Calculus Concepts: Velocity and Tangents

Calculus Concepts: Velocity and Tangents

Assessment

Interactive Video

Mathematics, Physics, Science

9th - 12th Grade

Hard

Created by

Patricia Brown

FREE Resource

This video introduces calculus, focusing on the tangent and velocity problems. It explains how to find the equation of a tangent line using secant lines as approximations and discusses average and instantaneous velocity. An example of a rock thrown on Mars illustrates these concepts. The lesson concludes with a summary of the key points.

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10 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What was one of the main reasons calculus was developed?

To simplify geometry

To address tangent and velocity problems

To improve arithmetic skills

To solve algebraic equations

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What two elements are needed to write the equation of a line?

A point and a curve

A slope and a point

A tangent and a secant

A derivative and an integral

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Why do we use a secant line to approximate a tangent line?

Because it is easier to calculate

Because it requires only one point

Because it provides an exact solution

Because it uses two points to estimate the slope

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

As the point Q moves closer to point P, what happens to the slope of the secant line?

It remains constant

It approaches the slope of the tangent line

It becomes zero

It becomes steeper

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What does the slope of the tangent line represent in terms of velocity?

Average velocity

Variable velocity

Instantaneous velocity

Constant velocity

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

How is average velocity calculated?

By dividing change in position by change in time

By finding the slope of the tangent line

By using the derivative of the position function

By dividing total distance by total time

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

In the Mars rock problem, what is the formula for the height of the rock?

y = 10t - 1.86t^2

y = 10t + 1.86t^2

y = 1.86t^2 - 10t

y = 10t^2 - 1.86t

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