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
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.
Yes, I had a typo sorry.
It was.
(2x^2+3)(x-1)
Is it 19 now?
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.
Zed
To find the distance between the object and its starting point after 15 seconds using the given velocity function, we need to integrate the velocity function with respect to time from the initial time (0 seconds) to the final time (15 seconds).
The velocity function is given as v(t) = 0.2t^2 + 2t + 10 meters/second.
To find the distance function s(t), we need to find the antiderivative or integrate the velocity function with respect to time.
∫ v(t) dt = ∫ (0.2t^2 + 2t + 10) dt
Using the power rule of integration, the indefinite integral is:
s(t) = 0.2 * (t^3/3) + 2 * (t^2/2) + 10t + C
where C is the constant of integration.
To find the value of C, we can use the initial condition s(0) = 0, which represents the starting point. Substituting t = 0 into the distance function s(t):
s(0) = 0.2 * (0^3/3) + 2 * (0^2/2) + 10 * 0 + C
0 = 0 + 0 + 0 + C
C = 0
Therefore, the distance function s(t) is:
s(t) = 0.2 * (t^3/3) + 2 * (t^2/2) + 10t
Now we can find the distance between the object and its starting point after 15 seconds:
s(15) = 0.2 * (15^3/3) + 2 * (15^2/2) + 10 * 15
= 0.2 * (3375/3) + 2 * (225/2) + 150
= 562.5 + 225 + 150
= 937.5 meters
Therefore, the distance between the object and its starting point after 15 seconds is 937.5 meters.
For the second question, to find the slope of the tangent line to the graph of the equation y = (2x^2 + 3)x - 1 at x = 2, we need to find the derivative of the function y with respect to x.
Given the function y = (2x^2 + 3)x - 1, we can find its derivative dy/dx using the power rule of differentiation:
dy/dx = d/dx (2x^2 + 3)x - d/dx (1)
= (2x^2 + 3) + (4x)
Now we substitute x = 2 into the derivative expression:
dy/dx | x=2 = (2(2)^2 + 3) + (4(2))
= 19
Therefore, the slope of the tangent line to the graph of the equation y = (2x^2 + 3)x - 1 at x = 2 is 19.