Advanced Higher Maths 2012-2013 : Discussion and Help Thread

Hey guys,

Would be so grateful if anybody could help at all with these questions? Missed a lot of the course so struggling a wee bit. It's question 2, 3 and 4 from the picture attached. I think I kind of have them, but some clarification would be helpful. Thanks a lot.

Reply 384

Original post by Bonzo10

Hey guys,

Would be so grateful if anybody could help at all with these questions? Missed a lot of the course so struggling a wee bit. It's question 2, 3 and 4 from the picture attached. I think I kind of have them, but some clarification would be helpful. Thanks a lot.

What in particular are you stuck on? For both 2 and 3 use the chain and product rules and for 4 use the quotient rule. You also need to know the standard derivatives for trig and e^x.

Reply 385

Original post by ukdragon37

What in particular are you stuck on? For both 2 and 3 use the chain and product rules and for 4 use the quotient rule. You also need to know the standard derivatives for trig and e^x.

Thanks for your response.

For 2, I got as far as:

dy/dx = -(sinx)^2.(secx)^2.e^-tanx +2e^-tanx.sinxcosx and can't figure out where to go from there.

For 3, I got as far as dy/dx = x(3xsec^2(3x) + 2tan(3x)), and the 0<x<pi/6 is throwing me.

For 4, I got as far as e^x(2x-1)/(1+2x)^2, however I have a feeling that this is wrong.

Thank you again :smile:

Reply 386

Original post by Bonzo10

Thanks for your response.

For 2, I got as far as:

dy/dx = -(sinx)^2.(secx)^2.e^-tanx +2e^-tanx.sinxcosx and can't figure out where to go from there.

For 3, I got as far as dy/dx = x(3xsec^2(3x) + 2tan(3x)), and the 0<x<pi/6 is throwing me.

For 4, I got as far as e^x(2x-1)/(1+2x)^2, however I have a feeling that this is wrong.

Thank you again :smile:

For 2 that is correct and it is f'(x). To find f'(pi/4) you just need to substitute pi/4 for x in your answer and simplify.

For 3 that is also correct and you are done. The limits are just being pedantic in that they are making it "technically" correct.

For 4 that is also the correct derivative. You just need to make it = 0 and solve for x.

Reply 387

Original post by ukdragon37

For 2 that is correct and it is f'(x). To find f'(pi/4) you just need to substitute pi/4 for x in your answer and simplify.

For 3 that is also correct and you are done. The limits are just being pedantic in that they are making it "technically" correct.

For 4 that is also the correct derivative. You just need to make it = 0 and solve for x.

Thank you, repped.

Reply 388

Zepploydath

Obviously when curve sketching you need to work our Stationary Points and their nature. When would I need to look for a non-horizontal POI? Just when there is a maximum TP and a minimum TP and not a horizontal POI? OR do I always attempt to look for a non-horizontal POI?

Reply 389

Original post by Zepploydath

Obviously when curve sketching you need to work our Stationary Points and their nature. When would I need to look for a non-horizontal POI? Just when there is a maximum TP and a minimum TP and not a horizontal POI? OR do I always attempt to look for a non-horizontal POI?

You never need to unless it explictly asks you to find all POI (althoguh remember that horizontal POI are stationary points so you need to find them in any case). If you draw all the stationary points and asymptotes then you'll get the shape about right regardless of any non-horizontal POI.

Reply 390

Ok, me again. Doing an A-C unit test and have a question I'm finding challenging. I hope yous don't mind helping

Unit Test

In the expansion of (1 + px)(1 +px)⁵, where p, q € R, and p and q are not = 0 or each other, the coefficient of x² is zero and the coefficient of x³ is -270.

Find the values of p and q.

I've got to the point where I have expanded and have this which, for a start I'm unsure if it is correct, is not allowing me to go further:

pq⁵x⁶ + (5pq⁴ +q⁵)x⁵ + (10pq³ +5q⁴)x⁴ + (10pq² + 10q³)x³ + (5pq + 10q²)x² + (p + 5q)x + 1

I have tried equating the coefficients of x² and x³ to their respective values in the question but I'm stuck. I'm not even sure if what I've done so far is right. Any help would be appreciated, thank you.

Reply 391

849582394

Original post by Bonzo10

Ok, me again. Doing an A-C unit test and have a question I'm finding challenging. I hope yous don't mind helping

I've got to the point where I have expanded and have this which, for a start I'm unsure if it is correct, is not allowing me to go further:

pq⁵x⁶ + (5pq⁴ +q⁵)x⁵ + (10pq³ +5q⁴)x⁴ + (10pq² + 10q³)x³ + (5pq + 10q²)x² + (p + 5q)x + 1

I have tried equating the coefficients of x² and x³ to their respective values in the question but I'm stuck. I'm not even sure if what I've done so far is right. Any help would be appreciated, thank you.

There's a better way than expanding it all (that I know some really cool dude was supposed to give you before he had to go scrape some dirt off plants), which I'll type out in a min, but the correct expansion of that is:

p q^5 x^6+5 p q^4 x^5+10 p q^3 x^4+10 p q^2 x^3+5 p q x^2+p x+q^5 x^5+5 q^4 x^4+10 q^3 x^3+10 q^2 x^2+5 q x+1

From there you can get the coefficients of the x² and x³ terms:

( 5 p q + 10 q^2 ) x^2

( 10 p q^2 + 10 q^3 ) x^3

Then equating coefficients:

5 p q + 10 q^2 = 0

5 q (p + 2q) = 0

p+2q = 0

(or 5q = 0, but we know q isn't 0)

p = -2q

and

10 p q^2 + 10 q^3 = -270

Now solve these as simultaneous equations. Sub in p=-2q to the second one:

10 (-2q) q^2 + 10 q^3 = -270

-10 q^3 = -270

q=3

and then that into the first one:

p = -2(3)

p = -6

(edited 11 years ago)

Reply 392

849582394

Other method:

(1 + px)(1 +qx)^5

General term for (1 + px):

\binom{1}{r} 1^{1-r}(px)^r

General term for (1 + qx)^5:

\binom{5}{r} 1^{5-r}(qx)^r

We know the coefficient of the x² term and the x³ term, and we can use the general terms to find those coefficients in terms of p and q without expanding the whole thing out.

If you imagine you've already expanded (1 +qx)⁵, you'd have a bunch of terms in there of varying powers of x - from x⁰ all the way up to x⁵. Since you've already done it the long way you should be able to see that there are a number of ways that x² terms come out the other end when you multiply the (1 + px) by the expansion of (1 +qx)⁵. First, you get the x⁰ term from the first bracket multiplying by the x² term from the big expansion; then you've got the x¹ term from the first bracket multiplied by the x¹ term from the big expansion. Both x⁰ * x² and x¹ * x¹ will give you an x² term. If you were doing it the long way you'd have seen two x² terms after multiplying it out and collected them together.

To represent that (for x²) with the general terms we do:

[\binom{1}{0} 1^1 (px)^0] \times [\binom{5}{2} 1^3 (qx)^2] + [\binom{1}{1} 1^0 (px)^1] \times [\binom{5}{1} 1^4 (qx)^1]

= [1] \times [10q^2 x^2] + [px] \times [5qx]

= 10q^2 x^2 + 5pqx^2

= (10q^2 + 5pq ) x^2

Which should look familiar from above. You can do the same for x³ and you'll get the same simultaneous equations out at the end.

(edited 11 years ago)

Reply 393

Original post by 849582394

Other method:

(1 + px)(1 +qx)^5

General term for (1 + px):

\binom{1}{r} 1^{1-r}(px)^r

General term for (1 + qx)^5:

\binom{5}{r} 1^{5-r}(qx)^r

We know the coefficient of the x² term and the x³ term, and we can use the general terms to find those coefficients in terms of p and q without expanding the whole thing out.

If you imagine you've already expanded (1 +qx)⁵, you'd have a bunch of terms in there of varying powers of x - from x⁰ all the way up to x⁵. Since you've already done it the long way you should be able to see that there are a number of ways that x² terms come out the other end when you multiply the (1 + px) by the expansion of (1 +qx)⁵. First, you get the x⁰ term from the first bracket multiplying by the x² term from the big expansion; then you've got the x¹ term from the first bracket multiplied by the x¹ term from the big expansion. Both x⁰ * x² and x¹ * x¹ will give you an x² term. If you were doing it the long way you'd have seen two x² terms after multiplying it out and collected them together.

To represent that (for x²) with the general terms we do:

[\binom{1}{0} 1^1 (px)^0] \times [\binom{5}{2} 1^3 (qx)^2] + [\binom{1}{1} 1^0 (px)^1] \times [\binom{5}{1} 1^4 (qx)^1]

= [1] \times [10q^2 x^2] + [px] \times [5qx]

= 10q^2 x^2 + 5pqx^2

= (10q^2 + 5pq ) x^2

Which should look familiar from above. You can do the same for x³ and you'll get the same simultaneous equations out at the end.

Remarkable. Thank you very much. Even more remarkable was your ability to find this and me.

I was a step away from completing it the first way, just didn't think to put the value of p=-2q into the x³ coefficient.

Thanks again.

Reply 394

Zepploydath

I'm having a little trouble with asymptotes. In my notes when finding out a vertical asymptote you are looking for when the denominator is 0. So the curve of 1/x as a vertical asymptote at x=0. However my notes then mentions negatives and positives and as x tends towards negative and positive infinity. I'm not sure about any of this and if you need to know it? My notes says + means it approaches from the right and - means it approaches from the left. But surely you would know this anyway from solving the asymptote? :s-smilie:

Reply 395

Hype en Ecosse

20

Original post by Zepploydath

I'm having a little trouble with asymptotes. In my notes when finding out a vertical asymptote you are looking for when the denominator is 0. So the curve of 1/x as a vertical asymptote at x=0. However my notes then mentions negatives and positives and as x tends towards negative and positive infinity. I'm not sure about any of this and if you need to know it? My notes says + means it approaches from the right and - means it approaches from the left. But surely you would know this anyway from solving the asymptote? :s-smilie:

You do need to know this. Numberphile actually did quite a good video addressing this when talking about division by zero.

Anyway, what you've got is a function

\frac{1}{x}

, and you want to sketch this function in its entirety. Since you can't divide by zero, there's obviously not a zero point, so we have to find out what happens when you tend to zero.

\displaystyle\lim_{x \to 0} \frac{1}{x}

.

Now, if you start off with x = 1, you get 1. x = 0.1, you get 10, x = 0.01, you get 100. As you get closer and closer from the positive direction, your number gets bigger and bigger.

\displaystyle\lim_{x \to 0^+} \frac{1}{x} = +\infty

However, if you start with x = -1, you get -1. x = -0.1, you get -10, and so on. As you get closer and closer from the NEGATIVE side, your number gets smaller and smaller.

\displaystyle\lim_{x \to 0^-} \frac{1}{x} = -\infty

So it does matter what side you approach it from.

It's not immediately obvious by "solving the asymptote" (I'm not sure what you mean by this, I assume you meant "finding"). As you could also approach positive infinity from the negative side in another function, or approach negative infinity from the positive side. What happens to a function as it tends to zero depends entirely on that function. So it's important to be able to work it out.

Reply 396

Zepploydath

Original post by Hype en Ecosse

You do need to know this. Numberphile actually did quite a good video addressing this when talking about division by zero.

Anyway, what you've got is a function

\frac{1}{x}

, and you want to sketch this function in its entirety. Since you can't divide by zero, there's obviously not a zero point, so we have to find out what happens when you tend to zero.

\displaystyle\lim_{x \to 0} \frac{1}{x}

.

Now, if you start off with x = 1, you get 1. x = 0.1, you get 10, x = 0.01, you get 100. As you get closer and closer from the positive direction, your number gets bigger and bigger.

\displaystyle\lim_{x \to 0^+} \frac{1}{x} = +\infty

However, if you start with x = -1, you get -1. x = -0.1, you get -10, and so on. As you get closer and closer from the NEGATIVE side, your number gets smaller and smaller.

\displaystyle\lim_{x \to 0^-} \frac{1}{x} = -\infty

So it does matter what side you approach it from.

It's not immediately obvious by "solving the asymptote" (I'm not sure what you mean by this, I assume you meant "finding"). As you could also approach positive infinity from the negative side in another function, or approach negative infinity from the positive side. What happens to a function as it tends to zero depends entirely on that function. So it's important to be able to work it out.

Yeah, I meant finding, sorry.

What I'm not sure about is how this is relevant when sketching the graph?

And for horizontal and slant asymptotes the process seems to be reversed and you are you are doing x tending towards positive and negative infinity. What does this mean when sketching the graph?

Reply 397

Hype en Ecosse

20

Original post by Zepploydath

Yeah, I meant finding, sorry.

What I'm not sure about is how this is relevant when sketching the graph?

And for horizontal and slant asymptotes the process seems to be reversed and you are you are doing x tending towards positive and negative infinity. What does this mean when sketching the graph?

I've attached a graph of 1/x. If you look at it, you can clearly see it launching off into

\pm \infty

. Now pretend you've never seen the graph of 1/x before, or pretend you've never sketched the function. You know that there is a vertical asymptote at x = 0, so you know the graph never touches that point.

However, there are two ways the graph can move closer and closer to the asymptote without ever touching it. It can move up, and it can move down. If you've never seen a graph of a function before, you have no way of telling if it goes up or down - a characteristic that's fundamental to the shape. So you have to find out what direction it goes so that you can sketch it properly. Look at the differences between the graphs of

\frac{1}{x}

and

-\frac{1}{x}

to see how this direction matters in sketching the shape of the graph.

As for with horizontal and slant asymptotes tending towards

\pm \infty

, just look at the graph of 1/x again. It shoots off into infinity to the left and right as well. :wink:

For another function with an oblique asymptote, it would just shoot off into infinity diagonally like here.!

(edited 11 years ago)

Reply 398

I am Ace

4

by u substitution,

u=x^3

Unparseable latex formula:

int\ {frac\{x^2}{x^6+1}}dx)

I got it to

\frac{1}{3x^6+3})

\frac{1}{3u^2+3}du

Then I tried to let

v=3u^2+3

And for the past hour I've went round in circles using u substitution getting nowhere

Reply 399