Question

1. The distance s(t) between an object and its starting point is given by the anti derivitive of the velocity function v(t). find the distance between the object and its starting point after 15 sec if v(t)=0.2t^2+2t+10 meters/seconds.

MY ANSWER: 600 meters.

2.Find the slope of the tangent line to the graph of the equation y= (2x^2+3)x-1) at x=2.

MY ANSWER: 19

Answer 1

1. S - S0 = (0.2/3)t^3 + t^2 + 10t
= 225 + 225 + 150 = 600 ok

2. The way you have written y with parentheses is equivalent to
y = 2x^3 +3x -1

y' = 6x^2 + 3, which is not 19 when x = 2. Perhaps you interpreted y(x) differently or mistyped it. You have a missing left-facing parenth.

Answer 2

Yes, I had a typo sorry.
It was.

(2x^2+3)(x-1)

Is it 19 now?

Answer 3

f(x) = (2x^2+3)(x-1) = 2x^3 -2x^2 +3x -3
f'(x) = 6x^2 -4x +3
At x=2, that equals 24 -8 +3 = 19, yes.

Answer 4

Zed