Solutions — Instantaneous Rates of Change and Tangents

Q1 — Evaluating f′(x) for y = x² − 4x

Differentiate: f′(x) = 2x − 4

(a) f′(1) = 2(1) − 4 = −2. The function is decreasing at x = 1.

(b) f′(3) = 2(3) − 4 = 2. The function is increasing at x = 3.

(c) f′(0) = 2(0) − 4 = −4. The function is decreasing at x = 0.

Q2 — Equations of tangents

(a) y = x² at x = 3:
f′(x) = 2x → f′(3) = 6. f(3) = 9. Point: (3, 9).
Tangent: y − 9 = 6(x − 3) → y = 6x − 9

(b) y = 2x³ at x = 1:
f′(x) = 6x² → f′(1) = 6. f(1) = 2. Point: (1, 2).
Tangent: y − 2 = 6(x − 1) → y = 6x − 4

(c) y = x² − 2x + 1 at x = 2:
f′(x) = 2x − 2 → f′(2) = 2. f(2) = 4 − 4 + 1 = 1. Point: (2, 1).
Tangent: y − 1 = 2(x − 2) → y = 2x − 3

(d) y = 4⁄x = 4x⁻¹ at x = 2:
f′(x) = −4x⁻² = −4⁄x² → f′(2) = −4⁄4 = −1. f(2) = 2. Point: (2, 2).
Tangent: y − 2 = −1(x − 2) → y = −x + 4

Q3 — Equations of normals

(a) y = x² + 1 at x = 2:
f′(x) = 2x → f′(2) = 4. f(2) = 5. Point: (2, 5).
Normal gradient = −1⁄4.
Normal: y − 5 = −¹⁄₄(x − 2) → y = −x⁄4 + 11⁄2

(b) y = x³ − x at x = 1:
f′(x) = 3x² − 1 → f′(1) = 2. f(1) = 1 − 1 = 0. Point: (1, 0).
Normal gradient = −1⁄2.
Normal: y − 0 = −¹⁄₂(x − 1) → y = −x⁄2 + 1⁄2

Q4 — Tangent parallel to y = 2x − 1 on y = x² − 6x

Parallel lines have the same gradient. Target gradient = 2.

f′(x) = 2x − 6. Set equal to 2: 2x − 6 = 2 → 2x = 8 → x = 4.

f(4) = 16 − 24 = −8.

Answer: The point is (4, −8).

Q5 — Tangent to y = x³ − 3x through the origin

Let the point of tangency be (a, f(a)) where f(a) = a³ − 3a.

f′(x) = 3x² − 3, so tangent gradient at x = a is f′(a) = 3a² − 3.

Tangent equation: y − (a³ − 3a) = (3a² − 3)(x − a)

Simplify: y = (3a² − 3)x − 3a³ + 3a + a³ − 3a = (3a² − 3)x − 2a³

For the tangent to pass through (0, 0): 0 = (3a² − 3)(0) − 2a³ → −2a³ = 0 → a = 0.

Wait — re-examine: substituting x = 0, y = 0: 0 = −2a³ → a = 0 gives the point (0, 0) itself on the curve. But also check a where f(a)/a = f′(a):

f(a)⁄a = (a³ − 3a)⁄a = a² − 3. Set equal to f′(a) = 3a² − 3:

a² − 3 = 3a² − 3 → −2a² = 0 → a = 0, giving point (0, 0).

Also check y-intercept approach: tangent y = (3a²−3)x − 2a³ passes through origin when −2a³ = 0, so a = 0, OR check separately — a³ − 3a = (3a²−3)·a → a³ − 3a = 3a³ − 3a → −2a³ = 0 → a = 0.

But we should also check a = 0 is valid: f(0)=0, f′(0)=−3. Tangent: y = −3x. This passes through the origin. Point of tangency: (0, 0).

Note: Check if there are other real solutions by reconsidering: a³ − 3a = (3a² − 3)a gives 0 = 2a³, confirming a = 0 is the only solution.

Q6 — Equal gradients on y = x² and y = −x² + 4x

Gradient of y = x²: dy⁄dx = 2x

Gradient of y = −x² + 4x: dy⁄dx = −2x + 4

Set equal: 2x = −2x + 4 → 4x = 4 → x = 1.

Answer: At x = 1, both curves have gradient 2.

Check: y = x² gives gradient 2(1) = 2; y = −x² + 4x gives −2(1) + 4 = 2. ✓

Q7 — Projectile velocity h = −5t² + 20t

(a) v(t) = h′(t) = −10t + 20 m/s

(b) v(2) = −10(2) + 20 = 0 m/s. At t = 2 s the ball is at its maximum height — instantaneously neither rising nor falling.

(c) v(t) = 0 when −10t + 20 = 0 → t = 2 s.

Q8 — Finding a in y = ax² given tangent

f(x) = ax², so f′(x) = 2ax.

At x = 2: gradient = f′(2) = 4a. The given tangent y = 8x − 8 has gradient 8.

So 4a = 8 → a = 2.

Verify: f(2) = 2(4) = 8. Tangent: y − 8 = 8(x − 2) → y = 8x − 8. ✓

Q9 — Tangent and normal to y = x² at x = 1

f(x) = x², f′(x) = 2x.

At x = 1: gradient = 2, point = (1, 1).

Tangent: y − 1 = 2(x − 1) → y = 2x − 1

Normal: gradient = −1⁄2. y − 1 = −¹⁄₂(x − 1) → y = −x⁄2 + 3⁄2

Intersection (other than (1,1)):

The tangent and normal only intersect at the point of tangency (1, 1), since they are two distinct lines meeting at exactly one point. They cannot intersect again — two distinct non-parallel lines meet at exactly one point.

Set equal: 2x − 1 = −x⁄2 + 3⁄2 → 2x + x⁄2 = 3⁄2 + 1 → 5x⁄2 = 5⁄2 → x = 1.

Answer: The tangent and normal intersect only at (1, 1).

Q10 — y = x³ + bx through (2, k) with gradient 15

f(x) = x³ + bx, so f′(x) = 3x² + b.

Condition 1 — gradient at x = 2 is 15:
f′(2) = 3(4) + b = 12 + b = 15 → b = 3

Condition 2 — point (2, k) lies on the curve:
k = f(2) = 2³ + 3(2) = 8 + 6 = 14

Answer: b = 3, k = 14.