How do you prove something is tangetnt to something else? (1 Viewer)

[pomsky]

http://imgur.com/RqTjgFG

 

[pomsky]

Also in Q 8a.)
http://imgur.com/2lqcSbP
But the answer quotes the formula P = Ae^(kt)
http://imgur.com/b4B0CHO

Are you allowed to do that? And if you're not, how do you get rid of the C when integrating?
Then stuck. lol

 

[pomsky]

And in Q8b.) (lols soz for all the qs guys.)
Question: http://imgur.com/tqKRSCe

Answers: http://imgur.com/Q86nRgC

I get that for f(x) to be increasing, f'(x) >0, but why did they jump to the discriminant being less than 0? How do they know it's not a concave down curve?

TY Also how do I make these images actually images (to save people from clicking around) ? ahaha

 

[pomsky]

These are all from the 2010 BOSTES paper, btw.

 

[pomsky]

LOL
there's one more : How do you find max value of f(x) in Q9 bii? I know how to read the graph for x values, but not exactly sure how to for y values. The f(0)= 0 is probably there to help with the integration constant but there's no equation so I'm a bit stuck.
Q:
http://imgur.com/z0iYTel
Ans: f(x) = 4

 

[Crisium]

And in Q8b.) (lols soz for all the qs guys.)
Question: http://imgur.com/tqKRSCe

Answers: http://imgur.com/Q86nRgC

I get that for f(x) to be increasing, f'(x) >0, but why did they jump to the discriminant being less than 0? How do they know it's not a concave down curve?

TY Also how do I make these images actually images (to save people from clicking around) ? ahaha

There are two ways to prove that it is increasing for a certain value:

1) Use the first derivative method

2) Prove that it is a positive definite (i.e. The discriminant is zero but the coefficient of x^2 is positive - It will look like a concave up parabola above the x-axis with no roots)

 

[vitamin D]

There are two ways to prove that it is increasing for a certain value:

1) Use the first derivative method

2) Prove that it is a positive definite (i.e. The discriminant is zero but the coefficient of x^2 is positive - It will look like a concave up parabola above the x-axis with no roots)

so (triangle)>0 ?

 

[Crisium]

so (triangle)>0 ?

It will have no roots so triangle/discriminant < 0

Have a look at this:

https://www.google.com.au/search?q=...sSExMLQyAIVxJCUCh3ZBwBN#imgrc=IC8t-EQ5DvhGFM:

 

[qwert73]

for the tangent question you just equate the two equations to make x^2-4x+4=0. you then show that the discriminant is equal to 0 which means that there is only one root and therefor one point of intersection which means that the line is a tangent to the curve.

 

[vitamin D]

It will have no roots so triangle/discriminant < 0

Have a look at this:

https://www.google.com.au/search?q=...sSExMLQyAIVxJCUCh3ZBwBN#imgrc=IC8t-EQ5DvhGFM:

i always assumed (triangle)>0 was 2 solutions, (triangle)<0 was no solutions and (triangle)=0 was one solution

 

[pomsky]

i always assumed (triangle)>0 was 2 solutions, (triangle)<0 was no solutions and (triangle)=0 was one solution

What you assume is correct. It's also what Crisium is saying.

And gosh AHAHAHAHA forgot about the coefficient there. fanks

 

[pomsky]

for the tangent question you just equate the two equations to make x^2-4x+4=0. you then show that the discriminant is equal to 0 which means that there is only one root and therefor one point of intersection which means that the line is a tangent to the curve.

Which two equations? For some reason I got BT : y = 2x and when that equates with parabola, discriminant =4.
But good methodology TY TY TY

 

[pomsky]

bump for all the other Qs

 

[qwert73]

for question 8a it is like the questions when they say prove dP/dt = kP, you just work backwards to find the original equation and then do the question from there

for question 8b for get the first derivative and then you use quadratic equation to find the roots(or the discriminant which is the important part). if the curve has a discriminant greater than zero it means that there are 2 roots(x-intercepts) which we don't want because that means that at some values of x the curve is below the x axis which means it is decreasing at those points(as this is the 1st derivative). therefor we need the discriminant to be less than zero so that there are no intercepts which means that the graph is always above the x-axis and that it is always increasing.

 

[qwert73]

Which two equations? For some reason I got BT : y = 2x and when that equates with parabola, discriminant =4.
But good methodology TY TY TY

your 2 lines should be y=x^2(the curve) and y=4x-4(the line you just found)

 

[InteGrand]

Also in Q 8a.)
http://imgur.com/2lqcSbP
But the answer quotes the formula P = Ae^(kt)
http://imgur.com/b4B0CHO

Are you allowed to do that? And if you're not, how do you get rid of the C when integrating?
Then stuck. lol

 

[pomsky]

LOL
there's one more : How do you find max value of f(x) in Q9 bii? I know how to read the graph for x values, but not exactly sure how to for y values. The f(0)= 0 is probably there to help with the integration constant but there's no equation so I'm a bit stuck.
Q:
http://imgur.com/z0iYTel
Ans: f(x) = 4

Does anyone know about this q here?

 

[keepLooking]

Does anyone know about this q here?

You know the area of the the derivative graph is the total distance traveled. It is 4 because that is the largest (positive) area ?

 

[pomsky]

You know the area of the the derivative graph is the total distance traveled. It is 4 because that is the largest (positive) area ?

Woop, now I do. TY x)