Implicit differentiation and related rates

An implicit curve may have several branches

An equation such as x²+y²=25 describes a circle rather than one global function y=f(x). Away from the left and right extremes, its upper and lower arcs can be treated locally as functions of x. Implicit differentiation works on a differentiable local branch without first solving for y. Verify any supplied point in the original equation

Every y term carries its chain factor

When differentiating with respect to x, y=y(x), so d(y²)/dx=2y y′ and d(sin y)/dx=cos y·y′. In a product xyⁿ, the product rule gives yⁿ+nxyⁿ⁻¹y′. Constants have derivative0. Collect every y′ term before isolating the derivative

A finite tangent gradient

For x²+y²=25, differentiation gives2x+2y y′=0, so y′=−x/y wherever y≠0. The point(3,4) lies on the curve and has tangent gradient−3/4. The negative sign is consistent with the falling upper-right arc

For x²+y²=25, give the tangent gradient at(3,4)

Collect the product-rule contributions

For xy²+x²+2y=9 at(1,2), differentiate to y²+2xy y′+2x+2y′=0. Thus(y²+2x)+(2xy+2)y′=0. Substitution gives6+6y′=0 and y′=−1. The coefficient of y′ is nonzero, so this calculation gives a finite local gradient

For xy²+x²+2y=9, give y′ at(1,2)

Construct a tangent or normal from the actual point

For2(x−1)²+3(y+1)²=14 at(2,1), y′=−2(x−1)/[3(y+1)]=−1/3. The tangent is y=−x/3+5/3 and the normal y=3x−5. The normal uses a negative reciprocal only when the tangent gradient is finite and nonzero

For 2(x−1)²+3(y+1)²=14 at(2,1), give the normal’s y-intercept

A vertical tangent is not a zero gradient

For a regular implicit curve G(x,y)=0, collect Gx+Gy y′=0. If Gy≠0, y′=−Gx/Gy. If Gy=0 but Gx≠0, the curve may instead be represented locally by x as a function of y, with dx/dy=0: its tangent is vertical. If Gx=0 and Gy≠0, the tangent is horizontal

What are the tangent and normal at the marked point?

A singular point needs separate analysis

If Gx=Gy=0, the quotient gives0/0 and the regular-point test is inconclusive. At the origin, xy=0 is the union of two axes, with two crossing tangent directions. In contrast, x²+y²=0 has only an isolated real point and no curve arc to which a tangent can be assigned. Do not label every0/0 case vertical

Circular implicit differentiation uses radians

For sin y+2xy−x²=0, the derivative equation is cos y·y′+2y+2xy′−2x=0. Thus y′=(2x−2y)/(cos y+2x), where the denominator is nonzero. At(0,π), the original equation holds and the denominator is−1, so the gradient is2π

For sin y+2xy−x²=0, give M when the gradient at(0,π) is Mπ

Time is now the independent variable

For related rates, define all changing quantities as functions of the same time t. A length, area or volume at one instant is not a constant throughout the motion. Differentiate the governing relation with respect to t before inserting instantaneous values. Fixed lengths and imposed ratios can be constants; changing dimensions carry their own rates

A ladder links two changing lengths

A5m rigid ladder has its foot3m from the wall and top4m above the floor. If the foot moves outward at1/5m/s, then x²+y²=25 gives2x ẋ+2y ẏ=0 and ẏ=−3/20m/s. For the acute angle θ with the floor, x=5cosθ gives θ̇=−ẋ/y=−1/20rad/s. Both the top height and angle decrease at this instant

A 5m ladder has its foot 3m from the wall and its top 4m above the floor. If its foot moves outward at 1/5m/s, give the signed rate of change of the top height in m/s.

Forward and reverse rates for a sphere

For an ideal sphere, V=4πr³/3 and S=4πr². Differentiating in time gives V̇=4πr²ṙ and Ṡ=8πrṙ. When r=3cm and ṙ=−1/10cm/s, V̇=−18π/5cm³/s and Ṡ=−12π/5cm²/s. Given V̇ instead, divide by4πr² to recover ṙ; this requires positive radius

For a sphere with r=3cm and ṙ=−1/10cm/s, give the coefficient ofπ in its volume rate

Two independent cylinder dimensions need the product rule

For V=πr²h with both dimensions varying, V̇=π(2rh ṙ+r²ḣ). At r=2cm,h=3cm,ṙ=−1/10cm/s,ḣ=3/10cm/s, the contributions are−6π/5 and6π/5cm³/s, giving net rate0. Neither dimension rate is zero. No fixed similarity relation was assumed

For V=πr²h at r=2cm, h=3cm, ṙ=−1/10cm/s and ḣ=3/10cm/s, give the net volume rate in cm³/s.

In a fixed cone, which expression correctly converts a depth derivative into a time rate?

Check physical configuration and units

A ladder requires positive interior horizontal and vertical distances and maintained contacts; a sphere requires positive radius. A water-depth model must stay between the cone’s tip and rim. These models justify an instantaneous rate near the stated configuration, not arbitrary extrapolation. Rate signs, physical quantity positivity and model validity are separate checks

Derive mixed implicit rules, ungraded

Differentiate each relation on a regular local branch: x sin y+x²cos y=1 at(1,0); x e^y+y=1 at(1,0); and y/x+x/y=5/2 at(1,2). State domain exclusions, collect y′ terms and verify the supplied points before finding their gradients

Model: chain, product and quotient contributions

The circular relation gives y′=−(sin y+2x cos y)/(x cos y−x²sin y), hence−2 at(1,0). The exponential relation gives y′=−e^y/(x e^y+1), hence−1/2. For the quotient relation x,y must both be nonzero: (xy′−y)/x²+(y−xy′)/y²=0, giving−3/2+(3/4)y′=0 and y′=2 at(1,2). Each formula additionally requires its collected y′ coefficient to be nonzero

Explain the tangent exceptions, ungraded

Compare the circle x²+y²=25 at(5,0), the crossing xy=0 at(0,0), and the isolated real locus x²+y²=0. Explain why an undefined slope quotient does not by itself decide whether there is a vertical tangent, several tangent directions or no curve arc

Model: distinguish regular and singular geometry

At(5,0), the circle gradient(10,0) is nonzero, so the tangent is the vertical line x=5 and the normal horizontal. For xy=0, both partial derivatives vanish at the origin, and direct factorisation reveals two axes. For x²+y²=0, nonnegative squares force the single point(0,0); there is no neighbouring real curve arc. The last two cases require examining the original locus, not dividing0 by0.

Choose and justify a volume model, ungraded

Compare water in a fixed cone, a family of similar cylinders and a cylinder whose radius and height vary independently. State which quantities are fixed, which ratio if any remains fixed, the valid configuration interval and the time-derivative equation. Explain why an instantaneous radius cannot be substituted as a permanent constant before differentiation

Model: assumptions determine which chain is valid

In a fixed cone R,H are constant and r/h=R/H, so V=πR²h³/(3H²) and V̇=πR²h²ḣ/H² for0<h<H. For similar cylinders h/r=c remains fixed, so V=cπr³ and V̇=3cπr²ṙ. With independent cylinder dimensions use V̇=π(2rh ṙ+r²ḣ) and r,h>0. Setting r to its instantaneous value before differentiation would erase a genuine radius contribution. Whether these assumptions fit an actual situation remains a modelling judgement

Check a related rate with technology, ungraded

For the5m ladder example, model the foot locally by x(t)=3+t/5 and the top by y(t)=√[25−x(t)²]. In radian mode also use θ(t)=arccos[x(t)/5]. Compare central differences at t=0 using smaller positive time steps with−3/20m/s and−1/20rad/s. Keep both sample configurations physical and record your own calculations

Model: a local numerical check

With x(t)=3+t/5, y(t)=√[25−x(t)²] and θ(t)=arccos[x(t)/5], for sufficiently small |t| both lengths are positive and below5m, so the relations describe the same local configuration. Central differences of y and θ at t=0 tend to−0.15 and−0.05 respectively. Their units differ. This numerical comparison supports the differentiated equations; it does not justify carrying the motion through contact loss or establish that a learner executed the investigation

Exact values and complete explanations

Give implicit derivative coefficients, finite or axis-aligned tangent and normal lines, and exact signed time rates in the stated forms. Distinguish geometric derivatives with respect to length from time rates. Full derivations, arbitrary symbolic forms, complete sketches and contextual model validation remain ungraded