VCE Methods Question Thread!

[Reus]

Factorise the denominator and take out the negative in the factor (x-5) and you should end up with -1/6

Thanks so much for that, so much clearer now haha.
Could anyone assist me in this?

[Bluegirl]

Haven't been taught this so unsure on what to do/

Find the vertical distance between two points A and B defined as follows:
A is on the graph with equation y= x^2-3x-4
B is on the graph with equation y=-2x^2+3X+8
Both A and B have x-values of 0. Give an exact answer.


What are the steps to solve problems like these?

[swagsxcboi]

Haven't been taught this so unsure on what to do/

Find the vertical distance between two points A and B defined as follows:
A is on the graph with equation y= x^2-3x-4
B is on the graph with equation y=-2x^2+3X+8
Both A and B have x-values of 0. Give an exact answer.


What are the steps to solve problems like these?

We have the point and a point on the parabola will be represented by , which we can also write as . We also know that the distance between two points and is given by , so you can get that function in terms of only, then differentiate and equate the derivative to zero to minimise the function and substitute that value of back in to find the minimum value of

EDIT: Beaten.

sub in x = 0 into both equations to find out what A and B is.
then use distance between two points formula

[Bluegirl]

sub in x = 0 into both equations to find out what A and B is.
then use distance between two points formula

Thanks so much!

What about if it says 'both A and B have x-values of a. Give an answer in terms of a'?)

[keltingmeith]

It's the exact same process - however, since you won't be working with numbers (working in "a" instead), you won't get as nice of an answer.

[swagsxcboi]

Thanks so much!

What about if it says 'both A and B have x-values of a. Give an answer in terms of a'?)

A is on the graph with equation y= x^2-3x-4
B is on the graph with equation y=-2x^2+3X+8
Both A and B have x-values of 0. Give an exact answer.

just sub x=a
so your two co-ordinates will be A (a, a^2-3a-4)   B (a, -2a^2+3a+8)
then sub those into the distance formula!

[swagsxcboi]

The upper end of an 8 m ladder rests against a vertical wall with the lower end on the horizontal ground. The lower end of the ladder slips along the ground away from the wall at a rate of 6 m/s. Find the rate at which the upper end of the ladder is moving the instant the ladder is 4 m from the wall.

[Stevensmay]

The upper end of an 8 m ladder rests against a vertical wall with the lower end on the horizontal ground. The lower end of the ladder slips along the ground away from the wall at a rate of 6 m/s. Find the rate at which the upper end of the ladder is moving the instant the ladder is 4 m from the wall.

I'm defining the top point of the ladder on the vertical wall to be height H, and the distance from the wall to be L.
We also know the ladder will be constantly 8m long, and that it is moving away from the wall at a rate of 6m/s, therefore

The question is asking us to find the rate at which the height of the ladder is changing, or
Using the chain rule, we can get that
Since we can sorta pretend these are fractions, something will equal , which via cancelling will give us what we want.



All we need to do know is find an expression for , so it will involve H and L. First thing that comes to mind is Pythagoras's Theorem!


Taking the positive root as a length cannot be negative.

Now all we need to do is differentiate to get our




Going back to , we substitute in our derivatives we found to get





The question asks us to find the rate when the ladder is 4m from the wall, so L = 4.







So when the ladder is 4m away from the wall, the height of the ladder is decreasing at

[idontknow2298]

Please help! Thank you
find the quadratic function f such that f(x)=f(4)=0 and 7 is the greatest value of f(x)

[keltingmeith]

Okay, so we have a quadratic function, and we know that it has a maximum value of 7, and that f(4) = 0.

Well, since it's a quadratic function, it's highest point is either going to be infinity (if it's a positive quadratic) or at the turning point (if it's a negative quadratic). Since our maximum is 7, that must be the y-value of the turning point.

We also know that the y-value is 0 when x = 4 - we have an x-intercept when y equals 0, so we have two co-ordinates (assuming the f(x) in "f(x)=f(4)=0" is meant to have a co-ordinate. I'll henceforth refer to it as b):

(b, 0) and (4, 0). So, in intercept form:

f(x) = a(x - b)(x - 4)

We also know that the turning point will occur half-way between the two turning points - so the x-value of this turning point is (b + 4)/2 = c. So now we have one third point (c, 7). Plugging this into the equation we have so far will give you the value of a, and so you have found the quadratic.

[Frozone]

what are the sequence of transformations for this graph:

y= -2 sin(x/3) ?

[lzxnl]

what are the sequence of transformations for this graph:

y= -2 sin(x/3) ?

From y = sin x?

y = sin x => y = -sin x  or  -y = sin x       requires a reflection in the x axis. Think about it; we replace y with -y
y = -sin x => y/2 = -sin x  or  y = -2 sin x     requires a dilation by factor 2 from the x axis as we replace y with y/2
y = -2 sin x => y = -2 sin x/3   requires a dilation by factor 3 from the y axis as we replace x with x/3

[psyxwar]

13. A street light is mounted at the top of a 15-ft-tall pole. A man 6 ft tall walks away from the pole with a speed of 5 ft/s along a straight path. How fast is the tip of his shadow moving when he is 40 ft from the pole?

My answer was 10/3, actual answer was 25/3. Plus, the rate equation I got was a constant, which doesn't make any sense/ suggests it's wrong. The confusing thing is that the textbook I got this from had the same diagram in their answer with a different answer.

[Zealous]

13. A street light is mounted at the top of a 15-ft-tall pole. A man 6 ft tall walks away from the pole with a speed of 5 ft/s along a straight path. How fast is the tip of his shadow moving when he is 40 ft from the pole?

My answer was 10/3, actual answer was 25/3. Plus, the rate equation I got was a constant, which doesn't make any sense/ suggests it's wrong. The confusing thing is that the textbook I got this from had the same diagram in their answer with a different answer.

Another related rates question! Time for more fancy diagrams haha     You've done most of it correctly, but I've shown working for the whole question for others to see.



Also, you mentioned that your rate equation was constant which doesn't make any sense. It actually does make sense. He is walking away from the lamp post at a constant rate, so the rate that the tip of his shadow moves can also be constant and directly proportional to his walking rate.

[psyxwar]

Another related rates question! Time for more fancy diagrams haha     You've done most of it correctly, but I've shown working for the whole question for others to see.

(Image removed from quote.)

Also, you mentioned that your rate equation was constant which doesn't make any sense. It actually does make sense. He is walking away from the lamp post at a constant rate, so the rate that the tip of his shadow moves can also be constant and directly proportional to his walking rate.

Thanks!

I guess I was confused as to why they gave us the info if we dont use it lol. Interestingly, the textbook has the same diagram and ratios as mine, but your answer O___O

I think the question was worded pretty ambigiously when it talks about the 'tip of the shadow'. :S