Numerical solutions of equations?

Question

The parametricmequations are
x = t^3 - e^(-t), y = t^2 - e^(-2t)
(i) Find the equation to the curve at the point where t=0

The curve cuts the y-axis at the point A.
(ii) Show that the valu of t at A lies between 0 and 1.

yes i was akin for tangent

Answer 1

(i) Just plug in t=0.

x = 0³ - e^0 = 0 - 1 = -1
y = 0² - e^0 = 0 - 1 = -1

(-1 , -1)

Now that you've added that you want the tangent line, we need the slope.

dx / dt = 3t² + e^(-t) = 1
dy / dt = 2t + 2e^(-2t) = 2

The slope is dy / dx = (dy / dt) / (dx / dt) = 2

So the equation is:

y + 1 = 2(x+1)
y = 2x + 1

(ii) Here you want to plug in t=1 too:

x = 1³ - e^(-1) = 1 - 1/e
y = 1² - e^(-2) = 1 - 1/e²

approximately:
x = 0.632120559
y = 0.864664717
http://www.google.com/search?q=1-1/e
http://www.google.com/search?q=1-1/e^2

Now, we use the mean value theorem:
http://en.wikipedia.org/wiki/Mean_value_theorem

to say that since x<0 at t=0 and x>0 at t=1, there is some t between 0 and 1 such that x(t)=0. That is an intersection of the y axis.

Answer 2

I suppose you are asking for the tangent to the curve at t=0
dx/dt = 3t^2+e^-t
and dy/dt=2t+2e^-2t
and the slope is dy/dt /dx/dt which at t =0 is 1/2
x(0)= -1 and y(0) = -1
so the tangent is y+1=1/2(x+1)
at t=0 x(0)=-1 and at t= 1 x(1)= 1-1/e >0
As x(t) is continuos through [0,1] there exists at last one point
between 0,1 where x(c) =0 which lies on the y axis

Answer 3

As an engineer I prefer analytic solutions as I can see how the variables influence the result. I can see which variables can be ignored to simplify an equation. This cannot be done if I just get a numerical solution.