i=N? Typeset for clarity Oct 2 suggested approved edit on What is the formula of the summation of integers from i=0 to i=N? Oct 2 answered What is the best approach when things seem hopeless? Oct 2 comment Construct a monotone function which has countably many discontinuities What are the summation limits and index on the example? Oct 2 comment Rate of increase in the area of a square Good! If you have more questions, ask away. Don't forget to use the @_____ if you want to ask someone something particular about their explanation. Oct 2 comment Rate of increase in the area of a square @jordan Yes, kind of. In this case, $s\prime$ is somewhat ambiguous because the problem deals with rates of change . $s\prime(t)$ might be better notation. However, the best is what everyone answering your questions has been using: $\frac{ds}{dt}$. That way you know what you differentiating with respect to what. But yes, it is implicit differentiation. // Two notes of etiquette. One, If you're responding to someone in comments, use @ (their name) so they know you responded. Two, giving up is not they way to go. Everyone wants to help you learn and sometimes that means basic step-by-step. Oct 2 comment Velocity word problem @Jordan Precisely. Well, almost precisely. $15-2(1.86t) = v(t)$ is the velocity (derivative) at any time $t$. But, we have height $h$ as a function of time $t$ ($h = 15t-1.86t^2$) So, if we can just find $t$ when $h$ = 25, you can plug that t into the equation for velocity and you're home free. For a general quadratic equation $ax^2 + bx + c = 0$ the solutions are given by the equation: $$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ Here, your equation is: $$25 = 15t-1.86t^2$$ $$-1.86t^2 + 15t - 25 = 0$$ Given that, just plug the values into the general equation and you'll have solutions. Oct 2 comment Rate of increase in the area of a square @jordan, the derivative of $s^2$ with respect to s is $2s$. However, we want the derivative of $s^2$ with respect to t, so we have to multiply it by $\frac{ds}{dt}$. I know it seems like magic, but it follows from the chain rule for derivatives. Oct 2 comment Rate of increase in the area of a square @jordan Sorry, mistyped square as sphere. My bad, didn't mean to confuse you further. Oct 2 comment Velocity word problem @jordan As for solving for t in my hint, moving everything to the right hand side and it's a basic quadratic equation... Oct 2 comment Velocity word problem @jordan "I know that but if I take the derivative of 25=15t−1.86t2 I just get 0=15-2(1.86t) which I know doesn't help because that is the same as the derivative when it is at 0." That was your comment two hours ago. What you did taking the equation from 25=15t+1.86t2 to 0=15+2(1.86t) was finding the derivative. The only modification I made in my answer was to replace h=25 with just h. Then, $\frac{d}{dt}(h) = \frac{dh}{dt} = velocity = v$. The derivative ($\frac{d}{dt}$) of the RHS is simply what you stated in your comment above. Oct 2 answered Velocity word problem