I would probably fail this question because I'd be too busy cackling and run out of time.
lmao
also wouldn't the distance between them grow at a constant rate? am i an idiot? they're each moving at a constant speed in straight lines right?
Yes, this is an arithmetic, not a calculus problem.
I guess a little geometry
Rewriting this question to make it so they're accelerating at those speeds instead.
Isn't it still finding the rate of change with respect to time? You find the derivative of the function even if that result is a constant rate. That's differential calculus, no?
Maybe I'm wrong.
It has no relation to differential calculus, calculus relates to continuous change. Pythagoras could've worked this out about 2000 years prior to the discovery of calculus. You're not finding the gradient of an effectively infinitesimal point on an ever-changing curve.
The question is asking for the rate at which they're moving from one another, so the answer will be a number in ft/sec. The problem probably wants you to set it up as related rates and then do integration.
Also I haven't done calculus in like 6 years
I'm pretty sure my differential calculus professor did this question for my first exam over 10 years ago. The dude was chill as shit, and one of the best math professors i've ever had.
I never did this in school. What variable represents all the different choices you could have made along the way? delta squared?
That's when you add probabilistic variables and it becomes a nightmare
5 ft/s is an incredibly slow run. My boy is fake sprinting away while actually going at a slow walking pace.
And 1ft/sec is practically an impossible speed to walk at, she's like gone so insane she's pigeon-stepping away.
he's running through the tears and tripping and snotting and it's really pathetic
Because its DiffEq, I'm assuming these were intended to be accelerations rather than velocities. So, he starts walking away and then picks up into a sprint. She starts at a crawl and advances to a steady walk.
But I bet the person writing the prompt (or the app translating the question) fumbled the notation for s^2, so now (hopefully) the entire class is going to fail this question and get it curved out.
Agreed. But even then, who the hell walks away from a point with linear acceleration? That's really hard to picture.
I mean, its a math problem at its heart, not an engineering problem.
You can solve this without a single differentiation or integration. This is barely algebra.
Quite possibly the hardest I ever worked in in any class was my effort to pass AP calculus my senior year of high school. The goal of passing the exam and testing out of math in college was paradise on my horizon: just imagine it, no more math class, EVER! I danced with joy when I got the test results back and saw that my score was high enough to get out of the gen ed requirement. Eventually I realized that the gen ed requirement at my college was to take a lab and a non-lab science/math class, so I wouldn't necessarily have had to take math again regardless. Oh well.
Had the same experience with stats at CC. Got a C and as a 3.8 student it was my happiest moment. Then I learned I needed 3 more quarters of Stat for my major. Stats at university was a fucking breeze compared to CC.
Okay so by the Pythagorean theorem, a^2 + b^2 = c^2
If we take the boy's position at time t as a, then a = 5t, and the girl's position b = t
Thus their distance at a given time is sqrt((5t)^2 + t^2) = sqrt(26t^2) =sqrt(26)t
Taking the derivative of that in terms of t gives us sqrt(26) feet
I think, it's been like 10 years since I last studied calculus
I mean, precisely. It's Pythagoras and linear algebra, this is not calculus.
Huh? You took the derivative by taking 't' off the end, that ain't calculus.
Derivatives and integrals are calculus. Just because a derivative is easy doesn't mean it's not calculus, in the same way that 2 + 2 = 4 being easy doesn't make it not arithmetic.
I mean, sure, it's an arguable position, but an actively unhelpful use of language. You're arguing that anybody that knows about gradients (or even just like, how two still objects don't move) knows calculus, therefore most people must know calculus by like age 8.
You're not employing the tools of calculus in any meaningful or non-negligible way. You're just using basic newtonian laws and claiming you used complex quantum mechanics waveforms. Show your actual working out of the derivative just like you did with the basic algebra and I'll concede and give you the star sticker.
You’re arguing that anybody that knows about gradients (or even just like, how two still objects don’t move) knows calculus
I never said this. What are you talking about?
Show your actual working out of the derivative just like you did with the basic algebra and I’ll concede and give you the star sticker.
The solution to any derivative of the form d/dx(a*x) where a is a real number is a. It's like the first thing you learn when you study calculus.
I never said this. What are you talking about?
The gradient of any line is mx + c = y, that's grade school math. That's all you used to work it out and you're claiming you used calculus.
Alas, the answer for the coveted star was:
Multiply the equation with d/dt: (d/dt) * t*sqrt(26) = dt * sqrt(26) / dt.
dts cancel out, therefore the derivative is sqrt(26).
Still a very unnecessary (and simple) application of calculus, but at least it would've shown that's whatcha did.
I never used the slope equation. I took a derivative d/dt. I clearly indicated the step where I did that. At this point, I can only conclude that you're keeping this going out of some weird desire to get one over on me, so I'm gonna disengage.
yeah ur right. but the rate of change of the distance between them is always sqrt(26) ft/s. the problem is flawed since it doenst actually require calculus to solve, just plug in t=1 and you get sqrt(26)
the problem is flawed
I hate questions that have this confusing language element (rate of separating... after 5 seconds) when it isn't even clear if the person writing the question knew what they were asking. You're left second-guessing the test-prompt, because now you're double checking to see if the original 5 ft/s, 1 ft/s were actually accelerations rather than velocities. You're worried that the answer is too easy, because it doesn't involve the math you were prompted to study prior to the exam. You're wasting time on parsing the language rather than doing the math.
Its just an awful way to conduct an exam, because its rewarding confidence rather than competence.
As a math guy, the answer is that he's not going to move on for a while, while she already has.
I spent the semester of college I did not learning any of this and none of you motherfuckers is gonna make me start now.
