Repeat Exercise 16 for s(t)=5 t^3-3 t^5, t>0

Name: Problem 17, Chapter 3: Applied Calculus For Business, Economics, and Finance
Uploaded: 2021-04-02T03:25:13-08:00
Duration: 4 min 10 s
Channel: Meredith Murphy
Description: Repeat Exercise 16 for s(t)=5 t^3-3 t^5, t>0

Key Concepts

Differentiation of Polynomial Functions

This concept involves applying power rules to compute the derivative of polynomials. Differentiation is used to find slopes of curves at specific points, which in applications to motion problems can yield the velocity of an object from its position function.

Kinematics – Position, Velocity, and Acceleration

In kinematics, the position function represents an object's location over time. The first derivative of the position function gives the velocity, which is the rate of change of position. The second derivative provides the acceleration, indicating how the velocity changes over time.

Critical Points and Optimization

Critical points are values where the derivative is zero or undefined, possibly indicating local maximums, minimums, or points of inflection. Identifying these points is essential for analyzing the behavior of a function, such as determining when an object changes direction or reaches extreme positions, which is frequently assessed in motion and optimization problems.

Transcript

00:01 Question 17 would like you to repeat exercise 16 for the position function s of t equals 5t cubed, minus 3t to the fifth, where t is greater than zero.

00:13 So question 16, part a, that would be finding velocity and acceleration functions.

00:20 Velocity v of t is just a derivative of our position function, so 15t squared minus 15 t to the fourth.

00:30 And then acceleration a of t is equal to 30 t minus 60t cubed.

00:43 Part b then, they would like you to find the time at which velocity is zero and the time at which acceleration is zero.

00:51 So v of t is equal to 0, which is equal to 15 t squared minus 15 t to the fourth.

01:00 You can take out 15 t squared so multiply that by 1 minus t squared so then you have just t equals 0 and t equals 1 not negative 1 because t here is greater than 0 for a of t equals 0 that's equal to 30 t minus 60 t cubed again taking out 30 t so you have 1 minus 2, t squared, in this case then t is equal to 0 and also square 2 over 2.

01:41 Moving on to part c, they would like you to find the time at which the velocity is maximum.

01:50 So your v max is going to be when v prime of t is equal to 0.

01:57 That's either going to be a max or it could be a minimum.

02:00 So then you have to test the double prime of t if it's less than 0.

02:05 Zero, that means it is concave down and it's a maximum.

02:10 So from there, b double prime of t is equal to a prime of t, which is equal to 30 minus 180 t squared...

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