Geometrical Application of differention (1 Viewer)

[atar90plus]

Hello
I want to ask a few Q's relating to Differentiation (Geometrical App). First what is the difference between a POI or a HPOI? How do you tell which one is which?
Also a few problems I need help with

1. Consider the curve : y=x^4-8x^3
This question asks you to find the POI. The answers are (0,0) and (4,-256) I understand where the point (0,0) comes from but could you guys explain to me where the point (4,-256) is there

2. The line y=x+1 is a tangent to the curve y=x^2+3x+2 . What is the point of contact

4. A cubic curve has a minimum turning point at the point (-2,-15), a maximum turning point at (1,12) and it cuts the y-axis at 5. What is the equation of the cubic?

5. Find the smallest distance between the graphs of y=x^2-4x+12 and y=2x+1

Also could you guys please give an example question and solution of how to find a POI.

Thanks to whoever contributes

 

[Kimyia]

Hello
I want to ask a few Q's relating to Differentiation (Geometrical App). First what is the difference between a POI or a HPOI? How do you tell which one is which?
Also a few problems I need help with

1. Consider the curve : y=x^4-8x^3
This question asks you to find the POI. The answers are (0,0) and (4,-256) I understand where the point (0,0) comes from but could you guys explain to me where the point (4,-256) is there

2. The line y=x+1 is a tangent to the curve y=x^2+3x+2 . What is the point of contact

4. A cubic curve has a minimum turning point at the point (-2,-15), a maximum turning point at (1,12) and it cuts the y-axis at 5. What is the equation of the cubic?

5. Find the smallest distance between the graphs of y=x^2-4x+12 and y=2x+1

Also could you guys please give an example question and solution of how to find a POI.

Thanks to whoever contributes

1. You get the second derivative of 12x^2 - 48x. You can factorise that to 12x(x-4). Make that equal to zero to find your POIs and you arrive at the x-coordinates 0 and 4 as POIs.

 

[AAEldar]

1) Point of inflexions occur when the second derivative equals zero. After differentiating you get:





So here is where the x=4 comes from, then you sub that into the original equation and get the y value of -256.

2) Equate them:











Subbing into both equations we find that y=0, showing that it touches and only once.

4) Set up a general equation of a cubic, i.e:



We know that max and min occur when the first derivative equals zero, so we have to find the derivative:



We have two points, and four unknowns. For the max and min, y'=0 and we sub in the x values and get the 2 equations:









And we still need two more equations. We have x and y values we can sub into the original equation as we know it passes through those points, and can make them:









Then we can solve those simultaneously.



Have to go, hopefully these are correct and someone can give you help with the rest.

 

[asianese]

POI: When the concavity changes. . The first derivative need not be 0.
HPOI: When the concavity changes AND the gradient of the tangent is 0. That is At that point. This is also known as a saddle point (I'm pretty sure - not in the syllabus, though). It is the middle of where the concavity is changing and the gradient of the tangent is 0 at x=0, for example.

 

[Carrotsticks]

POI: When the concavity changes. . The first derivative need not be 0.
HPOI: When the concavity changes AND the gradient of the tangent is 0. That is At that point. This is also known as a saddle point (I'm pretty sure - not in the syllabus, though). It is the middle of where the concavity is changing and the gradient of the tangent is 0 at x=0, for example.

Saddle point isn't the best term to use, since it is more often used in Multi-Variable Calculus (with x,y,z).