Mathematics Question Thread

[jazzycab]

Hey . This question is from the Magaret Grove Extension Math Book, (chapter 2). I've been stuck on in for ages.
Grant is at point A on one side
of a 20 m wide river and needs to
get to point B on the other side
80 m along the bank.

Grant swims to any point on the
other bank and then runs along
the side of the river to point B. If
he can swim at 7 km/h and run
at 11 km/h, find the distance he
swims (x) to minimise the time
taken to reach point B. Answer to
the nearest metre.

If somebody could please help me with this solution that would be great! thank you

Note that I used speeds of 7000 m/hr and 11000 m/hr to ensure I was working with appropriate units, however, due to the fact that the thousandth is a factor of the derivative, it should work even if you use the units given.
Tip: Don't try and do this by specifying x as the run distance, use x as the swim distance as given in the question. It can be done, but the resultant quadratic is horrendous to solve algebraically.

[kauac]

Is there a solid formula for finding a focal chord's length in a parabola (not a latus rectum)?

When I was trying to work out how to find it, when the question gave one point where it intersects the parabola, I found the distance between this point and the focus, then multiplied it by the focal length (a) and got the right answer. Will this method work for all parabolas, or was this just a coincidence?

[cocopops201]

Hi guys,
So I'm currently stuck on this question.
I've figured out the x-coordinates: x=5 or -1
but i've totally forgotten how to find k the constant.

Thank you for anyone who replies!!

[jazzycab]

Hi guys,
So I'm currently stuck on this question.
I've figured out the x-coordinates: x=5 or -1
but i've totally forgotten how to find k the constant.

Thank you for anyone who replies!!

This is a discriminant question. The resultant quadratic obtained in the previous part of the question will have a discriminant equal to zero, because there is one solution to its equation

[RuiAce]

Is there a solid formula for finding a focal chord's length in a parabola (not a latus rectum)?

When I was trying to work out how to find it, when the question gave one point where it intersects the parabola, I found the distance between this point and the focus, then multiplied it by the focal length (a) and got the right answer. Will this method work for all parabolas, or was this just a coincidence?

Please send a screenshot, I got a bit lost reading this aha

Note that I used speeds of 7000 m/hr and 11000 m/hr to ensure I was working with appropriate units, however, due to the fact that the thousandth is a factor of the derivative, it should work even if you use the units given.
Tip: Don't try and do this by specifying x as the run distance, use x as the swim distance as given in the question. It can be done, but the resultant quadratic is horrendous to solve algebraically.

You don't need to worry about units in HSC maths. (Same cannot be said for HSC physics though.)

This question was done previously; the textbook answers just use the units provided.

[jamonwindeyer]

Is there a solid formula for finding a focal chord's length in a parabola (not a latus rectum)?

When I was trying to work out how to find it, when the question gave one point where it intersects the parabola, I found the distance between this point and the focus, then multiplied it by the focal length (a) and got the right answer. Will this method work for all parabolas, or was this just a coincidence?

Hey! As far as I know this would be a coincidence, I'm not aware of any theorem relating the distance from a point to a focus to the length of that focal chord! Would love to be corrected though - Do keep in mind that you'd likely need to show working in an exam situation anyway since such a formula isn't taught in the HSC

[RuiAce]

Hey! As far as I know this would be a coincidence, I'm not aware of any theorem relating the distance from a point to a focus to the length of that focal chord! Would love to be corrected though - Do keep in mind that you'd likely need to show working in an exam situation anyway since such a formula isn't taught in the HSC

Oh, decided to draw it out after your comment. Ya, you're right, it's a coincidence.

Only thing that can be quoted is the focus-directrix definition \(PS = PM\). Nothing can really be said about the length of a chord

[kauac]

With that in mind, what is the best way in approaching questions such as part c?

[RuiAce]

With that in mind, what is the best way in approaching questions such as part c?

Tbh I would've just brute forced this. A clever way is possible with 3U techniques but I don't even think that the clever method is worthwhile.

Step 1. Find the equation of the line though \( (0,-3) \) and \( \left(2, \frac13\right) \)
Step 2. Determine where that line re-intersects with the parabola through simultaneous equations
Step 3. Plug into the distance formula.

[kauac]

Step 1. Find the equation of the line though \( (0,-3) \) and \( \left(2, \frac13\right) \)
Step 2. Determine where that line re-intersects with the parabola through simultaneous equations
Step 3. Plug into the distance formula.

Great! Thanks!! 

[SophiePalmer26]

I'm struggling with this question: For what value of n is the sum of the arithmetic series 5+9+13+... equal to 152?
If anyone has an easy way to find the solution for these types of questions I would be grateful! 

[slinkybench]

A few difference of 2 cubes questions I was stuck on.

Simplify:
1. a^3 + b^3 + a + b
2. x^3 - 1 / x^2 - 1
3. (3 / a - 2) - (3a / a^2 + 2a + 4)

Thanks in advance!

[RuiAce]

I'm struggling with this question: For what value of n is the sum of the arithmetic series 5+9+13+... equal to 152?
If anyone has an easy way to find the solution for these types of questions I would be grateful! 

Presumably \(n\) is the number of terms, or else this question makes no sense.

\begin{align*}\frac{n}{2}[10 + 4(n-1)]&=152\\ n[5 + 2(n-1)]&=152\\ 2n^2 + 3n - 152 &= 0\\ n &= \frac{-3 \pm \sqrt{3^2 - 4(2)(-152)}}{2(2)}\\ &= -\frac{19}{2}, 8\end{align*}

[SophiePalmer26]

Presumably \(n\) is the number of terms, or else this question makes no sense.

\begin{align*}\frac{n}{2}[10 + 4(n-1)]&=152\\ n[5 + 2(n-1)]&=152\\ 2n^2 + 3n - 152 &= 0\\ n &= \frac{-3 \pm \sqrt{3^2 - 4(2)(-152)}}{2(2)}\\ &= -\frac{19}{2}, 8\end{align*}

Thank you so much, that actually helps so much considering that i've never been good at maths

[RuiAce]

A few difference of 2 cubes questions I was stuck on.

Simplify:
1. a^3 + b^3 + a + b
2. x^3 - 1 / x^2 - 1
3. (3 / a - 2) - (3a / a^2 + 2a + 4)

Thanks in advance!

Presumably you meant to factorise, as Q1 and Q2 are already fully simplified
\begin{align*}a^3+b^3 + a+b &= (a+b)(a^2-ab+b^2)+(a+b)\\ &= (a+b)\left[(a^2-ab+b^2)+1 \right]\\ &= (a+b)(a^2-ab+b^2+1)\end{align*}

If the transition was not clear enough from line 1 to line 2

\begin{align*}(a+b)(a^2-ab+b^2)+(a+b) &= uv + u\\ &= u(v+1)\\ &= (a+b)(a^2-ab+b^2+1)\end{align*}

I'm not sure what I can do with \( x^3 - \frac{1}{x^2} - 1\) and all I can do with \(\left(\frac{3}{a} - 2\right) - \left( \frac{3a}{a^2}+2a+4 \right) \) is just cancel out some \(a\)'s and get to \( \frac{3}{a} - 2 - \frac{3}{a} - 2a - 4 = -2a - 6 \)