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doonside technology answers? (1 Viewer)
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Fosweb
I could be your Doctor...
6:
i) Simply differentiate the equation of the parabola or parametrically differentiate P(2ap,ap^2) to get dy/dx and then shove into line equation formula thing.
ii) Use the formula for the intersection of two lines and their gradients (the tan@ = ABS( (m_1 - m_2)/(1+m_1*m_2)), since you know the two gradients (you just found one of them in part i, and the other one you find the same way)
iii) Well, this one is easy. Why that is even worth 2 marks i have no idea.
6b)
i)Simply integrate the equation of motion that they give you.
ii)Derive the cartesian equation using the results of part i, and shove in x = d, you will get the answer.
iii)Gravity acts with the same vertical force downwards on both the slug and the target... Ie, they fall at the same rate. I disagree again with the marks for this question.
Question 7 looks cool, I might actually do it.
i) Simply differentiate the equation of the parabola or parametrically differentiate P(2ap,ap^2) to get dy/dx and then shove into line equation formula thing.
ii) Use the formula for the intersection of two lines and their gradients (the tan@ = ABS( (m_1 - m_2)/(1+m_1*m_2)), since you know the two gradients (you just found one of them in part i, and the other one you find the same way)
iii) Well, this one is easy. Why that is even worth 2 marks i have no idea.
6b)
i)Simply integrate the equation of motion that they give you.
ii)Derive the cartesian equation using the results of part i, and shove in x = d, you will get the answer.
iii)Gravity acts with the same vertical force downwards on both the slug and the target... Ie, they fall at the same rate. I disagree again with the marks for this question.
Question 7 looks cool, I might actually do it.
Fosweb
I could be your Doctor...
Q7 is not as bad as it looks...
7A:
i) I cant do this. Must be missing something...
ii) You know the radius, so then you know that cos@=(x/2) so x = 2cos@
Shove @ in from part i, and you get x = 2cos(5t + pi/4)
Differentiate that twice to get x(dot) and x(doubledot) and then you will notice that its in the form x(dd) = -n<sup>2</sup>x which is SHM
iii) You've got everything you need from part ii now.
iv) Simply put at t=0 into the x(dot) equation from ii, and I got 10/sqrt(2)
7B:
i) You are finding the volume of the curve they gave you, so use
I{0->h} x<sup>2</sup> dy
You rearrange the curve given and x<sup>2</sup> = 4y<sup>1/2</sup>
Integrate and you will get the thing they ask you to show.
ii) What the question is asking is 'how fast is the height falling' (rate), so you need: <sup>dh</sup>/<sub>dt</sub>, which is the same as: <sup>dh</sup>/<sub>dV</sub>*<sup>dV</sup>/<sub>dt</sub>
You have dV/dt from the question (it is h^(1/2) ), and differentiate the volume to get dV/dh.
Multiply to get dh/dt, and you should end up with a constant (4pi), which shows that it is falling at a constant rate.
Thats all.
Anyone know how to show 7a, part i?
7A:
i) I cant do this. Must be missing something...
ii) You know the radius, so then you know that cos@=(x/2) so x = 2cos@
Shove @ in from part i, and you get x = 2cos(5t + pi/4)
Differentiate that twice to get x(dot) and x(doubledot) and then you will notice that its in the form x(dd) = -n<sup>2</sup>x which is SHM
iii) You've got everything you need from part ii now.
iv) Simply put at t=0 into the x(dot) equation from ii, and I got 10/sqrt(2)
7B:
i) You are finding the volume of the curve they gave you, so use
I{0->h} x<sup>2</sup> dy
You rearrange the curve given and x<sup>2</sup> = 4y<sup>1/2</sup>
Integrate and you will get the thing they ask you to show.
ii) What the question is asking is 'how fast is the height falling' (rate), so you need: <sup>dh</sup>/<sub>dt</sub>, which is the same as: <sup>dh</sup>/<sub>dV</sub>*<sup>dV</sup>/<sub>dt</sub>
You have dV/dt from the question (it is h^(1/2) ), and differentiate the volume to get dV/dh.
Multiply to get dh/dt, and you should end up with a constant (4pi), which shows that it is falling at a constant rate.
Thats all.
Anyone know how to show 7a, part i?
N
ND
Guest
@ = 5t+pi/4 cos it starts at pi/4, and moves around at 5 radians/sec.
Fosweb
I could be your Doctor...
Duh. Thanks. Thought I was missing something.