VCE Methods Question Thread!

[BubbleWrapMan]

The point of the table is to tell you the signs of the gradients, and therefore the shape of the curve, so you don't have to write specific gradient values in between the stationary points.

[Planck's constant]

We need to test a value between and , but we shouldn't just use imo. eg if , . Clearly . So, we would be testing the gradient on the other side of the other turning point

Without loss of generality, I let the 'middle' x value be the 'middle' of and , guaranteeing that it's between and a no matter what.
Hence the middle value is

Yes, this is by far the best method for this part of the problem

[datfatcat]

Yes, this is by far the best method for this part of the problem

Spoiler

I like your username

[shadows]

A piece of wire of length 1m is bent into the shape of a sector of a circle of radius a cm and sector angle of theta.
Let the area of the sector of A cm^2.


a) How do I find A in terms of a and theta

I got A  theta/(360)  x pi(a)^2
(This is not the answer) Answer is (1/2)a^2 x theta  (LOL EDIT: NVM I GOT IT =.=)

b) Find A in terms of theta

how i do this?


At noon the captain of a ship sees two fishing boats approaching. One of them is 10km due east and travelling west at 8km/h. The other is 6km due north, travelling south at 6km/h. At what time will the fishing boats be closest together and how far apart will they be?

[zvezda]

Q16 in the essentials textbook, exercise 12J
How would i go about doing this question?
Any help appreciated.
Thanks

[brightsky]

A piece of wire of length 1m is bent into the shape of a sector of a circle of radius a cm and sector angle of theta.
Let the area of the sector of A cm^2.


a) How do I find A in terms of a and theta

I got A  theta/(360)  x pi(a)^2
(This is not the answer) Answer is (1/2)a^2 x theta  (LOL EDIT: NVM I GOT IT =.=)

b) Find A in terms of theta

how i do this?


At noon the captain of a ship sees two fishing boats approaching. One of them is 10km due east and travelling west at 8km/h. The other is 6km due north, travelling south at 6km/h. At what time will the fishing boats be closest together and how far apart will they be?

for b) make use of the fact that a+a+theta*r=1.

for your second question, recall the general method for solving max/min problems. find D (where D is the distance between the boats) and determine when dD/dt = 0. D should be easy to find given the direction of motion of the first ship is at right angles to that of the second ship; all you need to use is pythagoras' theorem.

[Sanguinne]

probability question
excercise 10a in maths quest 3 & 4, question 33

Tatiana is trying out for a place on the high jump team. In order to qualify she must clear three of the four heights. She knows that she has a 70% chance of clearing the first height and a 65% chance of clearing any subsequent height. What is the probability to 4 decimal places that Tatiana:
a) clears the first, third and fourth heights only?
b) clears three heights?
c) clears three heights, give she did not clear the first height?

my answer for a is correct but my answer for b is wrong.
the book;s answer is 0.3929
help is appreciated

[b^3]

Since the probability for the first height is different from the others, we will need to split the problem up.
On the first height she can either clear it (0.7) or not clear it (0.65).
Now if she is successful on her first jump for her to clear a total of 3 heights, she needs to clear 2 heights of the remaining 3, if not successful she needs to clear all three remaining heights.

Let be the event that she clears the height.
Let be the event that she clears a height after the first height.
Let be the number of heights that she clears.

Now since the heights are independent of each other, and we have a single probability for each jump, we have a binomial distribution.
~

So firstly lets look at the situation where she clears the first height.
So we have (Use the binomial function on your calc if needed).

Now lets look at the situtation where she doesn't clear the first height.
So we have (Use the binomial function on your calc if needed).

Now the total probability that we are looking for here is

[jimmy22]

A long trough whose cross-section is parabolic is 1.5 metres wide at the top and 2 metres deep. Find the depth of water when it is half full.

[darklight]

Find the value of k for which y=-0.25x + k is a normal to y= 2x2 - 8x.

I worked out the right answer for k (-21/4) however I did this by saying gradient of tangent = 4 and equating it to the gradient of the curve to find the x value. The back of book says gradient of line is -0.25 hence gradient of the NORMAL is 4. Have they made a mistake or have I?

[b^3]

A long trough whose cross-section is parabolic is 1.5 metres wide at the top and 2 metres deep. Find the depth of water when it is half full.

Plot the cross sectional area (the parabola) onto a set of axes, making it symmetric about the axis. You can find out the equation of the parabola, from the intercepts and which you can get from the question (depends on how you draw it). Then integrate to find the area between the curve and the . Now if you half this area, the trough will be half full, so you can equate the integral, but now with the terminals in terms of to this value and solve for .

Find the value of k for which y=-0.25x + k is a normal to y= 2x2 - 8x.

I worked out the right answer for k (-21/4) however I did this by saying gradient of tangent = 4 and equating it to the gradient of the curve to find the x value. The back of book says gradient of line is -0.25 hence gradient of the NORMAL is 4. Have they made a mistake or have I?

The gradient of the normal for the line will be 4, this in turn will be tangential to the parabola at that point. So when you say the gradient of the tangent, you would be referring to the tangent of the parabola, while the book is referring to the normal of the straight line.

[nspire]

hey guys!

can someone please help me with this question:

Water is draining from a cone-shaped funnel at a rate of 500cm^3/min.
The cone has a base radius of 20cm and a height of 100cm.
Let h cm be the depth of water in the funnel at time t minutes.
What is the rate of decrease of h in cm/min?

thankyou~

[Zealous]

 

hey guys!

can someone please help me with this question:

Water is draining from a cone-shaped funnel at a rate of 500cm^3/min.
The cone has a base radius of 20cm and a height of 100cm.
Let h cm be the depth of water in the funnel at time t minutes.
What is the rate of decrease of h in cm/min?

thankyou~

First we can set up different rates from the information given.

(rate of volume with respect to time)

and we want to find:
(which is the rate of change in height with respect to time)

Using the chain rule, using the rate we know ()



Volume of a cone:


Have a look at the image here, we can use triangles to see how the radius changes in terms of the height, because we have the original dimensions. (using similar triangles)






Sub it back into volume of a cone.








Bring it all back into the chain rule from before:







Hopefully I'm correct, I haven't done related rates in a while.. huhhhehe

[shadows]

V= 6pi(r)^2 + (2/3)pi(r)^3     r=4


Show that a increase of q% in the radius will result in 2.3% of the volume.

Thank you.

[EspoirTron]

I have two questions, if anyone could help that would be greatly appreciated.

y= 2x-1/x=1 (after sketching the graph which was okay); 'Hence, or otherwise, solve the inequality 0<2x-1/x+1<2 (how would you solve this?)

Secondly: f(x) = x^3 - 8x^2 -4x + 32 (after showing that f(x)= (x-2)(x+2)(x-8) ) the question asks; 'Hence solve 2^3y - 2^(2y+3) -2^(y+2) -2^5 = 0, solve for y), how would you solve this?

If anyone could help that would be greatly appreciated, thank you in advance!