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Math 32....how was it? :/
Do you remember the question in which we had to find the stationary points first and then the nature of that point? It was question 6 or question 7, I guess :confused: -
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Math 32....how was it? :/
That's also correct, I presume. The final answer would've been 'y = (-1/2) ln (3 - e^2x) then. -
Math 32....how was it? :/
cosec 2x = sec x + cot x (1/sin 2x) = (1/cos x) + (cos x/sin x) (1/2 sin x cos x) = (sin x + cos^2 x)/(sin x cos x) 'sin x cos x' present in both the denominators get cancelled. (1/2 sin x cos x) = (sin x + cos^2 x)/(sin x cos x) (1/2) = (sin x + cos^2 x) (1/2) = sin x + (1 - sin^2 x) 1 = 2... -
Math 32....how was it? :/
y = x^(1/2) ln x Volume = pie ( y )^2 Volume = pie [ x^(1/2) ln x ]^2 Volume = pie [x (ln x)^2] Now use integration by parts to integrate 'x (ln x)^2' x (ln x)^2 (x^2/2)(ln x)^2 - [x^2 /2 * 2 ln x * 1/x] (x^2/2)(ln x)^2 - [x ln x] Now integrate 'x ln x' x ln x (x^2/2) ln x - [x^2/2 * 1... -
Math 32....how was it? :/
Answer to the volume of revolution question was pie/4(e^2 - 1). -
Math 32....how was it? :/
I am an A* student too and the mathematics coach at my school. This's how you had to do the differential question: dy/dx=e(2x+y) dy/dx=e^(2x) x e^( y ) [1/e^( y )] dy = e^(2x) dx e^(-y) dy = e^(2x) dx Integrate both the sides. -e^(-y) = [e^(2x)/2] + c Put x=0 and y=0 to find the value of... -
maths p32 answers
The question was dy/dx=e^(2x+y) right? dy/dx=e(2x+y) dy/dx=e^(2x) x e^( y ) [1/e^( y )] dy = e^(2x) dx e^(-y) dy = e^(2x) dx Integrate both the sides. -e^(-y) = [e^(2x)/2] + c Put x=0 and y=0 to find the value of 'c'. -e^(-y) = [e^(2x)/2] + c -e^(-0) = [e^(0)/2] + c -1 = (1/2) + c -1 -... -
Math 32....how was it? :/
What an easy paper it was except for the last part of the Vector question. Still wont be losing more than 4 marks InshAllah. -
Mathematics: Post your doubts here!
Yes, you've to switch your calculator to the radian mode when finding the angle of the sector or dealing with any question related to this chapter. -
Mathematics: Post your doubts here!
(i) Find the length of the side OT using the pythagoras theorem. OT^2 = OP^2 + PT^2 OT^2 = 5^2 + 12^2 OT^2 = 169 OT = 13 Next, we need to find the the arc length. For that, we'll first be needing to find the angle θ. Use the trigonometric property of 'tan θ = opp/adj' to find the value of... -
Mathematics: Post your doubts here!
What's the answer to this question? -
Mathematics: Post your doubts here!
I had accidentally mistyped that part. This is how it is: (√6) cos θ/2 + (√10)sin θ/2 = 3 4 cos(θ/2 − 52.23) = 3 θ/2 − 52.23 = (cos^-1)(3/4) θ/2 − 52.23 = -41.41 θ/2 = -41.41 + 52.23 θ/2 = 10.82 θ = 21.7° We've got 41.41 from (cos^-1)(3/4). As the angle could've been in the 2 quadrants... -
Mathematics: Post your doubts here!
No hard feelings but you've used the wrong range; albeit your answer is correct but the method isn't. The range which needs to be modified is '0 < θ < 360', not '0 < θ < 90'. The range '0 < θ < 90' is for 'a'. -
Mathematics: Post your doubts here!
I'll do the ii(a) part of this question first so that you can understand this method. (√6) cos θ + (√10)sin θ R=4 a=52.23 4 cos(θ − 52.23) In the ii(a) part of this question which is (√6) cos θ + (√10)sin θ = −4, we'll be needing to modify the range first. 0 < θ < 360 0 - 52.23 <... -
Mathematics: Post your doubts here!
(a/x)^2 (a^2)/(x^2) (a^2) (x^-2) Now integrate 'x'. (a^2) (x^-2) (a^2) [(x ^-1)/-1] - (a^2)/x -
Mathematics: Post your doubts here!
This is the formula used to find the volume when a region is rotated about the x-axis: Volume = π ∫ ( y )^2 The equation 'y = (a/x)' is already mentioned in the question, simply put it inside the formula and solve it. π ∫ ( y )^2 π ∫ ( a/x )^2 -
Mathematics: Post your doubts here!
π ∫ (a/x)^2 = 24 π π ∫ (a^2/x^2) = 24 π Both the pi's get cancelled so we are left with: ∫ (a^2/x^2) = 24 Integrate with respect to 'x': ∫ (a^2)(x^-2) = 24 (a^2) ∫ (x^-2) = 24 (a^2) (-1/x) = 24 Plug in the upper limit (x=3) and the lower limit (x=1) in the integrated equation. (a^2)... -
Mathematics: Post your doubts here!
(x + 1/x)^2 Open the square bracket. x^2 + 2(x)(1/x) + (1/x)^2 x^2 + 2 + 1/(x^2) Now integrate with respect to 'x'. x^2 + 2 + x^-2) (x^3)/3 + 2x - 1/x + c Therefore, the answer is '(x^3)/3 + 2x - 1/x + c'. -
Mathematics: Post your doubts here!
Refer to the above diagram and solve the horizontal and the vertical components of forces. Horizontal Component: T cos θ + T sin θ = 15.5 T cos θ = 15.5 - T sin θ Vertical Component: T cos θ + 8 = T sin θ 15.5 - T sin θ + 8 = T sin θ 23,5 = 2 T sin θ 11.75 = T sin θ Therefore, T sin θ =...