VCE Methods Question Thread!

[Redoxify]

Why is it that when we find the equation of tangents, we first find the gradient at the x-value, then find the corresponding y-value and substitute them into y=mx+c?

so we are able to use the equation
we sub x into original equation to get the y value so we have (x,y)
we find the derivative of the function, we place the x value we have inside the function so that it finds the gradient at the point,
thus we can input the information we have to find the equation of the tangent

[cosine]

so we are able to use the equation
we sub x into original equation to get the y value so we have (x,y)
we find the derivative of the function, we place the x value we have inside the function so that it finds the gradient at the point,
thus we can input the information we have to find the equation of the tangent

So you're saying we find the exact equation of the tangent, that is, the line that touches the specific x-value?

What about the normal, why is it -1/m ?

[keltingmeith]

So you're saying we find the exact equation of the tangent, that is, the line that touches the specific x-value?

What about the normal, why is it -1/m ?

Because the normal is perpendicular to the tangent. I.e.,

[cosine]

Find the general solutions to 2sin(3x)=1 over [0, pi]

[cosine]

Find the general solutions to 2sin(3x)=1 over [0, pi]

How can I sketch x^4e^-x without a CAS? Would this ever be examinable, or is it too time consuming?

Cheers

[Alter]

Find the general solutions to 2sin(3x)=1 over [0, pi]

The wording of this question is a bit weird. It should either be 'find the general solution', 'find the general solution to [equation], and hence the x-intercepts in the domain [0, pi]', or just "find the solutions to [equation]". Which part are you having trouble with?

And the x^4e^-x graph would never be examinable without a CAS imho. Way too difficult and I don't think that's on the methods course.

[knightrider]

How come when you solve for y on the CAS you get and

whereas by hand if you solved for y you would get

Why does the CAS get 2 answers and how?

[cosine]

The wording of this question is a bit weird. It should either be 'find the general solution', 'find the general solution to [equation], and hence the x-intercepts in the domain [0, pi]', or just "find the solutions to [equation]". Which part are you having trouble with?

And the x^4e^-x graph would never be examinable without a CAS imho. Way too difficult and I don't think that's on the methods course.

Im having troubles with the actual general solution part, i can solve for the x values but not sure how to put it into general solutions. Thanks

[silverpixeli]

Im having troubles with the actual general solution part, i can solve for the x values but not sure how to put it into general solutions. Thanks

Full marks for general solutions questions:

~sine and you~

<find first 2 solutions like it's a normal question>

... working out ...

x=solution1, 1=solution2

<general solution part, add nP to both solutions to make them general>

x=solution1+nP, x=solution2+nP,

where P is the graph's period.



~tangent~

<one solution per period instead of 2, so just find the first one>

... working ...

x=solution1

<general solution part: add nP to solution>

x=solution1+nP,

again, P is the period of the trig function



That's the ONLY difference between finding the first 2 solutions and finding every solution (general solution). The idea is, you're giving them ALL the possible solutions to the equation, in a succinct form. To get the first two solutions, the let n=0. To get the next ones, they add n=1 periods to each. To get the next next ones, they can add n=2 periods. To get the ones before the first cycle, they can add n=-1 periods. By saying +nP, you are listing every possible integer number of n that they can choose and hence every possible amount of periods to add and thus every possible solution to the equation.

Feel free to ask for clarification on this, it's a weird concept to grasp since there's really nothing else this 'general' in the course.

How come when you solve for y on the CAS you get and

whereas by hand if you solved for y you would get

Why does the CAS get 2 answers and how?

So you probably introduced the log a step too early to see the second solution.








compare with the equivalent working










which reduces to what the calculator got. In the negative case, you leave it all inside the log, and in the positive case, there's no annoying minus sign so you can bring the 1/4 out the front of the log.


HOWEVER, the first solution from CAS is a bit fishy imo,



, like a square root, will never be negative, because even roots only return positive powers or are undefined if their argument is negative.

is bound to be undefined. I dont even thing this second answer is valid.

[silverpixeli]

How can I sketch x^4e^-x without a CAS? Would this ever be examinable, or is it too time consuming?

Cheers

assuming not

You could try using multiplication of ordinates, but your graph won't be very precise without a CAS. multiplication of ordinates can give you the rough shape though, like you'll notice that it touches the x and y axis at (0,0) and has an asymptote in the positive region (probably after a bump where it has a local maximum which you could find with the product rule if you felt inclined), and grows stupidly fast in the negative region.

Find the point(s) on the arc of the parabola y = x^2, x E [0, 1] which are nearest to the point (0, m).

Thanks

There are two ways to approach minimising distance equations.

One is to set up an equation of the distance between the two points:
-(0,m) is your first point
- is your second point, and you'll be looking for a solution in the range
(do you see why the second point is (x,x^2)?)

the distance between two points in 2D is:

from pythagoras' theorem

so for us:

(it doesn't matter which is point 1 or 2)



This is d, a function of x and m. Since m is just some constant and we're interested in changing x, we can minimise this distance using calculus;

, find and make sure it's in
(maybe i should have used capital D for distance, dd/dx looks silly but hopefully you can see what i mean)

Once you find this minimum's x value, you have your answer



The other way is to find when the gradient of the curve is perpendicular to the gradient of the line between the two points. Draw a diagram and you'll see that if the curve is the closest/furthest it can be from a point, these gradients MUST be perpendicular. If they are not, that suggests that if you more a little one way you will become closer/further so you're not yet at a max/min of distance.
tl;dr: the closest a point gets to a curve is when the gradient of the curve is perpendicular to the gradient to the point.

The gradient of the curve is

The gradient of the line between the curve and the point (0,m) is



again it doesn't matter which we pick to be point 1 or two as long as we're consistent for x and y

To solve the problem, let these two gradients be perpendicular (m1*m2=-1 or m1=-1/m2):



Solve that for x in the range 0 to 1 should work just as well as the first method

Good luck!

[99.90 pls]

There are two ways to approach minimising distance equations.

One is to set up an equation of the distance between the two points:
-(0,m) is your first point
- is your second point, and you'll be looking for a solution in the range
(do you see why the second point is (x,x^2)?)

the distance between two points in 2D is:

from pythagoras' theorem

so for us:

(it doesn't matter which is point 1 or 2)



This is d, a function of x and m. Since m is just some constant and we're interested in changing x, we can minimise this distance using calculus;

, find and make sure it's in
(maybe i should have used capital D for distance, dd/dx looks silly but hopefully you can see what i mean)

Once you find this minimum's x value, you have your answer



The other way is to find when the gradient of the curve is perpendicular to the gradient of the line between the two points. Draw a diagram and you'll see that if the curve is the closest/furthest it can be from a point, these gradients MUST be perpendicular. If they are not, that suggests that if you more a little one way you will become closer/further so you're not yet at a max/min of distance.
tl;dr: the closest a point gets to a curve is when the gradient of the curve is perpendicular to the gradient to the point.

The gradient of the curve is

The gradient of the line between the curve and the point (0,m) is



again it doesn't matter which we pick to be point 1 or two as long as we're consistent for x and y

To solve the problem, let these two gradients be perpendicular (m1*m2=-1 or m1=-1/m2):



Solve that for x in the range 0 to 1 should work just as well as the first method

Good luck!

Thanks Pixeli! I get all of that, but here's my dilemma... I'm told that after the magnitude of m exceeds a certain point (i.e. It gets too large or too small), the endpoints of the parabola become the closest points. I'm just having a hard time figuring out at which values of m this occurs.

[cosine]

Full marks for general solutions questions:

~sine and you~

<find first 2 solutions like it's a normal question>

... working out ...

x=solution1, 1=solution2

<general solution part, add nP to both solutions to make them general>

x=solution1+nP, x=solution2+nP,

where P is the graph's period.



~tangent~

<one solution per period instead of 2, so just find the first one>

... working ...

x=solution1

<general solution part: add nP to solution>

x=solution1+nP,

again, P is the period of the trig function



That's the ONLY difference between finding the first 2 solutions and finding every solution (general solution). The idea is, you're giving them ALL the possible solutions to the equation, in a succinct form. To get the first two solutions, the let n=0. To get the next ones, they add n=1 periods to each. To get the next next ones, they can add n=2 periods. To get the ones before the first cycle, they can add n=-1 periods. By saying +nP, you are listing every possible integer number of n that they can choose and hence every possible amount of periods to add and thus every possible solution to the equation.

Feel free to ask for clarification on this, it's a weird concept to grasp since there's really nothing else this 'general' in the course.







So you probably introduced the log a step too early to see the second solution.








compare with the equivalent working










which reduces to what the calculator got. In the negative case, you leave it all inside the log, and in the positive case, there's no annoying minus sign so you can bring the 1/4 out the front of the log.


HOWEVER, the first solution from CAS is a bit fishy imo,



, like a square root, will never be negative, because even roots only return positive powers or are undefined if their argument is negative.

is bound to be undefined. I dont even thing this second answer is valid.

Thanks so much!

So to state the general solutions for a graph, we have to use P as the period of the graph, instead of 2pi? Also what about its just a straight forward question just asking us to state the general solutions, do we also add 2pi to every solution or does it depend on the period ??

[silverpixeli]

Thanks Pixeli! I get all of that, but here's my dilemma... I'm told that after the magnitude of m exceeds a certain point (i.e. It gets too large or too small), the endpoints of the parabola become the closest points. I'm just having a hard time figuring out at which values of m this occurs.

Ah good question. I didn't think of that, but you're definitely right! In both methods I've kind of assumed that your answer will be in the range [0,1].

for example whenever m is less than 0, the point (0,m) will be closest to the endpoint (0,0) (you can see this on a graph if you have for example (0,-3).
Graphically, even if it's not picked up by the formulas because of a zero-division, the line connecting (0,-3) and (0,0) is a vertical line (undefined gradient) which is perpendicular to the gradient of the curve at (0,0), namely 0
So when m is less than 0, the lower end point is the closest.


In the other extreme, we're looking at when m gets so large that the closest point would be at some value of x beyond 1. Graphically, you can imagine x so high that the closest point is just the highest point on the curve (1,1) and the shape doesn't even matter any more. This corresponds to the case where you try a value of m and get an x value of 1. If you try any m greater than that, you'll get a value of x that is greater than 1, which clearly isn't allowed, so the closest you can be to the point must actually be (1,1)

Thanks so much!

So to state the general solutions for a graph, we have to use P as the period of the graph, instead of 2pi? Also what about its just a straight forward question just asking us to state the general solutions, do we also add 2pi to every solution or does it depend on the period ??

You dont add 2pi unless the period is 2pi. Think about it using a graph. The period is defined as the horizontal shift before the cycle repeats itself. If you therefore move your solutions sideways by a multiple of the period, you're guaranteed to land on another value that satisfies the equation you're solving. If you just go around adding 2pi to every solution, you may miss some intercepts or not even land on solutions at all. For example, consider the graph y=sin(pix/6), and say we want to find when this graph = 1/2

sketch

This corresponds to the trig equation 1/2=sin(pix/6)
Solving normally for the solutions in the first cycle, we find that x=1 and x=5

But the next two solutions aren't at 1+2pi and 5+2pi, they're at 1+12=13 and 5+12=17 because the period is 2pi/(pi/6)=12

Likewise, the next next solutions would be at 1+2*12 and 5+2*12, and you can keep going

In general, they're at 1+n*12 and 5+n*12, for all different integers n

We had to find the first two solutions like it was a normal question, and then look for more




You may be finding confusion because there's another way to solve these questions that involves adding 2pi to something. But it's at a different stage. If you're on the way to finding your values of x, at one point you may get pix/6=something1 and pix/6=something2
If you add 2pi*n to both solutions at this point, and then solve for x later, you're going to transform that +2pi*n into a +P*n and it all works out

[lzxnl]

assuming not

You could try using multiplication of ordinates, but your graph won't be very precise without a CAS. multiplication of ordinates can give you the rough shape though, like you'll notice that it touches the x and y axis at (0,0) and has an asymptote in the positive region (probably after a bump where it has a local maximum which you could find with the product rule if you felt inclined), and grows stupidly fast in the negative region.





There are two ways to approach minimising distance equations.

One is to set up an equation of the distance between the two points:
-(0,m) is your first point
- is your second point, and you'll be looking for a solution in the range
(do you see why the second point is (x,x^2)?)

the distance between two points in 2D is:

from pythagoras' theorem

so for us:

(it doesn't matter which is point 1 or 2)



This is d, a function of x and m. Since m is just some constant and we're interested in changing x, we can minimise this distance using calculus;

, find and make sure it's in
(maybe i should have used capital D for distance, dd/dx looks silly but hopefully you can see what i mean)

Once you find this minimum's x value, you have your answer



The other way is to find when the gradient of the curve is perpendicular to the gradient of the line between the two points. Draw a diagram and you'll see that if the curve is the closest/furthest it can be from a point, these gradients MUST be perpendicular. If they are not, that suggests that if you more a little one way you will become closer/further so you're not yet at a max/min of distance.
tl;dr: the closest a point gets to a curve is when the gradient of the curve is perpendicular to the gradient to the point.

The gradient of the curve is

The gradient of the line between the curve and the point (0,m) is



again it doesn't matter which we pick to be point 1 or two as long as we're consistent for x and y

To solve the problem, let these two gradients be perpendicular (m1*m2=-1 or m1=-1/m2):



Solve that for x in the range 0 to 1 should work just as well as the first method

Good luck!

First note that the function decays to 0 for large x and that it blows up to infinity for negative x. Then, differentiate the function to find turning points. Classify them. That should give you a pretty solid idea of what the graph looks like.

[99.90 pls]

Thanks silverpixeli!