Computer Science Canada

Fun Calculus Problem
Author:  Panphobia [ Fri May 10, 2013 12:02 pm ]
Post subject:  Fun Calculus Problem
So on my test there were a couple pretty fun questions
code:
There is a function y = ax^2 + bx + c , if the x intercepts are 0 and 8, and the slope of the tangent at x =2 is 16, what are a, b, c
I got a = -4, b = 32, c =0. The second one
code:
The function y=x^3+3 has a tangent at (1,4) this tangent also intersects at another point P, find point P
that one was pretty easy but fun, my friend was convinced that I got it wrong, but I am pretty sure I am right saying that, P = (-2,-5) because this is what I did
code:
The derivative function d/dx = 3x^2 so the slope of the line is 3(1)^2, then we know that y = mx + b so you sub (1,4) into that, and also you sub the slope, so 4 = 3 + b, and b = 1, so y = 3x + 1, now to get the intersection point you make y = y,
3x + 1 = x^3 + 3
0 = x^3 - 3x + 2
0 = (x-1)(x-1)(x+2)
since we already know x=1 is an intersection point we find that -2 is the only different one
so we sub -2 into y = x^3 +3 and get y = -5 so P = (-2,-5)
am I right?
Author:  Raknarg [ Sat May 11, 2013 11:55 am ]
Post subject:  RE:Fun Calculus Problem
Yeah I got the same answers for both
Author:  Panphobia [ Mon May 27, 2013 12:08 pm ]
Post subject:  Re: Fun Calculus Problem
Ok so I wrote another calculus test recently, and the thinking question was similar to that of the previous here it is
code:
The function f(x) = ax^3 + bx^2 + c goes through point (2,-3) and has an inflection point at (1,1),with this information what are the values of the constants a,b,c
I got answers that
code:
a = 2, b =  -6, and c = 5
I was just wondering if I got it right, thank you and that is all.
Author:  d310 [ Mon May 27, 2013 2:52 pm ]
Post subject:  Re: Fun Calculus Problem
I agree with the answer.
Author:  Panphobia [ Mon May 27, 2013 6:00 pm ]
Post subject:  RE:Fun Calculus Problem
How the hell would I answer a question like this
code:
Consider the function f(x) = ax^3 + bx^2 + cx + d
what is the equation to the tangent of the only inflection point?
How would I get one single answer for this? Without it being crazy with variables all over?
Author:  Dreadnought [ Mon May 27, 2013 6:24 pm ]
Post subject:  Re: Fun Calculus Problem
Well first you solve for the inflection point (you should know how to do this).

Then you find the slope of the tangent at that point. You probably also know that the slope is the first derivative evaluated at the inflection point.
So substitute in the value of x at the inflection point and simplify it into something relatively nice (2 terms).

Then finding the equation of the tangent line should be easy (you know the slope and the y-intercept is pretty obvious).
Author:  Panphobia [ Mon May 27, 2013 6:31 pm ]
Post subject:  RE:Fun Calculus Problem
I mean, I know how to solve it easily with all those variables in the equation, but I mean to solve for a definite line, is that possible?
Author:  Dreadnought [ Mon May 27, 2013 6:51 pm ]
Post subject:  Re: Fun Calculus Problem
Well the line does depend on the values of a,b,c, and d so you can't get a single line for all values (that would imply all tangents to a cubic are a single line). But you can give the tangent line in a form that depends on a,b,c and d. Then if you knew a,b,c and d you would just substitute in the values and have a nice equation.
Author:  Panphobia [ Mon May 27, 2013 6:53 pm ]
Post subject:  RE:Fun Calculus Problem
yea thats what I did, because it said find an equation of the tangent to that function. The question on the test was worded weird but yea, I did it that exact way. and the equation of the tangent to the inflection point does not depend on d, only a, b and c Razz. This is because when you take the derivative(the tangent to any point on the graph), the d value disappears because it is like f(x) = dx^0 f'(x) = 0dx^-1
Author:  Dreadnought [ Mon May 27, 2013 7:00 pm ]
Post subject:  Re: Fun Calculus Problem
The slope of the tangent line does not depend on d but the equation does. Remember that to define a line in the Cartesian plane, you need both a slope and a point on the line (and this point will depend on d).

EDIT:
Maybe it's this point that's bugging you.
You might be used to giving the equation of the tangent line in the form y = mx + b (m is the slope, b is the y intercept), however b isn't necessarily nice in this case. So I give you the linear approximation to f at the point a. L(x) = f'(a)*(x-a) + f(a) where a is point around which you want to approximate (in this case the inflection point).

L(x) is the tangent line to f(x) at the point x = a.
Does this make it easier?

EDIT to the edit: You'll notice that this is actually what you would get if you solved for b and substituted into y = mx + b.
EDIT 3: Ok I'm feel I need to show you where L(x) comes from.
code:
You want a line with a slope of f'(a) going through f(a) where a is the inflection point.
We find the y-intercept of this line, b = f(a) - f'(a)*a   <--- comes from f(a) = f'(a)*a + b since f(a) is on the line y = f'(a)*x + b
Substitute into L(x) = f'(a)*x + b
L(x) = f'(a)*x + f(a) - f'(a)*a
L(x) = f(a) + f'(a)*(x-a)


Edit 4: Does this help?
Author:  Panphobia [ Mon May 27, 2013 7:30 pm ]
Post subject:  RE:Fun Calculus Problem
Yea I understand it, but I got the question right on the test anyway Razz
Author:  Dreadnought [ Mon May 27, 2013 7:33 pm ]
Post subject:  Re: Fun Calculus Problem
Oh, ok well good then! (I needed something to distract me from boring homework anyway)
Author:  Panphobia [ Mon May 27, 2013 7:52 pm ]
Post subject:  RE:Fun Calculus Problem
this unit was probably the unit where I had to do some work, next unit is optimization, I think it is probably going to be the easiest unit of the semester Razz have you taken it already?
Author:  Dreadnought [ Mon May 27, 2013 10:04 pm ]
Post subject:  Re: Fun Calculus Problem
Oh yes I've taken it, a couple years ago... I recently finished second year math at U of Waterloo. Razz

Optimization is a great application of the calculus you learn in high school. It's much easier than anything you've learned prior (like completing the square to find the vertex of a parabola).

Considering you were asking about a tricky differential equation the other day I'd say you should be fine for optimization. (if you teacher gave you that question on a test I find that kinda evil...)
Author:  Panphobia [ Mon May 27, 2013 10:09 pm ]
Post subject:  RE:Fun Calculus Problem
my teacher is evil, he puts like university integration questions on the tests, when in reality integration isn't a part of the course. Anything that is not in the course syllabus you are not allowed to assess on. Sooooooo,....
Author:  Dreadnought [ Mon May 27, 2013 10:22 pm ]
Post subject:  Re: Fun Calculus Problem
Well it's not bad to learn more. In my opinion the best teachers I had in high school were the ones that gave me the most work. My calculus teacher, for example taught integration near the end of the course.

Nevertheless, I still feel that question was evil for a high school test.
Author:  Panphobia [ Sat Jun 08, 2013 4:39 pm ]
Post subject:  RE:Fun Calculus Problem
I just had another test in calculus and this is my last before the exam, it is in optimization, I found my thinking question too easy so I asked another class for theirs and this is what it was
code:
You are given 60 cm of wire, some of this wire is to be used for an equilateral triangle and the rest should be used for a circle, what are the dimensions fof both the circle and the rectangle such that their combined area is a minimum?
I got the radius = 0.454 cm and one side of the triangle = 19.05, did I get it right?
Author:  Dreadnought [ Sat Jun 08, 2013 5:47 pm ]
Post subject:  Re: Fun Calculus Problem
I'm pretty sure the answer is x = 12.4642 and r =3.5981 where x is the length of one side of the equilateral triangle and r is the radius of the circle.
Author:  Panphobia [ Sat Jun 08, 2013 7:32 pm ]
Post subject:  Re: Fun Calculus Problem
How did you get that? This is what I did
code:
Since we know the wire is 60cm that is one constraint, so
60 = 2(PI)r + 3a
I am going to be using the equation for the area of a triangle that only needs the three sides so here that is
area of triangle = sqrt(p(p-a)^3)
p = perimiter /2 = 3a/2
area of triangle = sqrt((3a/2)(3a/2-a)^3) = sqrt((3a/2)(a^3/8)) = sqrt(3a^4/16)
area of circle = (PI)r^2
A = sqrt(3a^4/16)+(PI)r^2

from there just sub in and differentiate, please tell me if I did something wrong, and I know you can use Pythagorean theorem for the height of the triangle also, but I chose to do it this way, anyway if I did do anything wrong could you show me your solution?
Author:  Dreadnought [ Sat Jun 08, 2013 8:53 pm ]
Post subject:  Re: Fun Calculus Problem
The formula you give for the area is correct and it has a minimum at a = 12.4642.

What do you differentiate with respect to, a or r?

EDIT: If we substitute for r, here's a plot of the area (this url doesn't play nice with the tags apparently...)

http://www.wolframalpha.com/input/?i=plot+sqrt%283%29*%28x%5E2%29%2F4+%2B+pi*%5B%2860+-+3*x%29%2F%282*pi%29%5D%5E2
Author:  Panphobia [ Sat Jun 08, 2013 9:18 pm ]
Post subject:  RE:Fun Calculus Problem
yea I differentiated wrong, haha when it came around to it I didn't do it right, doh! I also have a similar worded question and I am not sure what the second part is saying so
code:
You are given 100 cm of wire to create a circle and a square, what are the dimensions of the circle and square such that their total area is a minimum? a maximum?
I got a minimum but when he said maximum, did he just mean to put all the wire into the circle or all the wire into the square? Oh and by the way I think I got like 7 cm as the radius and 14 cm as one side to the square, forgot, did this a while ago.
Author:  Dreadnought [ Sat Jun 08, 2013 10:21 pm ]
Post subject:  Re: Fun Calculus Problem
Panphobia wrote:
yea I differentiated wrong, haha when it came around to it I didn't do it right, doh!

Happens a lot, when I know my answer isn't right, I usually take a break then start again from scratch later and compare.
Panphobia wrote:

I got a minimum but when he said maximum, did he just mean to put all the wire into the circle or all the wire into the square?

You're right, the answer is to use all the wire for one shape, but which shape? The second part can be done with any differentiation if you know how to compute perimeters and areas of circles and squares.

EDIT: Also 7 and 14 looks good to me.
Author:  Panphobia [ Sat Jun 08, 2013 10:33 pm ]
Post subject:  RE:Fun Calculus Problem
Oh thank the lord, that was my test question, my other one was pretty damn easy it was basically
code:
two different lines
y' = -x/3 + 30
y = x
you want to find the maximum area box you can make where the width is along the y axis and the two other vertices touch the two lines
that was the best I could do without a picture but basically what I did was Area = xy and that the width is equal to y' - y sooooo A = x(-x/3 + 30 - x), is that correct at all?
Author:  Dreadnought [ Sat Jun 08, 2013 11:31 pm ]
Post subject:  Re: Fun Calculus Problem
Well, if my understanding of how the box is formed is correct then that looks like it would be the area.

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