Instantaneous Velocity of Particle at t=1 with s(t) = 2t2-4t

[Speedking96]

Homework Statement

Estimate the instantaneous velocity of a particle with position function s(t) = 2t²−4t at t = 1 using the four intervals [0.9, 1], [0.99, 1], [1, 1.01], and [1, 1.1].

2. The attempt at a solution

At t=1, y = -2

Slope of a line: a = (y2 - y1)/(x2 - x1)

= (y2 + 2) / (x2 - 1)
= (2t²−4t +2) / (x2 -1)

At t = 0.9

= (2(0.9)²−4(0.9) +2) / (0.9 -1)
= -0.2


At t = 0.99

= (2(0.99)²−4(0.99) +2) / (0.99 -1)
= -0.02

At t = 1.01
= (2(1.01)²−4(1.01) +2) / (1.01 -1)
= 0.02

At t = 1.1
= (2(1.1)²−4(1.1) +2) / (1.1 -1)
= 0.2


From the graph of the parabola, it's pretty obvious that the equation of the tangent line is y= -2

But, I don't know where I'm going wrong.

 

[ehild]

Homework Statement

Estimate the instantaneous velocity of a particle with position function s(t) = 2t²−4t at t = 1 using the four intervals [0.9, 1], [0.99, 1], [1, 1.01], and [1, 1.1].

2. The attempt at a solution

At t=1, y = -2

Slope of a line: a = (y2 - y1)/(x2 - x1)

= (y2 + 2) / (x2 - 1)
= (2t²−4t +2) / (x2 -1)

The independent variable is t instead of x.

At t = 0.9

= (2(0.9)²−4(0.9) +2) / (0.9 -1)
= -0.2


At t = 0.99

= (2(0.99)²−4(0.99) +2) / (0.99 -1)
= -0.02

At t = 1.01
= (2(1.01)²−4(1.01) +2) / (1.01 -1)
= 0.02

At t = 1.1
= (2(1.1)²−4(1.1) +2) / (1.1 -1)
= 0.2


From the graph of the parabola, it's pretty obvious that the equation of the tangent line is y= -2

But, I don't know where I'm going wrong.

You need the instantaneous velocity , that is, the slope of the tangent line. What is the slope of the line y=-2? So what is the instantaneous velocity at t=1?

ehild

 

[Speedking96]

Woops. The slope is obviously zero.