Mathematics Advanced • Year 11 • Module 3 • Lesson 3
The Derivative as the Gradient of a Tangent
Apply first principles to multi-step problems involving projectile velocity, tangent lines to parabolic profits, drug concentration rates and intersection of tangent and normal.
Problem 1 — Instantaneous velocity of a thrown ball
A ball is thrown vertically upward. Its height above the ground (in metres) at time t seconds is modelled by
s(t) = 20t − 5t², 0 ≤ t ≤ 4.
Set up: What are we solving for?
(i) From first principles, derive the instantaneous velocity function v(t) = s ′(t). Show every step. 3 marks
(ii) Find v(0), v(1) and v(2), and interpret each in one sentence (faster up / slower up / at rest / falling). 2 marks
(iii) Find the time at which the ball is momentarily at rest (v(t) = 0). What is the maximum height reached? 2 marks
Stuck? Revisit lesson § Worked Example 1 — first principles for a quadratic.Problem 2 — Tangent to a parabolic profit curve
A small business' daily profit (in dollars) for selling x units is modelled by P(x) = 100x − 2x², for 0 ≤ x ≤ 50.
Set up: What are we solving for?
(i) Differentiate P from first principles to obtain P ′(x). 3 marks
(ii) Find the gradient of the tangent to the profit curve at x = 10 and at x = 30. Compare in one sentence (where is profit growing faster, and what does that mean economically?). 2 marks
(iii) Find the value of x at which the tangent to the profit curve is horizontal (gradient = 0). State the corresponding profit. 2 marks
Problem 3 — Tangent equation from first principles
Consider the curve y = x² + 2x.
Set up: What are we solving for?
(i) Use first principles to find dy/dx. 3 marks
(ii) Hence find the equation of the tangent to the curve at the point where x = 1. 2 marks
(iii) Find the x-coordinate of any point at which the tangent to the curve is parallel to the line y = 6x − 5. 2 marks
Stuck on (iii)? Tangent parallel to a line ⇔ same gradient. The gradient of y = 6x − 5 is 6.Problem 4 — Rate of drug concentration change
The concentration of a drug in a patient's bloodstream (in mg/L) is modelled by C(t) = 12t − t², where t is in hours, 0 ≤ t ≤ 12.
Set up: What are we solving for?
(i) Differentiate C from first principles to obtain the instantaneous rate of change C ′(t). 3 marks
(ii) Calculate C ′(2), C ′(6) and C ′(10). For each, state whether the concentration is rising, falling or stationary at that moment. 3 marks
(iii) At what time is the concentration at its peak? Justify using your derivative. 2 marks
Problem 5 — Where tangent and normal cross the x-axis
Consider the parabola y = x² at the point P(2, 4).
Set up: What are we solving for?
(i) Find the equation of the tangent to y = x² at P(2, 4). 2 marks
(ii) Find the equation of the normal to y = x² at P(2, 4). 2 marks
(iii) Find the x-intercept of the tangent and the x-intercept of the normal. Comment in one sentence on whether the tangent intercept is to the left or right of P, and similarly for the normal. 3 marks
Stuck? x-intercept: set y = 0 in each line and solve for x.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Thrown ball s(t) = 20t − 5t²
Set up. Find an exact velocity function from first principles, then evaluate it at specific times and find where it equals zero (turning point at the top of the trajectory).
(i) s(t + h) = 20(t + h) − 5(t + h)² = 20t + 20h − 5(t² + 2th + h²) = 20t + 20h − 5t² − 10th − 5h². s(t + h) − s(t) = 20h − 10th − 5h². Divide by h: 20 − 10t − 5h. limh→0 (20 − 10t − 5h) = v(t) = 20 − 10t m/s.
(ii) v(0) = 20 m/s (moving up fast — the launch speed). v(1) = 10 m/s (still rising, slowing). v(2) = 0 m/s (momentarily at rest — top of trajectory).
(iii) v(t) = 0 when 20 − 10t = 0, i.e. t = 2 s. Maximum height: s(2) = 40 − 20 = 20 m.
Problem 2 — Profit P(x) = 100x − 2x²
Set up. Differentiate to get the marginal profit function, evaluate at sample x-values to compare growth rates, then find the maximum.
(i) P(x + h) = 100(x + h) − 2(x + h)² = 100x + 100h − 2x² − 4xh − 2h². P(x + h) − P(x) = 100h − 4xh − 2h². Divide by h: 100 − 4x − 2h. limh→0 = P ′(x) = 100 − 4x.
(ii) P ′(10) = 100 − 40 = $60/unit (each extra unit adds about $60 to profit at this scale). P ′(30) = 100 − 120 = −$20/unit (each extra unit now loses $20 — production has passed the sweet spot). Profit grows faster at x = 10; by x = 30 the business is losing on each additional unit.
(iii) P ′(x) = 0 when 100 − 4x = 0, i.e. x = 25 units. Maximum profit: P(25) = 2500 − 1250 = $1,250.
Problem 3 — Tangent to y = x² + 2x
Set up. Differentiate from first principles, use the derivative to find the tangent gradient at x = 1, then solve for where the tangent has a prescribed gradient.
(i) f(x + h) = (x + h)² + 2(x + h) = x² + 2xh + h² + 2x + 2h. f(x + h) − f(x) = 2xh + h² + 2h. Divide by h: 2x + h + 2. limh→0 = dy/dx = 2x + 2.
(ii) At x = 1: gradient = 2(1) + 2 = 4. Point: (1, 1² + 2·1) = (1, 3). Tangent: y − 3 = 4(x − 1) ⇒ y = 4x − 1.
(iii) Set 2x + 2 = 6 ⇒ x = 2. (At this x-coordinate, the tangent to y = x² + 2x has gradient 6, parallel to y = 6x − 5.)
Problem 4 — Drug concentration C(t) = 12t − t²
Set up. The derivative gives instantaneous rate of concentration change. Where it is zero, the concentration has a turning point (peak/trough).
(i) C(t + h) = 12(t + h) − (t + h)² = 12t + 12h − t² − 2th − h². C(t + h) − C(t) = 12h − 2th − h². Divide by h: 12 − 2t − h. limh→0 = C ′(t) = 12 − 2t mg/L per hour.
(ii) C ′(2) = 12 − 4 = 8 mg/L per h (concentration rising). C ′(6) = 12 − 12 = 0 mg/L per h (stationary, instantaneously). C ′(10) = 12 − 20 = −8 mg/L per h (falling, drug being cleared).
(iii) Peak occurs when C ′(t) = 0, i.e. at t = 6 hours. (Confirmed by sign of the derivative: rising before t = 6, falling after.)
Problem 5 — Tangent and normal to y = x² at (2, 4)
Set up. Find tangent gradient at the point using dy/dx = 2x, then write tangent and normal equations, then find x-intercepts.
(i) dy/dx = 2x, so at x = 2 the tangent gradient = 4. Tangent: y − 4 = 4(x − 2) ⇒ y = 4x − 4.
(ii) Normal gradient = −1/4. Normal: y − 4 = −1/4 · (x − 2) ⇒ y = −x/4 + 9/2 (equivalently 4y + x = 18).
(iii) Tangent x-intercept: 4x − 4 = 0 ⇒ x = 1 (to the left of P). Normal x-intercept: −x/4 + 9/2 = 0 ⇒ x/4 = 9/2 ⇒ x = 18 (well to the right of P). The tangent dives back to the x-axis just past x = 1, while the very shallow normal slope means it travels far before hitting the x-axis at x = 18.