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Tangent Lines and the Physical Meaning of the Derivative
This test develops the geometric and physical interpretations of a derivative. Students form tangent equations, interpret velocity and acceleration and connect the sign of a derivative with the behaviour of a function.
Questions: 10Estimated time: 17 minutes
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Question 1out of 10
What does a negative derivative of position mean at an instant?
Velocity is the position derivative v(t) = s′(t), and acceleration is a(t) = v′(t). For s(t) = At² + Bt, v(t) = 2At + B. Negative velocity means the position coordinate decreases, not necessarily negative acceleration.
Question 2out of 10
Position is s(t) = 3t² − 6t. Find velocity at t = 3.
Velocity is the position derivative v(t) = s′(t), and acceleration is a(t) = v′(t). For s(t) = At² + Bt, v(t) = 2At + B. Negative velocity means the position coordinate decreases, not necessarily negative acceleration.
Question 3out of 10
Position is s(t) = 6t² + 8t. Find velocity at t = 6.
Velocity is the position derivative v(t) = s′(t), and acceleration is a(t) = v′(t). For s(t) = At² + Bt, v(t) = 2At + B. Negative velocity means the position coordinate decreases, not necessarily negative acceleration.
Question 4out of 10
Find the tangent equation to f(x) = 4x² at x = 1.
The derivative f′(a) is the instantaneous rate of change and tangent slope. Tangent equation: y − f(a) = f′(a)(x − a). Find f(a) and f′(a) separately. If f′(a) = 0, the tangent is horizontal, but an extremum is not guaranteed.
Question 5out of 10
How are position and velocity related in a motion problem?
Velocity is the position derivative v(t) = s′(t), and acceleration is a(t) = v′(t). For s(t) = At² + Bt, v(t) = 2At + B. Negative velocity means the position coordinate decreases, not necessarily negative acceleration.
Question 6out of 10
Velocity is v(t) = 3t + 0. Find acceleration.
Velocity is the position derivative v(t) = s′(t), and acceleration is a(t) = v′(t). For s(t) = At² + Bt, v(t) = 2At + B. Negative velocity means the position coordinate decreases, not necessarily negative acceleration.
Question 7out of 10
Find the tangent equation to f(x) = 2x² at x = 4.
The derivative f′(a) is the instantaneous rate of change and tangent slope. Tangent equation: y − f(a) = f′(a)(x − a). Find f(a) and f′(a) separately. If f′(a) = 0, the tangent is horizontal, but an extremum is not guaranteed.
Question 8out of 10
For f(x) = 2x² + 7x, find the tangent slope at x = 2.
Use (axⁿ)′ = anxⁿ⁻¹, the derivative of a constant is 0 and derivatives add across sums. Thus (ax² + bx + c)′ = 2ax + b. Substitute the point x for a derivative value; for f′(x) = r, solve 2ax + b = r.
Question 9out of 10
Which formula describes the tangent line at x = a?
The derivative f′(a) is the instantaneous rate of change and tangent slope. Tangent equation: y − f(a) = f′(a)(x − a). Find f(a) and f′(a) separately. If f′(a) = 0, the tangent is horizontal, but an extremum is not guaranteed.
Question 10out of 10
What can be stated if f′(a) = 0?
The derivative f′(a) is the instantaneous rate of change and tangent slope. Tangent equation: y − f(a) = f′(a)(x − a). Find f(a) and f′(a) separately. If f′(a) = 0, the tangent is horizontal, but an extremum is not guaranteed.
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Sample questions
- For f(x) = ___x² + (___)x, find the tangent slope at x = ___.
- Find the tangent equation to f(x) = ___x² at x = ___.
- Position is s(t) = ___t² + (___)t. Find velocity at t = ___.
- Velocity is v(t) = ___t + (___). Find acceleration.
- What does a negative derivative of position mean at an instant?
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