2U Maths Question Thread: Ask Your Questions Here!

[jamonwindeyer]

Hey everyone! It would be awesome to have a place to collaborate and help each other out on tricky problems, we have a community full of brainiacs. So here is that space! Have at it! Ask any questions you have on your mind below. I'll do the rounds once a day, and if it is a question which I feel should be shown to lots of people, I'll even start a new thread for it to share the knowledge 

[Sabrina Spellman]

Where is the function x^2e^x decreasing is the question, so how does this proof work??

[jamonwindeyer]

Where is the function x^2e^x decreasing is the question, so how does this proof work??

Hey Sabrina!

Right, I'm going to assume that you can get the derivative. To look for where a function is decreasing, we look for where it's gradient is negative. We can do this by looking at where the first derivative is negative, i.e.:



Now we know, as a property of the exponential function, that for all real x, so we can ignore that part of the expression, it has no effect on the sign.



Right, we consider this a number of ways, but it is LOTS easier to consider it graphically, which is what the proof omits. Do a rough sketch of this parabola (see below) and we see it has intercepts at 0 and -2. The quadratic is negative between these intercepts. Hence, the answer is -2<x<0.



Does that make sense?

[Sabrina Spellman]

Hey Sabrina!

Right, I'm going to assume that you can get the derivative. To look for where a function is decreasing, we look for where it's gradient is negative. We can do this by looking at where the first derivative is negative, i.e.:



Now we know, as a property of the exponential function, that for all real x, so we can ignore that part of the expression, it has no effect on the sign.



Right, we consider this a number of ways, but it is LOTS easier to consider it graphically, which is what the proof omits. Do a rough sketch of this parabola (see below) and we see it has intercepts at 0 and -2. The quadratic is negative between these intercepts. Hence, the answer is -2<x<0.

(Image removed from quote.)

Does that make sense?

it does, thanks!!

[hsceebz]

i need integration explanations please

1) find the exact area bounded by the parabola y=x^2 and the line y=4-x
2) find the volume of the solid formed when the curve y=(x+5)^2 is rotated about the y-axis from y=1 and y=4

[cosine]

i need integration explanations please

1) find the exact area bounded by the parabola y=x^2 and the line y=4-x
2) find the volume of the solid formed when the curve y=(x+5)^2 is rotated about the y-axis from y=1 and y=4

1). We need to find the area that is bounded by the two black dots on the diagram embedded. We can do this by finding the intersection, that is, the values of the two black dots. Then we need to find the area of the blue graph, 4-x, and then minus area of the parabola.

Intersection:






Area =

Plug this straight into your calculator for the area, and remember to express as

Sorry though, I cannot help you much with the second one. I tried it, kept getting a negative volume, so yeah...

[jamonwindeyer]

i need integration explanations please

1) find the exact area bounded by the parabola y=x^2 and the line y=4-x
2) find the volume of the solid formed when the curve y=(x+5)^2 is rotated about the y-axis from y=1 and y=4

Hey hey! Thanks to cosine for the awesome explanation for the first half of that explanation. I'll tag in for the second half 

This one is tricky! More like what you would see in an Extension One Paper.  I'll also mention that this question wouldn't be asked in the HSC, as it isn't specific enough. Specifically, we don't know whether the question wishes us to find the volume enclosed by the parabola, and the lines y=1 and y=4, or otherwise. A question specifying a single arm is a possibility though. But since I'm a sucker for punishment, I'm going to assume it wants the area between the arms, as it is also the most nasty route.  I've actually never done a question like this before, so here goes!

The first thing we need to do is rearrange the given function to make x the subject:



The plus/minus represents the two arms of the parabola. If they want the area enclosed by the curve and the two lines, we need to use both arms. The arm furthest from the y axis will give us our first volume, this happens to be the minus. The other will give us a smaller volume, as it is closer to the y axis. The larger volume, minus the smaller volume, gives us the volume enclosed by the parabola. Hopefully that makes sense! If you draw a diagram, you will see that the arm represented by is further from the y axis than the other. To find the area enclosed, we take this larger area, minus the smaller area under the closer arm, denoted by .

Actually, never mind, here is a diagram! Larger volume in red, smaller in blue! The area we want is red-blue, simple!



I'll denote the arm which will give the larger volume as , and the smaller as .

The volume can be derived as follows.



I'll stress that, unless this is a 4 unit thing (which I am not familiar with), this will definitely NOT be asked in the HSC. If it was, it would lead you through the process, rather than hang you out to dry like that. And definitely would never go near a 2 unit paper, simply too intricate. Hopefully that answers the question, and hopefully it is correct! If anyone spots an error please shout out and I'll make amendments, or open the floor to someone more knowledgeable than I, this was new for me too! PS- If you wanted to do it with just one specific arm, you would simply find the volume for that arm only. Much easier.

EDIT: I'll try your question on Induction in the 3U forum tomorrow morning!

[christmaybe]

Hey I was wondering if you could explain the 2013 maths hsc questions 8 and 10.

[jamonwindeyer]

Hey I was wondering if you could explain the 2013 maths hsc questions 8 and 10.

Sorry for the delay, given the low traffic I've not checked the forum for a bit, but indeed, I can 

Question 8 (MC): The graph shows four points, and asks at which point and .

  means that the first derivative is positive, and thus the gradient is positive. The graph will be increasing. This leaves point B and D (A is decreasing, C is a stationary point).
  means that this point is also a point of inflexion, where the concavity changes. This description only suits B. The answer is B.

Question 10 (MC): The two terms given represent the velocity and the acceleration of the particle along the x-axis. It would perhaps be easier and less confusing then to instead say and .

Now, the velocity is negative, meaning it is moving in the negative x direction (think if the speedo in a car had negative readings for reverse... come to think of it, does the speedo still go up in reverse gear?). Regardless, it is moving to the left. However, the acceleration is positive, i.e., directed to the right. This means that, while it is travelling left, the acceleration is acting in the opposite direction. So it slows down. The answer is D .