2

A curve is given by:
$$x = 2t + 3 $$ $$y = t^3 - 4t$$

The point $A$ has parameter $t = -1$.
Line l is a tangent to the curve at $A$.
Line l cuts the curve at point $B$.

Find the value of $t$ at $B$.

I've worked out that the equation for l is: $2y + x = 7$ (which is correct). To work out the value of t at B, I rearranged the parametric equations to eliminate t and make y the subject. I did the same for the equation of line l (made y the subject). I equated the two together and formed an equation of a cubic curve: $x^3 - 9x^2 + 15x - 7 = 0$ (which is correct). The problem is that I do not know (or am expected to know) a method to find the roots of a cubic curve. I am wondering whether this is the best/correct approach, and whether there is an easier/alternative route to finding the answer.

Amzoti
  • 56,093
Wk_of_Angmar
  • 247

3 Answers3

3

We wish to find where our parameterized curve intersects the line $2y+x=7$. Let's parameterize that line by $x=2t+3,$ so we'll need $y=-t+2$. Under this parameterization, the $x$-values of our line and curve will be the same at any $t$, so we need only determine where the $y$-values are the same--i.e. find $t$ for which $$t^3-4t=-t+2\\t^3-3t-2=0.$$ Recall, though, that we already know they meet where $t=-1$, so we know we have a factor of $t+1$ in our cubic (that's why we went with this method, instead of yours). In particular, $$t^3-3t-2=(t+1)(t^2-t-2)=(t+1)^2(t-2),$$ so the cutting point occurs when $t=2$.

Cameron Buie
  • 102,994
3

Hint: Besides to @Cameron's parametric approach, you may eliminate the $t$ between two equations to find a united equation $y=f(x)$. Then you can solve the problem by working on a converted problem with new corresponding assumptions. In fact, $$(1/8)x^3-(9/8)x^2+(11/8)x+21/8 = y,~~~ A|_1^3$$

Mikasa
  • 67,374
1

An alternative method (and easier in my opinion) is to keep the equation $2y+x=7$ in cartesian form.

Now substitute $x=2t+3$ and $y=t^3-4t$ in to the equation for $l$:

$2(2t+3)+(t^3-4t)=7$

If you expand the brackets and simplify you get:

$t^3-3t-2=0$

You are given that $t=-1$, hence $(t+1)$ is a factor of $(t^3-3t-2)$.

By factorising (see here for details) you get:

$t^3-3t-2=(t+1)(t^2-t-2)=(t+1)^2(t-2)$

So the two solutions are $t=-1$ and $t=2$. Note that you are given $t=-1$ in the question so the value of $t$ at $B$ must be $t=2$.

Note also that $t=-1$ is a repeated root (i.e. it just touches the axis at that point rather than crossing it).

Community
  • 1
Max Goodridge
  • 151
  • It is certainly true that your approach (2 substitutions, 2 easy additions, and 2 easy multiplications) is easier than mine (2 substitutions, 4 easy additions, and 3 easy multiplications). Still the thrust (assume $2y+x=7,$ $x=2t+3,$ and $y=t^3-4t$ all hold, derive a polynomial equation in $t$ only, and factor) is identical. One wonders what substantive improvement you feel this answer offers almost three years after the question was posted. – Cameron Buie Nov 24 '15 at 13:53
  • 1
    The question was asking if there was an easier/alternative method, hence I shared one as this page is still used as a reference today that people can learn from. So I do believe my answer is still valid. – Max Goodridge Nov 24 '15 at 22:37
  • It's absolutely valid. – Cameron Buie Nov 25 '15 at 03:46