The distance between point A and B is 1km. A dog and man start walking at the same time from point A toward point B. They will keep walking (if reached B then walk back toward A immediately, and if reached A then walk back toward B immediately, so on).
1.Given that the dog walks twice as fast...
Yes that is how I wrote it. How should I solve the second part? I've been looking everywhere... The only thing I could find was that it's supposed to be the anti derivative.. Well 3x is the antiderivative of 3 isn't it?
All right, well this is probably wrong but here is what I got!
I took out the 2, which I didn't I could do which is probably what confused me the most..
I'm having trouble with the symbols right now but I essentially put the 2 in front of the integral of the function and substracted the...
\int^{6}_{3} f(x)dx = 42 - 12 = 30
Is this correct? If so I don't know how 2f(x)-3 will affect the integral without knowing what the actual function is... I must be missing something
That's probably what the question is and I'm assuming it was a typo. I will ask tomorrow for sure but in the mean time how would I solve if the question were indeed what gb7nash thinks?
I have looked through all of my notes and can't find anything to get me started...
Mark44 - I copy and pasted the actual question. However, my prof is chinese so maybe that's why the question seems confusing? I really don't understand what he wants me to find...
The function isn't given so I'm not quite sure how to do this. We've only learned how to calculate area beneath the curve with simple geometry and to use the Riemann sum.
We are just starting integrals right now and I'm having trouble with this problem. I know how to solve a definite integral but I don't quite understand what is being asked here? Just wondering if anyone could let me know where I should start?
if∫0,3f(x)dx=12, ∫0,6f(x)dx=42,find...
Here is how I did it in case it might help
lim->0 (e^x)' - 1' - x' - (x^2/2)' /x^3'
= lim->0 (e^x' - 1' - x')/ 3x^2'
= lim->0 (e^x' - 1') / 5x'
= lim x->0 (e^x) /5
Homework Statement
Evaluate limx->0 (e^x - 1- x - (x^2/2))/x^3
The Attempt at a Solution
I can't remember how to solve this limit. Do I need to evaluate each part seperately? I plugged in the 0 to find that the limit does exist. I just can't seem to figure out what to do next.
I also found:
If there is another object nearby, the time the cylinders take to orbit that object can be used to determine mass. (Kepler's law of planetary motion)
Use a spring to measure a pulling force on each cylinder and use Newton's second law
I don't want answers, just to be pointed in the right direction...
I think I found 2 ways.
To apply a force to each cylinder to measure the velocity and find the acceleration. Divide force by acceleration.
Use a small centrifuge and determine rpm and ''radius''. Using a=v2/r, find...
When a car is moving on a circular track which is sloped away from the centre of the track, is the static friction keep that car on the circle (constant speed) towards the outside of the circle, opposite the acceleration?
I'm having trouble seeing this because to me, the car would want to...
I actually do know how to subtract vectors, but my professor had said that we did not have time to cover that section, so we wouldn't see vector addition/subtraction.
So I assumed we wouldn't have any assignments on it, and thought there was another way to do this problem.
Anyways, I ended...
OK so I used the pythagorea theorem but that gave me the resulting momentum when I only need the difference in momentum.
It's probably really simple, I'm just not seeing it right now...
OK, so what would the variation of momentum be ?
P = mv
I have (4.17 j - 5.45 i) kgm/s But my problem is that I can't subtract j's and i's.
I need either two horizontal vectors or two vertical vectors
Then I just need to divide the change in momentum by 3/1000
Ohh, I thought I was looking for V. Vo = 29.43 j m/s
So I found Pi= 5.448i kgm/s
and Pf= 4.17j kgm/s
I need to find the change in momentum so Pf-Pi, how do I go about subtracting those vectors?