Recent content by Bill_Nye_Fan

Update: I just noticed that the formula ##a=v\cdot \frac{dv}{dx}## works a lot better in this scenario. Almost all the previous questions I was doing were using ##a=2\int _{ }^{ }f\left(x\right)dx## so I kind of got tunnel vision of only using that formula. ##a=\frac{\sqrt{x-1}}{v^2}##...

Thanks for the advice :) I'll start working through some trial functions

Homework Statement A body of mass 1 kg moves so that at time ##t## seconds, its displacement is ##x## meters from a fixed origin. The body is acted upon by a force of ##\frac{\sqrt{x-1}}{v^2}## Newtons, where ##v## is the velocity in m/s. The body is initially at rest 1 meter North of the...

Thank you both for your help :) I knew that ##\int _{ }^{ }f\left(x\right)d\left(f\left(x\right)\right)## has the general solution of ##\frac{1}{2}\left(f\left(x\right)\right)^2## regardless of what the function actually is. I was curious as to whether there would be a way to apply this while...

Hello all, I was just wondering if there is any rules for integrating a function with respect to it's own derivative. That is to say ##\int _{ }^{ }f\left(x\right)d\left(f'\left(x\right)\right)## or ##\int _{ }^{ }yd\left(\frac{dy}{dx}\right)## Thank you in advance for your time :)

Homework Statement A truck of mass 1.5 tonnes (1500 kg) is moving at a speed of 36 km/h down a hill of slope 1 in 6. The driver applies the breaks and the truck comes to rest after 2 seconds. Find the breaking force, which is assumed to be constant, ignoring any other resistant forces...

Ah yes I see what you're saying. Integrating a function (or in this case a derivative function) with respect to itself is the same as integrating a singular variable with respect to itself. I've done some quick working out and this is what I got: Do you think this looks right or am I way off...

Homework Statement A function of y, ##f(y)##, is known to be equal to the second derivative of function ##y(x)##. ( i.e. ##\frac{d^2y}{dx^2}=f\left(y\right)## ) Given that ##\int _{ }^{ }f\left(y\right)dy=x-1##, and function ##y(x)## has a stationary point at ##x=1## and an x-intercept at...

I'll give you a small hint to get you started: the first step is to solve a couple things simultaneously. You shouldn't have too much trouble with the rest from there :)

Okay thank you I'll just do it that way

I think my main problem is that I still don't know how to calculate the area between two curves properly. Say I was just given y=x-2 and y=-sqrt(4-x) and had to find the area enclosed between them between x=0 and x=2 Isn't the correct way of doing this to just set up an integral over the given...

Thank you all for your help, but I feel like some of this stuff is a little beyond my current course. We've only just started looking at definite integrals, and have spent even less time with areas between two curves. I highly doubt we're expected to be using horizontal strips instead of...

Ah yes that is a much better idea thank you. But still, my method should still be getting the same answer but it's not. Could you please explain to me where I'm going wrong?

Homework Statement Please find attachedHomework Equations [/B] Definite integrals and area between two curvesThe Attempt at a Solution [/B] Also find attached. The answer in the back of the book says that part c should be 10/3, but that would mean that (8sqrt (2))/3 would need to cancel out...

Thank you so much that has helped heaps