Continuity and Differentiability · Lesson 6 of 8
Derivatives in Parametric Form
“Follow two coordinates controlled by one parameter and obtain their slope without eliminating that parameter.”
• Recognise a parametric description x=f(t), y=g(t). • Derive dy/dx as the ratio of the coordinate rates using the chain rule. • Differentiate trigonometric, polynomial, and logarithmic parametrizations. • Explain why dx/dt must be nonzero for the standard derivative ratio. • Choose and use a helpful parametrization of an implicit relation. • Keep the parameter’s domain and any exceptional inputs in the final conclusion.
One parameter can control two coordinates
A relation between x and y need not be given by a formula y=f(x) or a single equation mixing x and y. Instead, a third variable t can tell us both coordinates: x=f(t) and y=g(t). As t changes, the point (x,y) traces a curve. We can find its slope by comparing how fast its two coordinates change.
A relation written as x=f(t), y=g(t) is in parametric form, and t is the parameter. The parameter may also be written as an angle θ. Each allowed parameter value supplies one point with coordinates x and y.
For example, x=a cosθ and y=a sinθ give x²+y²=a², so the points lie on a circle when a>0. The parameter is the angle selecting a point. In some problems t can be removed to give an explicit or implicit equation, but elimination may be difficult or unnecessary. Differentiation can be done directly from the parametric equations.
Deriving the ratio of rates
Think of t as the input driving both x and y. If y is locally a differentiable function of x, the chain rule relates its change with respect to t to its change with respect to x. Solving that relationship produces the standard parametric derivative formula.
The chain rule gives dy/dt=(dy/dx)(dx/dt). Where dx/dt is nonzero, divide by it to get dy/dx. Thus the rate of vertical coordinate change is divided by the rate of horizontal coordinate change. If the horizontal coordinate is increasing rapidly, a given vertical rate corresponds to a smaller slope; if it is decreasing, the sign of the ratio reflects that direction.
A good calculation has three visible steps: compute dx/dt, compute dy/dt, then divide in the correct order. Do not replace either derivative with the coordinate itself. The quotient y/x is a coordinate ratio, whereas (dy/dt)/(dx/dt) compares coordinate rates.
Problem
Find dy/dx if x=a cosθ and y=a sinθ, with a>0.
- 1.Differentiate each coordinate with respect to θ: dx/dθ=−a sinθ and dy/dθ=a cosθ.
- 2.Where sinθ≠0, the horizontal rate is nonzero. Form the quotient (a cosθ)/(−a sinθ).
- 3.Cancel the nonzero a to obtain dy/dx=−cotθ.
- 4.For example, at θ=π/4 the point is (a/√2,a/√2) and the slope is −1. From the implicit circle equation we would obtain y′=−x/y, which gives the same value and the same expression when y≠0.
Problem
Find dy/dx if x=at² and y=2at, where a>0.
- 1.The coordinate rates are dx/dt=2at and dy/dt=2a.
- 2.For t≠0, their quotient is (2a)/(2at)=1/t.
- 3.Elimination provides a check: t=y/(2a), so y²=4ax. Implicit differentiation gives 2yy′=4a, or y′=2a/y=1/t where y≠0.
- 4.At t=0 the horizontal rate is zero. The ratio cannot supply a finite slope there. The two branches of the parabola have increasingly steep slopes as the vertex is approached.
Problem
Find dy/dx for x=a(θ+sinθ), y=a(1−cosθ), where a>0.
- 1.Differentiate to get dx/dθ=a(1+cosθ) and dy/dθ=a sinθ.
- 2.When 1+cosθ≠0, the ratio is sinθ/(1+cosθ).
- 3.Using sinθ=2sin(θ/2)cos(θ/2) and 1+cosθ=2cos²(θ/2), simplify to tan(θ/2).
- 4.Thus dy/dx=tan(θ/2) at the allowed parameter values. Leaving the answer in θ is appropriate: the parameter already identifies the point and its slope.
A parametric derivative may remain a function of θ or t. Eliminating the parameter is useful for a check or for a requested answer in x,y, but it is not a compulsory step. The purpose of this method is often to avoid that elimination.
Choosing a parametrization that matches an identity
A useful parameter can be introduced even when the original relation is implicit. Look for a familiar identity that resembles the equation. The identity sin²θ+cos²θ=1 suggests a convenient description when fractional powers combine in that pattern.
Problem
For x⁽²⁄³⁾+y⁽²⁄³⁾=a⁽²⁄³⁾, a>0, use a parameter to find dy/dx at ordinary points with nonzero x and y.
- 1.Choose x=a cos³θ and y=a sin³θ. Their two-thirds powers are a⁽²⁄³⁾cos²θ and a⁽²⁄³⁾sin²θ, which add to a⁽²⁄³⁾.
- 2.The coordinate derivatives are dx/dθ=−3a cos²θ sinθ and dy/dθ=3a sin²θ cosθ.
- 3.Where the needed factors are nonzero, divide to obtain dy/dx=−tanθ.
- 4.Since y/x=tan³θ, the real cube root gives tanθ=∛(y/x). Thus dy/dx=−∛(y/x) at these ordinary points.
- 5.The parametrization remains meaningful at its axis points, but the cancellation and slope formula need separate checks there. We have stated the nonzero conditions rather than divide by a vanishing factor.
The cube-root sign in the last result matters. A real cube root preserves the sign of y/x, including in quadrants where that ratio is negative. By contrast, a square root is chosen nonnegative. Keeping those conventions straight connects this calculation to the sign decisions made in inverse trigonometric differentiation.
Combining the method with earlier derivative rules
Parametric form changes how we assemble the final derivative, but each coordinate is still differentiated with the familiar rules. Products, logarithms, inverse trigonometric functions, and roots may all occur inside x(t) and y(t). Check their domains before dividing the two rates.
Problem
Find dy/dx for x=a cosθ, y=b cosθ, with a≠0.
- 1.The derivatives are dx/dθ=−a sinθ and dy/dθ=−b sinθ.
- 2.For sinθ≠0, divide to obtain dy/dx=b/a.
- 3.This is a straight-line relation: y=(b/a)x. It is not the circle parametrization, because both coordinates use cosθ rather than one using sine.
- 4.At sinθ=0 the parameter’s two coordinate rates vanish. The quotient method does not apply directly, although the line relation describes the same slope on its traced domain. A zero coordinate rate can reflect the chosen parameter rather than a failure of the underlying relation.
Problem
Find dy/dx for x=a[cos t+log tan(t/2)], y=a sin t, with a>0 and 0<t<π, away from t=π/2.
- 1.The logarithm is defined because tan(t/2)>0 on 0<t<π. Its derivative is sec²(t/2)/[2tan(t/2)]=1/sin t.
- 2.Thus dx/dt=a[−sin t+1/sin t]=a cos²t/sin t, using 1−sin²t=cos²t.
- 3.The other coordinate gives dy/dt=a cos t. Where cos t≠0, divide the rates to get dy/dx=sin t/cos t=tan t.
- 4.The restriction t≠π/2 is needed because the horizontal rate vanishes there. We retained that restriction while simplifying the ratio.
Problem
Let x=√(exp[(sin⁻¹t)log a]) and y=√(exp[(cos⁻¹t)log a]), where a>0, a≠1, and −1<t<1. Show that dy/dx=−y/x.
- 1.The positive exponential quantities allow us to write x=exp[(sin⁻¹t)log a/2] and y=exp[(cos⁻¹t)log a/2].
- 2.Using the exponential and inverse-sine derivatives gives dx/dt=x log a/[2√(1−t²)].
- 3.Similarly dy/dt=−y log a/[2√(1−t²)], because the inverse-cosine derivative is negative.
- 4.Since a≠1, log a≠0; x is positive and the horizontal rate is nonzero. Dividing cancels the common factors and leaves dy/dx=−y/x.
- 5.If a=1 both coordinates are constantly 1. The parameter then supplies only one point, so the derivative ratio would be 0/0 and would not establish a slope.
The formula (dy/dt)/(dx/dt) requires dx/dt≠0 at the point where it is used. If both rates vanish, cancelling a common factor gives a formula for nearby valid parameters, not an automatic value at the exceptional parameter. Examine the relation or a limit separately if a conclusion there is required.
The same attention applies to parameter-dependent square roots. For example, coordinates containing √(cos2t) in a denominator require cos2t>0. A logarithm of tan(t/2) requires a positive tangent. Such conditions belong to the coordinates before any derivative is taken; a simplified slope cannot expand the original traced domain.
Check your understanding
Find the two coordinate derivatives before choosing a ratio. These questions also distinguish a formula expressed in a parameter from one that is valid at every parameter value. In the practice, avoid eliminating the parameter unless it is requested as a check.
Quiz
If x=f(t), y=g(t), and f′(t)≠0, which gives dy/dx?
For x=a cosθ and y=a sinθ, a>0 and sinθ≠0, what is dy/dx?
For x=at² and y=2at, a>0 and t≠0, what is dy/dx?
Which condition is required for the standard parametric derivative ratio at a point?
For x=a(θ+sinθ), y=a(1−cosθ), with a>0, what is the derivative where 1+cosθ≠0?
For x=a cosθ, y=b cosθ with a≠0 and sinθ≠0, what is the slope?
What does dx/dt=dy/dt=0 at one parameter value establish by itself?
Why is a≠1 needed when dividing the rates for x=√(exp[(sin⁻¹t)log a]), y=√(exp[(cos⁻¹t)log a])?
Practice Problems
- Without eliminating t, find dy/dx for x=2at², y=at⁴, assuming a≠0 and t≠0.
- Find dy/dx for x=a cosθ, y=b cosθ. Explain why the same trigonometric factor gives a straight-line relation.
- Find dy/dx for x=sin t, y=cos2t, and for x=4t, y=4/t. State the restrictions needed in each calculation.
- Find dy/dx for x=cosθ−cos2θ, y=sinθ−sin2θ. Keep track of both coordinate derivatives before simplifying.
- Find dy/dx for x=a(θ−sinθ), y=a(1+cosθ), a>0. State where dx/dθ is nonzero.
- Find dy/dx for x=sin³t/√(cos2t), y=cos³t/√(cos2t). Begin by recording cos2t>0 and then use product or quotient rules.
- For x=a[cos t+log tan(t/2)], y=a sin t, derive dy/dx on 0<t<π with dx/dt≠0.
- Find dy/dx for x=a secθ, y=b tanθ, assuming a,b>0 and the required trigonometric expressions and coordinate ratio are defined.
- Find dy/dx for x=a(cosθ+θsinθ), y=a(sinθ−θcosθ), a>0. Explain the cancellations in dx/dθ and dy/dθ.
- For x=√(exp[(sin⁻¹t)log a]), y=√(exp[(cos⁻¹t)log a]), prove dy/dx=−y/x under the conditions a>0, a≠1, and −1<t<1.
- Parametrize x⁽²⁄³⁾+y⁽²⁄³⁾=a⁽²⁄³⁾ using x=a cos³θ, y=a sin³θ. Verify the equation and find the slope at a point with x,y both nonzero.
- Give an example where dx/dt and dy/dt both vanish at a parameter value but the traced relation is a straight line. Explain why the quotient formula alone is insufficient at that parameter.
Key Takeaways
• A parameter assigns both coordinates of a point and may describe a curve without solving for y. • The chain rule gives dy/dx=(dy/dt)/(dx/dt) wherever dx/dt is nonzero. • Differentiate both coordinates with respect to the same parameter before forming their ratio. • An answer in t or θ is often the natural final form; elimination is optional unless requested. • Trigonometric identities can suggest useful parametrizations of implicit relations. • Keep the coordinates’ original domain and the nonzero horizontal-rate condition through every simplification.