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.
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)
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.
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.
I did take a derivative though
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.
I never said this. What are you talking about?
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.
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)
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.
yeah western academia is a joke :(