Implicit and Parametric Differentiation and Connected Rates of Change

When y is not isolated — for example x² + y² = 25 — differentiate both sides with respect to x, remembering that y is a function of x, so every term containing y is multiplied by dy/dx (chain rule).

Worked examples

1. x² + y² = r²: 2x + 2y·y′ = 0 ⟹ y′ = −x/y. At (3, 4) on x² + y² = 25 the slope is −3/4, so the tangent is 3x + 4y = 25 and the normal is 4x − 3y = 0.
2. x³ + y³ = 6xy (the folium of Descartes): 3x² + 3y²y′ = 6y + 6xy′ ⟹ y′(3y² − 6x) = 6y − 3x² ⟹ y′ = (2y − x²)/(y² − 2x).
3. xy + y² = 5: y + xy′ + 2yy′ = 0 ⟹ y′ = −y/(x + 2y).
4. sin(x + y) = x ⟹ cos(x + y)(1 + y′) = 1 ⟹ y′ = sec(x + y) − 1.

Products require the product rule first: d/dx(xy) = x·y′ + y.

To find d²y/dx² differentiate the expression for y′ again with respect to x, using the quotient rule, and substitute the value of y′ obtained before.

Worked example

For x² + y² = 25, y′ = −x/y. Then

• y″ = −[y − x·y′]/y² = −[y − x(−x/y)]/y² = −(x² + y²)/y³ = −25/y³ (using x² + y² = 25).
• At (3, 4): y′ = −3/4 and y″ = −25/64 < 0, so the curve is concave down there — consistent with the upper half of a circle.

Horizontal and vertical tangents

• Horizontal tangent where dy/dx = 0: the numerator vanishes (and the denominator does not).
• Vertical tangent where dy/dx is undefined: the denominator vanishes (and the numerator does not).
• On x² + y² = 25 the numerator x = 0 gives (0, ±5) with horizontal tangents, and the denominator y = 0 gives (±5, 0) with vertical tangents.

If x = f(t) and y = g(t), then by the chain rule

• dy/dx = (dy/dt) ÷ (dx/dt), provided dx/dt ≠ 0.
• d²y/dx² = [d/dt (dy/dx)] ÷ (dx/dt) — a frequent error is to divide d²y/dt² by d²x/dt², which is wrong.

Worked example 1 — a circle

x = 5 cos t, y = 5 sin t. dx/dt = −5 sin t, dy/dt = 5 cos t ⟹ dy/dx = −cot t. At t = π/3: dy/dx = −cot(π/3) = −1/√3 at the point (2.5, 4.33).

Worked example 2 — a parabola

x = t², y = t³ − 3t. dy/dx = (3t² − 3)/(2t). At t = 1 the slope is 0 (horizontal tangent at (1, −2)); the tangent is y = −2.

• d/dt(dy/dx) = d/dt[(3/2)(t − t⁻¹)] = (3/2)(1 + t⁻²); at t = 1 this is 3, and dx/dt = 2 ⟹ d²y/dx² = 3/2 > 0, so (1, −2) is a local minimum.

Motion reading

With t as time, dx/dt and dy/dt are the velocity components, the speed is √[(dx/dt)² + (dy/dt)²], and the slope dy/dx is the direction of motion.

Method: (1) draw and label the variables as functions of time; (2) write a relation linking them; (3) differentiate both sides with respect to t; (4) substitute the values at the instant asked for; (5) solve and give the sign and units.

Worked examples

1. Expanding circle: A = πr² ⟹ dA/dt = 2πr·dr/dt. When dr/dt = 3 cm/s and r = 10 cm: dA/dt = 60π ≈ 188.5 cm²/s.
2. Ladder sliding: a 5 m ladder leans against a wall; the foot moves away at 1 m/s. x² + y² = 25 ⟹ x·dx/dt + y·dy/dt = 0. At x = 3, y = 4: 3(1) + 4(dy/dt) = 0 ⟹ dy/dt = −0.75 m/s (the top descends).
3. Filling cone: water in a conical tank with r = h/2, V = (1/3)πr²h = (π/12)h³ ⟹ dV/dt = (π/4)h²·dh/dt. With dV/dt = 2 m³/min at h = 3: dh/dt = 8/(9π) ≈ 0.283 m/min.
4. Two ships: one sails north at 20 km/h from a port, the other east at 15 km/h. After 2 h, x = 30, y = 40, s = 50 and s·ds/dt = x·dx/dt + y·dy/dt ⟹ 50·ds/dt = 30(15) + 40(20) = 1250 ⟹ ds/dt = 25 km/h.
5. Shadow / angle: a plane flies horizontally at 600 km/h at 5 km altitude; when its distance over the ground is 13 km the angle θ of elevation satisfies tan θ = 5/x ⟹ sec²θ·dθ/dt = −5x⁻²·dx/dt; with sec²θ = (13/5)², dθ/dt = −0.05 rad/min ≈ −2.87°/min.

1. Find dy/dx for x² + 3xy + y² = 7. — (2x + 3y) + (3x + 2y)y′ = 0 ⟹ y′ = −(2x + 3y)/(3x + 2y).
2. Find dy/dx for y³ = x² + y. — 3y²y′ = 2x + y′ ⟹ y′ = 2x/(3y² − 1).
3. For x = t³ − t, y = t² + 1 find dy/dx at t = 2. — (2t)/(3t² − 1) = 4/11 ⟹ 4/11.
4. Find the tangent to x²y + y³ = 8 at (2, 2). — y′ = −(2xy)/(x² + 3y²) = −4/(4 + 12) = −1/4 ⟹ x + 4y = 10.
5. A spherical balloon grows at 10 cm³/s. How fast is the radius increasing when r = 5? — dV/dt = 4πr²·dr/dt ⟹ dr/dt = 10/(100π) = 1/(10π) ≈ 0.0318 cm/s.
6. A 10 m ladder's top slides down at 0.5 m/s when the foot is 6 m from the wall. Find the foot's speed. — x·dx/dt + y·dy/dt = 0 with y = 8: 6·dx/dt = 8(0.5) ⟹ 2/3 m/s.
7. For the ellipse x²/9 + y²/4 = 1 find the points with a horizontal tangent. — y′ = −(4x)/(9y) = 0 ⟹ x = 0 ⟹ (0, ±2); vertical tangents at (±3, 0).