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Dmitriy789 [7]
3 years ago
12

In the previous part, we obtained dy dx = 3t2 − 27 −2t . Next, find the points where the tangent to the curve is horizontal. (En

ter your answers as a comma-separated list of ordered pairs.)
x = t^3 - 3t, y = t^2 - 4
Mathematics
1 answer:
mina [271]3 years ago
8 0

Answer:

(27.55, 7.22), (-11.3, 3.21).

Step-by-step explanation:

When is the tangent to the curve horizontal?

The tangent curve is horizontal when the derivative is zero.

The derivative is:

\frac{dy}{dx} = 3t^{2} - 2t - 27

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

ax^{2} + bx + c, a\neq0.

This polynomial has roots x_{1}, x_{2} such that ax^{2} + bx + c = a(x - x_{1})*(x - x_{2}), given by the following formulas:

x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}

x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}

\bigtriangleup = b^{2} - 4ac

In this question:

3t^{2} - 2t - 27 = 0

So

a = 3, b = -2, c = -27

Then

\bigtriangleup = b^{2} - 4ac = (-2)^{2} - 4*3*(-27) = 328

So

t_{1} = \frac{-(-2) + \sqrt{328}}{2*3} = 3.35

t_{2} = \frac{-(-2) - \sqrt{328}}{2*3} = -2.685

Enter your answers as a comma-separated list of ordered pairs.

We found values of t, now we have to replace in the equations for x and y.

t = 3.35

x = t^{3} - 3t = (3.35)^{3} - 3*3.35 = 27.55

y = t^{2} - 4 = (3.35)^2 - 4 = 7.22

The first point is (27.55, 7.22)

t = -2.685

x = t^{3} - 3t = (-2.685)^3 - 3*(-2.685) = -11.3

y = t^{2} - 4 = (-2.685)^2 - 4 = 3.21

The second point is (-11.3, 3.21).

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