Image 1: If it didn't say a is a positive real constant, could a be -root 3 or positive root3? Or can a never be -root3?
Image 2: a) What's the difference between finding r(2)-r(0) and doing the usual antidiff, etc? What does r(2)-r(0) represent?
b) Do we need to let velocity=0 to check for a turn or something? Don't we usually do that?
Image 3: Do we get consequential marks in spesh?
I got one of the solutions for 4a) wrong, so I got one solution for b) wrong too. Since b) is only one mark, that's the ans mark and there would be consequential marks, right? If b) was two marks though, would I get one mark even though my final ans is wrong?
Image 4: I did integral from -3 to 5, for 1/2*(y+3) dy and multiplied by pi. I get 17 pi but the ans is 16 pi
1. Yes otherwise it can be negative, as only the modulus of all sides are equal |-sqrt(3)| = |sqrt(3)|
2.Because when you integrate the velocity vector, you get an arbitrary constant. Doing the usual anti-diff (and then subbing in your cakes for t=0) is the same as r(2) + r(0). r(2) -r(0) is incorrect as it assumes the origin is at the place of r(0) (note: the particle doesn't start at the origin)
(I am on my phone, so I may mistakes in my working out)
r(t) = (2t^2 -3t)i + t^2j -5tk + c
r(0) = i - 2k = c
Therefore r(t) = (2t^2 -3t +1)i + t^2j - (5t+2)k
r(2) = 3i + 4j - 12k
|r(2)| = sqrt(169) = 13 (distance from origin)
b) that is to find the TOTAL distance travelled. We are calculating the displacement (distance from the origin).
3) You might. Not entirely sure of that.
4) I think you made a mistake, because I am getting 16pi as well (can't show full working out right now).
Q7b: I understand how the ans drew cosex(2x)<cosec(x) but I drew sin(2x)>sin(x)
Should get the same ans but I got (0,pi/3) and the actual ans is (0,pi/3)union (pi/2, pi)
Q8bii) How am I supposed to know that f(X) and f-1(x) intersect only once?
Q9c) The examiners report says you can use a vector method- how do you do it?
7. You have to consider the reciprocal. Otherwise you will be neglecting one side of the solution. Draw the graph of both to find out (use Desmos if you want). Sin(x) is different to cosex(x) as there are also asymptotes and turning points.
8. It's a one to one function. It will intersect twice on both sides of the origin.
9. Can't show full working out right now. But what I would do is find the gradient of the tangent. Write down the corresponding vectors (ie. gradient of 1 represents the vector i + j). Then use cos(x) = a.b/(|a||b|)
Sorry for all the questions, I left starting exams until too late!
1b: Why do you use the unit vector of a, not a?
And when do you know when to use cosine + dot OR tan to find an angle?
6b) The examiner's report said we can't use partial fractions (3rd pic) Why not? (I know we use our ans from part a)
7a) How do you do this? Multiplication of ordinates by drawing out y=3x and y=arctan(2x) separately?
1. You can use either, it doesn't matter. Using the unit vector just makes it slightly easier.
6b) when the degree of the numerator is equal or greater than the degree of the denominator, you'll have to use long division to break it up and than integrate it.
7a) it's best to draw out both functions, and you'll see that the range is R as x can take any real value. (There would be no asymptotes)
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