Got it! Got the same answer, 78.138 . That was a strange integral lol. I can't thank you enough!:smile:
However, when the brick is dragged up, the angle could be calculated, because we now the length of the sides (1, 8, 10.1 ) by pythagorean theorem. Should I take that into account? Since 10N...
Not sure how to do that. At the point θ to dθ, I take dθ/dx and get -\csc^2x dx. Isn't it equivalent to when I use u-substitution, and I express the term I differentiate still using x, with the point of cancelling the already existing x-terms? So same here, I replaced θ with x-terms.
And also...
I miscalculated something in the post above, the integral of \sin u du Is equal to
14.4 not negative. I missed a negative sign from the csc^2. But it's still not 78.14.
Let θ= cotx , dθ=-csc^2xdx so we have
-10\int_{2}^{10} cos(cotx)csc^2x dx
Let u = cotx, du=-csc^2x dx
-10\int_{x=2}^{x=10} cos(u) dx = -10(sin(cotx)) from 2 to 10
Is that negative? No idea how I would calculate \sin(\cot x) , it's a composite function but..
10\int_{2}^10 \frac{x}{\sqrt{1+x^2}} dx Let u=1+x^2, du=2x dx so we have
5\int_{x=2}^{x=10} \frac{1}{\sqrt{u}} du
10\sqrt{u} = 10\sqrt{1+x^2}
From x = 2 to 10
\approx 78.14
Yes, that was just a typo on my part - I did integrate using dθ with cos (I did get a number above, which I assume is wrong?) and dx using the x-expression, which I haven't calculated yet, looks like a tricky integral, trig substitution maybe?
Btw about the limits of integration, from the...
Ah right - above I just used dθ , but It depends on the integrand, so dx means I need the integrand to be in terms of x . So I should use the expression \frac{x}{\sqrt{1+x^2}} , thus 10\int_{10}^2 \frac{x}{\sqrt{1+x^2}} dx
Clearly, since 10\int_2^{10}cosθ gives a negative number... I have no clue about physics, so I might be missing something about the work?
In the text, the brick is moving from 10 to 2, so maybe i should do:
10\int_{10}^{2}cosθ \approx 14.53
How's that?
Alternatively, I know that W = F \times...
If a brick is pulled across the floor by a rope thruogh a pulley, 1 meter above the ground - and work = W, where W = 10N , (in Newton).Show that the horizontal component of W, which is pulling the brick has the size
\frac{10x}{\sqrt{1+x^2}} (*)
Use this to calculate the amount of work needed...
I would love a subforum like that! There are actually very few economics forums out there - which of course could be because of the lack of demand, but if we were to follow Say's law; supply creates its own demand! So I think this is a great idea.
We must also not forget the amount of math...
That course is actually taught by the math department - the professor told me they never really see any econ students taking it (semi-small uni), it's set up for math majors. But I take your point, it is useful for econ and finance for sure! And you're right about the second LA course - which...