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The question asks to find a cubic equation given point A (0/18) slope here is 0. Point B only given the x=20 and slope is -0.6.

Please help, been trying to solve it for so long now. :(

1.) y=ax^3+bx^2+cx+d

2.) y'=3ax^2+2bx+c

-0,6=1200a+40b+1c

18=0a+0b+0c+1d

priyanka
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  • Welcome to Math.SE. Please show us what you have tried so far and where you are stuck and in what context you have encountered the problem , i.e edit your post with this information. This site uses MathJax formatting – JKnecht Mar 12 '16 at 13:25
  • Just to be clear, you are told that $y(0) = 18$, $y'(0) = 0$, and $y'(20) = -0.6$. Is that correct? If so, that's not enough information to uniquely define a cubic, but you could find some cubic. – bubba Mar 12 '16 at 13:41
  • 20 is the distance from Point A and on that point the slope is -0,6. (German native speaker, so excuse my bad english please) – priyanka Mar 12 '16 at 13:52
  • the equation should be y(x)=0,0005x^3-0,03x^2+18 – priyanka Mar 12 '16 at 13:55

2 Answers2

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You have too little information, but the general method is:

Write the general equation for a cubic, differentiate it, and plug in the information you have in those two expressions. That will give you a set of equations to solve.

But the information you mention will only give three equations with four unknonwns.

And you haven't used that the slope at $0$ is $0$.

Henrik supports the community
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What you wrote is not very clear. But, if I understand it correctly, you are only given three pieces of information -- one function value and two slopes. From this, you won't be able to calculate unique values for the four unknowns $a,b,c,d$.

But, as the comment from @Justin points out, you could find some values of $a,b,c,d$ that will work: \begin{align} y(0) &= 18 \; \Rightarrow \; d = 18 \\ y'(0) &= 0 \; \Rightarrow \; c = 0 \\ y'(20) &= -0.6 \; \Rightarrow \; 1200a+40b+c = -0.6 \end{align} From the third equation, we get $1200a+40b = -0.6$. We can just set $a=1$, arbitrarily, and then we get $b = -1200.6/40 = -30.015$.

So, a suitable cubic (but not the only one) is: $$ y = x^3 -30.015x^2 + 18 $$

bubba
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  • thank you, but the answer should be y(x)=0,0005x^3-0,03x^2+18 ?! – priyanka Mar 12 '16 at 14:07
  • There is no the answer. There are an infinite number of cubics that will work. Mine is just as good as yours :-). Check it, and you'll see that it works. – bubba Mar 12 '16 at 14:09
  • how is a=1? why is b -1200,6/40 ?? – priyanka Mar 12 '16 at 14:14
  • You can choose any value you like for $a$. The only requirement is that $1200a+40b = -0.6$. I just picked $a=1$ to make things easy. In your solution, they picked $a = 0.0005$. Both are correct. – bubba Mar 12 '16 at 14:16
  • why is b -0,03 in my equation shouldn't it be the same then ? – priyanka Mar 12 '16 at 14:32
  • No. It's always true that $b = (-0.6 - 1200a)/40$. Changing $a$ will require that you change $b$, also. – bubba Mar 12 '16 at 14:39