- Difficulty
- Standard
- Suggested time
- 5 min
- Marks
- 3
The definition of the derivative practice question 17
A particle moves along a line, and its position at time seconds is metres. Use the definition of the derivative to find its velocity at time .
Rules such as the power rule earn no marks on this question.
This one is usually lost in expanding the cube. How many terms does have when written out in full?
Expand before subtracting .
After subtracting, every term left in the numerator has a factor of . Divide it out, then let .
A formula in t. Use ^ for powers and brackets for every input: sin(2x), e^(x^2), log(x+1). Give exact values: 1/6, not 0.167.
Theory UBC Math 100 §3.2CLP-1 §1.2Key results
Worked solution
Strategy
Velocity is the derivative of position. Form the difference quotient of s, expand, cancel the h, and only then let h approach 0. The result is a formula that holds at every time t.
Working
Replace every t with t + h and expand the cube in full.
The terms without h cancel, as they must.
Divide every term by h.
Let h approach 0; every term with an h vanishes.
Answer
The answer is v of t equals 6 t squared minus 5, in metres per second.
Check
Near t = 0, s ≈ −5t moves at −5 m/s, and for large t, s ≈ 2t³ grows at about 6t² m/s: 6t² − 5 fits both.
Where this goes wrong
Expanding (t + h)³ as t³ + h³, which loses the cross terms and leaves v = −5. The cube of a sum has four terms, and 3t²h is the one that gives the velocity its t².
Leans on
Expanding binomials, from Functions, graphs and inverses
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