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This question seemed easy at first, except I keep getting stuck.
A curve with equation of the form y= ax^3 + bx^2 + cx + d has zero gradient at the point (1/3,4/27), and also touches, but does not cross, the x-axis at the point (1,0). Find a,b,c,d.
I suppose you differentiate and substitute the coordinates in, except there never seems to be enough information supplied.
Thanks! :)
I believe you have enough information. You have 4 unknowns and thus need 4 independent equations relating the unknowns:
1. The point P1 = (1/3,4/27) is on the curve
2. The gradient at P1 is zero: y' = 0 @ P1
3. The point P2 = (1,0) is on the curve
4. The gradient at P2 is zero because it touches the x-axis but doesn't cross: y' = 0 @ P2
With that I've found 4 equations:Quote:
Originally Posted by VBAhack
4 = a + 3b + 9c + d
0 = a + b + 3c
0 = a + b + c + d
0 = 3a + 2b + c
I'm not really sure how to put them together to solve for the unknowns. I don't know what to substitute into what, or what to subtract from what. I seem to be going in circles!
Hmmm, there are a couple of ways to approach it. One is to use matrix algebra. The coefficients of a, b, c, d yield a 4 x 4 matrix. Pre-multiplying the vector [4, 0, 0, 0] by the inverse of the matrix will yield the values of a, b, c, d. If you haven't had linear/matrix alebra, then old fashioned elimination will be needed. For example, subtracting the 2nd equation from the 3rd will result in an equation with 2 unknowns (c and d). Combining other equations will eventually allow you to find the value of 1 variable, then another, then another, until you have them all.
This might help (google search on 'simultaneous linear equations'):
http://www.themathpage.com/alg/simul...-equations.htm
P.S. If you have Excel, the matrix algebra approach is a snap, as it has built in matrix functions (MINVERSE, MMULT). If this is for a class, then your book must have a section on how to solve systems of equations. Looks like you had another post for a similar problem. Same methodology applies.
http://www.vbforums.com/showthread.php?t=383293
Good luck.
yikes, vectors aren't part of the course i'm doing. i know how to do simultaneous equations, except i can't seem to achieve anything by using them.
You may also wish to remember that if you do a second differential it can tell you whether you have a maximum, minimum or point of inflexion. As this is a cubic, you'll have one maximum and one minimum...
The idea of solving systems of equations using elimination (I believe it is attributed to Gauss) is to combine equations (add or subtract multiples if needed) to progressively eliminate variables until you can solve for one of the variables. Then, successive back substitutions will yield values for the other variables.Quote:
Originally Posted by freswood
In this case it is quite easy because you can eliminate 2 variables in one step. If you subtract equation (3) from (2) you get an expression relating d and c. Substituting this expression for d (thus eliminating d) back into the 4 original equations will result in 3 equations (2 of the 4 will be identical) in a, b, c.
If you then subtract the new equation (2) from the new equation (1), you get an expression for b in terms of c. Substituting this expression for b (thus eliminating b) back into the 3 equations yields 2 equations (2 of the 3 will be identical) in a and c.
With 1 of the 2 remaining equations, you can find c in terms of a. Substituting this expression for c (thus eliminating c) in the 2nd equation will give you the value for a. Knowing a you can find c from the 1st of the 2 remaining equations. Knowing a and c will allow you to find b and d using the previous equations.
This is pretty tedious stuff, but it does work. :thumb:
Thanks so much for your help! :) You're right, it is tedious. It'd be awful coming across something like this in an exam.