Implicit differentiation and rates of change flashcards

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Implicitly defined curve
A curve given by an equation in x and y in which y is not the subject, e.g. x² + xy + y² = 12.
Implicit differentiation
Differentiating every term of an equation in x and y with respect to x, treating y as a function of x (chain rule on y terms), then solving for dy/dx.
Stationary point of an implicit curve
A point on the curve where dy/dx = 0, so the tangent is horizontal (parallel to the x-axis).
Differential equation
An equation that contains a derivative, such as dN/dt = kN.
Connected rates of change
Two rates linked through a formula, e.g. dV/dt and dr/dt linked by V = (4/3)πr³ and the chain rule.
“A is proportional to B”
A = kB for some constant k.
d/dx of g(y)
g′(y) × dy/dx (chain rule). (No — learn it)
d/dx(yⁿ)
n yⁿ⁻¹ dy/dx. (No — learn it)
d/dx(xy)
y + x dy/dx (product rule). (No — learn the product rule)
d/dx(aˣ), a > 0
aˣ ln a. (No — derive it)
Chain rule for connected rates (volume and radius)
dV/dt = dV/dr × dr/dt. (No — learn it)
aˣ written as a power of e
aˣ = e^(x ln a). (Yes)
On an implicit curve, where is the tangent horizontal?
Where the numerator of dy/dx is zero (denominator non-zero): solve that relationship together with the curve.
On an implicit curve, where is the tangent vertical?
Where the denominator of dy/dx is zero, i.e. dx/dy = 0: solve together with the curve.
Tangent and radius of a circle at the same point
Perpendicular: the gradient from implicit differentiation × the radius gradient = −1.
Water in an inverted cone of top radius R and height H
Similar triangles: surface radius r = (R/H)h at depth h, so V = (1/3)π(R/H)²h³.
Shape of y = aˣ for a > 1 and its gradient
Increasing, through (0, 1); gradient aˣ ln a is proportional to the height, ln a at x = 0.
Why does d/dx(y²) need a factor dy/dx?
y depends on x, so y² is a composite function of x; the chain rule gives 2y × dy/dx.
Why does an implicit dy/dx usually contain y?
One x-value can correspond to several points with different gradients, so the gradient depends on y too.
How do you prove d/dx(aˣ) = aˣ ln a?
ln y = x ln a, so (1/y) dy/dx = ln a, so dy/dx = y ln a = aˣ ln a.
Why does a decreasing quantity get a minus sign in a differential equation?
With k > 0 the rate must be negative when the quantity falls, e.g. dm/dt = −km.
Water flows in at rate a and leaks out at rate kh. What is dV/dt?
dV/dt = a − kh: the net rate is rate in minus rate out.
Gradient of x² + y² = 25 at (3, 4)?
2x + 2y dy/dx = 0, so dy/dx = −x/y = −3/4.
Gradient of x²y − 2y² + 6 = 0 at (2, 3)?
2xy + x² dy/dx − 4y dy/dx = 0, so dy/dx = 2xy/(4y − x²) = 12/8 = 3/2.
dy/dx for eʸ + xy = 1?
eʸ dy/dx + y + x dy/dx = 0, so dy/dx = −y/(eʸ + x).
Differentiate y = 2^(3x).
ln y = 3x ln 2, so dy/dx = 3 ln 2 · 2^(3x).
Stationary points of x² − xy + y² = 3?
dy/dx = (2x − y)/(x − 2y) = 0 ⇒ y = 2x; then 3x² = 3 ⇒ (1, 2) and (−1, −2).
Sphere: dr/dt = 0.1 cm s⁻¹ when r = 5 cm. Find dV/dt.
dV/dr = 4πr² = 100π, so dV/dt = 100π × 0.1 = 10π ≈ 31.4 cm³ s⁻¹.
A square of side x has area increasing at 12 cm² s⁻¹. Find dx/dt when x = 3.
dA/dx = 2x = 6, so dx/dt = 12/6 = 2 cm s⁻¹.
Cooling: dθ/dt = −k(θ − 20). At θ = 60 it falls at 1.5 degrees per minute. Find k.
−1.5 = −k(40), so k = 0.0375 per minute.
Differentiating a constant on the right-hand side
It gives 0. Differentiate both sides of the equation, then collect the dy/dx terms.
Mixed term such as 3xy²
Product rule: 3y² + 6xy dy/dx. Missing either term gives the wrong derivative.
Gradient at a point on an implicit curve
Substitute both coordinates; if only x is given, find y from the curve first.
After setting the numerator of dy/dx to zero…
…you have a relationship (e.g. y = −2x), not the points: substitute it into the curve.
Connected rates: when do you substitute the given value?
After differentiating. Substituting first turns the formula into a constant with derivative 0.
y = xˣ: which rule?
Neither the power rule nor the aˣ rule. Take logs: ln y = x ln x, so dy/dx = xˣ(1 + ln x).

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