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Implicit and Parametric Differentiation and Connected Rates of Change

Mathematics

Implicit and Parametric Differentiation and Connected Rates of Change

Overview

Overview

  • Differentiate an equation in x and y without solving for y, using d/dx of every term.
  • Find dy/dx and d2y/dx2 for a curve given parametrically.
  • Write the tangent and the normal to implicit and parametric curves at a given point.
  • Model connected (related) rates of change and evaluate the required rate at an instant.

🌳 Concept tree

Beyond explicit differentiation

├── Implicit differentiation

  • │ ├── Differentiate term by term, treating y as a function of x
  • │ ├── Every y-term is multiplied by dy/dx (chain rule)

│ └── Collect dy/dx and isolate it

├── Second derivative

  • │ ├── d2y/dx2 from implicit dy/dx (quotient rule again)
  • │ └── Horizontal tangent dy/dx = 0 · Vertical tangent: denominator = 0

├── Parametric equations x = f(t), y = g(t)

│ ├── dy/dx = (dy/dt) ÷ (dx/dt)

│ ├── d2y/dx2 = [d/dt(dy/dx)] ÷ (dx/dt)

  1. │ └── Tangent
  2. normal
  3. area
  4. and motion interpretation

└── Connected rates

├── Relation linking the variables

├── Differentiate with respect to time t

└── Substitute the instant’s values, then solve

Implicit and Parametric Differentiation and Connected Rates of Change · 1 / 7
Implicit differentiation

Implicit differentiation

  • 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).
Implicit and Parametric Differentiation: a circular track and a tangent rod at (3, 4) show implicit and parametric differentiation: the slope is -3/4.
Implicit and Parametric Differentiation: a circular track and a tangent rod at (3, 4) show implicit and parametric differentiation: the slope is -3/4.
x2 + y2 = r22x + 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.
x3 + y3 = 6xy(the folium of Descartes): 3x2 + 3y2y′ = 6y + 6xy′ ⟹ y′(3y2 − 6x) = 6y − 3x2 ⟹ y′ = (2y − x2)/(y2 − 2x).
xy + y2 = 5y + xy′ + 2yy′ = 0 ⟹ y′ = −y/(x + 2y).
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.
Implicit and Parametric Differentiation and Connected Rates of Change · 2 / 7
The second derivative implicitly

The second derivative implicitly

  • 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

  1. For x2 + y2 = 25
  2. 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.
Comparison aspectMain ideaKeywords
Horizontaltangent where dy/dx = 0: the numerator vanishes (and the denominator does not)tangentnumeratorvanishesdenominator
Verticaltangent where dy/dx is undefined: the denominator vanishes (and the numerator does not)tangentundefineddenominatorvanishes
•
  • 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
numeratorgiveshorizontaltangents
Implicit and Parametric Differentiation and Connected Rates of Change · 3 / 7
Parametric differentiation

Parametric differentiation

  1. If x = f(t) and y = g(t)
  2. 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

  1. With t as time
  2. dx/dt and dy/dt are the velocity components
  3. the speed is √[(dx/dt)2 + (dy/dt)2]
  4. and the slope dy/dx is the direction of motion.
Implicit and Parametric Differentiation and Connected Rates of Change · 4 / 7
Connected (related) rates of change

Connected (related) rates of change

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.
Comparison aspectMain ideaKeywords
Expanding circleA = πr2 ⟹ dA/dt = 2πr·dr/dt. When dr/dt = 3 cm/s and r = 10 cm: dA/dt = 60π ≈ 188.5 cm2/s
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)
leansagainstwallfoot
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
waterconicaltankπr2h
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
sailsnorthporteast
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
planeflieshorizontallyaltitude
Implicit and Parametric Differentiation and Connected Rates of Change · 5 / 7
Practice with answers

Practice with answers

  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).
Implicit and Parametric Differentiation and Connected Rates of Change · 6 / 7
Exam traps

Exam traps

Comparison aspectMain ideaKeywords
Forgetting dy/dx on y-terms.
  • d/dx(y2) = 2y·y′
  • not 2y. This single slip invalidates the whole answer
singleslipinvalidateswhole
Second parametric derivative.
  • d2y/dx2 = [d/dt(dy/dx)]/(dx/dt)
  • not (d2y/dt2)/(d2x/dt2)
Sign of the rate.A decreasing quantity has a negative derivative (the ladder top: −0.75 m/s). State “decreasing at 0.75 m/s” only if words are asked fordecreasingquantitynegativederivative
Use the instant, not a general value.
  • Substitute x
  • y and the given rate at that moment
  • the geometry relation (Pythagoras
  • volume formula) supplies the missing length
Substitutegivenratemoment
Units.
  • cm2/s
  • m3/min
  • rad/min — mixing them is penalised separately
mixingthempenalisedseparately
Implicit curve points must satisfy the equation.Check the given point lies on the curve before differentiatingCheckgivenliesbefore
Implicit and Parametric Differentiation and Connected Rates of Change · 7 / 7