Why can't this be done? HSC 2004 paper (1 Viewer)

[hanif]

http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2004exams/pdf_doc/maths_04.pdf

Q8, b ii)



Why can't we find the gradient of the line BA and then take the negative reciprocal and derive the line of the tangent to the parabola this way? The tangent to line BA AND the parabola touches at point C, so I did it the aforementioned way but I came up with an answer of:

y-4x+15=0

which is wrong (tried to substitute point C and != 0 in part iii)

Q5 b)


Also, I'd like to understand the way Q5 b is solved, the differentiation is easy but I have a hard time trying to solve the likes of:

sin (2t) = 0

What I did was jam in values of 0, PI, PI/2, 3PI/4 etc.. and then see if it results in 0. Is there a much more intuitive way to do this?

Last question, I'm a bit confused with part iv) does this mean I have to find the second derivative and check to see if it is < 0 or can we find this from looking at the graph in ii) ? Thanks in advance!

 

[leesh95]

http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2004exams/pdf_doc/maths_04.pdf

Q8, b ii)

View attachment 28835

Why can't we find the gradient of the line BA and then take the negative reciprocal and derive the line of the tangent to the parabola this way? The tangent to line BA AND the parabola touches at point C, so I did it the aforementioned way but I came up with an answer of:

y-4x+15=0

which is wrong (tried to substitute point C and != 0 in part iii)

Q5 b)
View attachment 28836

Also, I'd like to understand the way Q5 b is solved, the differentiation is easy but I have a hard time trying to solve the likes of:

sin (2t) = 0

What I did was jam in values of 0, PI, PI/2, 3PI/4 etc.. and then see if it results in 0. Is there a much more intuitive way to do this?

Last question, I'm a bit confused with part iv) does this mean I have to find the second derivative and check to see if it is < 0 or can we find this from looking at the graph in ii) ? Thanks in advance!

I can't understand what you mean in question 8ii)

It does not state in the question that line BA and AC are perpendicular to each other. If it did then you could take the reciprocal. The tangent to the parabola is AC not BA! and remember tangent to a parabola has same gradient where it touches the parabola.

Sin 2t=0 for 0<t<pi

You need to know the values for which sin is 0. This is an integral part of the 2u course. Remember the unit triangles?
SO sin is 0 at t=pi
but we have 2t in the question so
2t=pi
t=pi/2
For the second part you have to find the second derivative and equate it to 0 to check the maximum velocity

 

[omgiloverice]

Q8 b) ii) because the two lines aren't perpendicular to each other.

5) b) you can graph it, you can memorise the identity sin (pi - x) = sin (x)