An object at rest begins moving
horizontally on the x-axis, starting at position x(t) = 0 feet at time t = 0 seconds.Suppose, after exactly 2 seconds we measure the position of the
object to be 3 feet to the right of the origin AND measure the object to be moving to the leftward at a speed of 8 feet per second.
Caution: We do NOT know anything about the position or motion of the object, other than what we know at times t = 0
seconds and t = 2 seconds. So do NOT simply assume the object is moving at constant speed the whole time.
Goal of the Problem: Estimate the position of the object after 2.5 seconds:
(1) by writing a solution as if you were carefully and clearly
explaining it (without calculus) to someone who has NOT had
calculus AND
(2) by writing a solution (with calculus) using the ideas of
linearization.
(1) Solution without calculus:
To estimate the position of the object after 2.5 seconds, let's consider the information we have. At t = 0 seconds, the object is at position x(t) = 0 feet. At t = 2 seconds, the object is 3 feet to the right of the origin and moving to the leftward at a speed of 8 feet per second.
Now, the object is moving to the leftward at a speed of 8 feet per second. This means that in 1 second, the object will move 8 feet to the left. Therefore, after 2.5 seconds, the object would have moved 2.5 * 8 = 20 feet to the left.
However, at t = 2 seconds, the object is 3 feet to the right of the origin. So, after 2.5 seconds, the object will be at a position that is 20 feet to the left of 3 feet.
Position after 2.5 seconds = 3 feet - 20 feet = -17 feet.
Therefore, the estimated position of the object after 2.5 seconds is -17 feet.
(2) Solution with calculus (using linearization):
To estimate the position of the object after 2.5 seconds using linearization, we can use the concept of the tangent line.
We know that at t = 2 seconds, the object is at position x(t) = 3 feet. We also know that the object is moving to the leftward at a speed of 8 feet per second.
The tangent line to the curve at t = 2 seconds represents the instantaneous rate of change of position at that point, which is the velocity. So, the equation of the tangent line can be written as:
x(t) = mx + b,
where m is the slope of the tangent line (velocity) and b is the y-intercept.
We know that the velocity is -8 feet per second (negative because the object is moving to the leftward) at t = 2 seconds. Therefore, the equation of the tangent line becomes:
x(t) = -8t + b.
To find the value of b, we substitute the known point (t = 2, x = 3) into the equation:
3 = -8(2) + b.
b = 19.
So, the equation of the tangent line is:
x(t) = -8t + 19.
Now, to estimate the position of the object after 2.5 seconds, we substitute t = 2.5 into the equation:
x(2.5) = -8(2.5) + 19,
x(2.5) = -20 + 19,
x(2.5) = -1.
Therefore, the estimated position of the object after 2.5 seconds is -1 feet.