VCE Methods Question Thread!

[keltingmeith]

They're claiming the answer to be k = 4.95 *10^-3

... m8, they're the same thing LOL.

hey
can someone please help me with this...
how do you determine if the point of inflection is vertical or horizontal- after using calculus to find stationary points and
nature of the graph?

thanks in advance 

Check the first derivative at that point - if it's 0, it's a horizontal point of inflection. If it's undefined, it's a vertical point of inflection.

[Stewart98]

OMG LOL, i've been entering 2^-1/140 instead of ln2^-1/140.
Rookie mistake 

[Maz]

Check the first derivative at that point - if it's 0, it's a horizontal point of inflection. If it's undefined, it's a vertical point of inflection.

Thankyou 

[Maz]

hello humans...again
i have this question...and i don't know why for some reason when i do it it isn't working...
determine the gradient function, given that y=(x-3)^3 (3x+7) by using the product and chain rule and give you answer in
a factorised form
i've done tons of these but i can't seem to get this one? i'd really appreciate any help 

[keltingmeith]

OMG LOL, i've been entering 2^-1/140 instead of ln2^-1/140.
Rookie mistake 

S'all good. In cases like these, I recommend leaking the exponent in front of the log - it looks a bit neater and helps you avoid these kinds of mistakes.

hello humans...again
i have this question...and i don't know why for some reason when i do it it isn't working...
determine the gradient function, given that y=(x-3)^3 (3x+7) by using the product and chain rule and give you answer in
a factorised form
i've done tons of these but i can't seem to get this one? i'd really appreciate any help 

Let u=(x-3)^3 and v=3x+7, then:

dy/dx = uv' + vu'

To find u', you'll then need to use the chain rule.

[Maz]

Let u=(x-3)^3 and v=3x+7, then:

dy/dx = uv' + vu'

To find u', you'll then need to use the chain rule.

hey...firstly thanks...i still can't get it though
dy/dx= ((x-3)^3*3) + (3x+7* (3(x-3)^2(1)) is what i got...but it's not equalling the derivative of the y function

Check the first derivative at that point - if it's 0, it's a horizontal point of inflection. If it's undefined, it's a vertical point of inflection.

question regarding this... for y=x^4
dy/dx= 4x^3 and the double derivative= 12x^2...makeing that equal 0 would give x=0 and thus point (0,0)
when that is put in the first derivative the answer is 0 (4*0^3) but the answer says that it isn't a point of inflection? could you please help clarify this? 

[keltingmeith]

hey...firstly thanks...i still can't get it though
dy/dx= ((x-3)^3*3) + (3x+7* (3(x-3)^2(1)) is what i got...but it's not equalling the derivative of the y function

Change that to:
dy/dx= ((x-3)^3*3) + (3x+7)*(3(x-3)^2(1))
and now it's right. Remember that you need to consider each term as a full term, hahah.

question regarding this... for y=x^4
dy/dx= 4x^3 and the double derivative= 12x^2...makeing that equal 0 would give x=0 and thus point (0,0)
when that is put in the first derivative the answer is 0 (4*0^3) but the answer says that it isn't a point of inflection? could you please help clarify this? 

Yup - you see, double derivative=0 does not ALWAYS been there's a point of inflection, it just means that the gradient is at a maximum/minimum there (in this case, it's a minimum). A quick check of the shape should let you know which it is! (however, in all cases I can think of, if it's not f(x)=(x-h)^n, then it's a point of inflection).

[Maz]

Change that to:
dy/dx= ((x-3)^3*3) + (3x+7)*(3(x-3)^2(1))
and now it's right. Remember that you need to consider each term as a full term, hahah.

Yup - you see, double derivative=0 does not ALWAYS been there's a point of inflection, it just means that the gradient is at a maximum/minimum there (in this case, it's a minimum). A quick check of the shape should let you know which it is! (however, in all cases I can think of, if it's not f(x)=(x-h)^n, then it's a point of inflection).

Thankyou again

[keltingmeith]

A new town was built 10 years ago to house the workers of a woollen mill established in a remote country area. Three years after the town was built, it had a population of 12000 people. Business in the wool trade steadily grew, and eight years after the town was built the population had swelled to 19240.

Assuming the population growth can be modelled by a linear relationship, find a suitable relation for the population, p, in terms of t, the number of years since the town was built.

I used the points (3,12000) and (8,19240) and found p = 1448t - 4336, but the answer is p= 1448t + 7656. wtf?

What do you mean wtf? You got the gradient right, so I won't bother with that calculation, but from there, if you use gradient point form, you get:

y-12,000=1448(x-3)
y=1448x+12,000-4344
y=1448x+7656

which is what they've put as the answer.

[exit]

Question a was correct but you made a few errors with question b.
When we calculate the value of a for the line PR, we already know the coordinate of P (3,0) and we know the coordinate of R (2a,a+2). With this information, we can find the gradient and solve for a. We have also been given the gradient of PR which is -2.
(a+2-0)/(2a-3)=-2
a+2/2a-3=-2
a+2=-2*(2a-3)
a+2=-4a+6
rearrange and collect like terms
a+4a=-2+6
5a=4
a=4/5

oops.. I got the original equation wrong as its y=-2x/3+2, in that case a is probably also wrong. But if you apply the same steps as I did, you should get the right answer

[eth-dog]

Hey guys, just wondering if anyone has bought Wolfram Alpha pro? Is it worth it?

[qazser]

Hey guys, just wondering if anyone has bought Wolfram Alpha pro? Is it worth it?

Use a fake email. Pi has a thread in the resources sticky

[upandgo]

hi all 

is there a trick to simplifying surds? say, if i had square root of 180, is there an easier method to simplify it to 6 sq. root of 5?

[Syndicate]

hi all 

is there a trick to simplifying surds? say, if i had square root of 180, is there an easier method to simplify it to 6 sq. root of 5?

you just have to keep simplifying until you can't

Firstly, start with a four (because sq. root 4 = 2)

180/ 4 = 45 ==> 2 sq. root 45

as you can see that 45 is a multiple of 9, you would then divide by 9

45/ 9 = 5 (sq. root 9 = 3) ==> 3x2 sq. root 5. Therefore, your final answer would be:



Well, I would say the trick in simplifying surds is that you must be good with your division.

Good Luck

[MightyBeh]

hi all 

is there a trick to simplifying surds? say, if i had square root of 180, is there an easier method to simplify it to 6 sq. root of 5?

Pretty much what syndicate said, but using prime decomposition/factorization is generally a pretty fool proof method.

Here's what I mean:

When you factorize a number with a factor tree, you will find every prime number that makes it up. You can look for perfect squares (4, 9, 16, 25, 36, etc.), but working all the way down usually doesn't take that much longer. From the tree, we can see that:


Or


So we can take the root of 180 by taking out the squares: