Pure 7.5 - Implicit and parametric differentiation

Pure 7.5 - Implicit and parametric differentiation

Find a curve's gradient even when its equation does not give yy directly in terms of xx. Differentiate implicit relations and parametric equations, then use the resulting first derivative to find tangents and normals, including horizontal and vertical cases.

Hidden Curves and First Derivatives

How can you find the tangent at (1,2)(1,2) on x2+xy+y2=7x^2+xy+y^2=7 without first isolating yy? Substituting the point gives 1+2+4=71+2+4=7, so it is on the curve. Rearranging the entire relation into explicit branches would be possible but awkward. We can differentiate the relation itself.

You already know

The chain rule differentiates a function of a changing input: (u2)′=2uu′(u^2)'=2uu'. The product rule gives (uv)′=uv′+vu′(uv)'=uv'+vu'. A tangent uses the curve's derivative; a normal is perpendicular.

An implicit relation can describe more than one yy-value for a single xx-value. Near a suitable point, one branch still has a gradient.

[DIAGRAM: asset_slug: p1_7_5_differentiate_simple_functions_and_relations_defined_implicitly_or_parametrically__diagram_01; asset_name: A tangent to an implicit curve; caption: Differentiate the relation to find the local slope without isolating y; description: Equal-scale axes show x squared plus xy plus y squared equals 7, P(1,2), tangent slope minus four fifths and normal slope five quarters.]
Differentiate the relation to find the local slope without isolating y

Implicit Relation

An implicit relation links xx and yy in one equation, such as x2+xy+y2=7x^2+xy+y^2=7, rather than already giving y=f(x)y=f(x). When finding dy/dx\mathrm{d}y/\mathrm{d}x, treat yy as a function of xx along the selected branch.

Consequently,

ddx(y2)=2ydydx,ddx(xy)=xdydx+y\frac{\mathrm{d}}{\mathrm{d}x}(y^2)=2y\frac{\mathrm{d}y}{\mathrm{d}x}, \qquad \frac{\mathrm{d}}{\mathrm{d}x}(xy)=x\frac{\mathrm{d}y}{\mathrm{d}x}+y

Both are applications of the recalled rules, which are not printed in the booklet. A bare 2y2y would be the derivative with respect to yy, not with respect to xx. The same chain factor makes d(y3)/dx=3y2 dy/dx\mathrm{d}(y^3)/\mathrm{d}x=3y^2\,\mathrm{d}y/\mathrm{d}x.

Key idea

Differentiate every term with respect to the same variable. Terms involving yy must include the way yy changes with xx.

The first derivative is our target here. Once it gives the tangent gradient, straight-line geometry supplies the tangent or normal.

Implicit Differentiation

Differentiate the whole relation, collect every term containing dy/dx\mathrm{d}y/\mathrm{d}x, then solve for it. Substitute a point afterwards: inserting numbers first destroys the changing quantities.

Worked example 1: the opening implicit tangent

Question. Find the tangent to x2+xy+y2=7x^2+xy+y^2=7 at P(1,2)P(1,2).

Plan. Use product rule for xyxy, chain rule for y2y^2, then collect the derivative terms.

2x+xdydx+y+2ydydx=0differentiate every term(x+2y)dydx=−(2x+y)collect and factordydx=−2x+yx+2yx+2y≠0mP=−2+21+4=−45substitute the point\begin{aligned} 2x+x\frac{\mathrm{d}y}{\mathrm{d}x}+y+2y\frac{\mathrm{d}y}{\mathrm{d}x}&=0 && \text{differentiate every term}\\ (x+2y)\frac{\mathrm{d}y}{\mathrm{d}x}&=-(2x+y) && \text{collect and factor}\\ \frac{\mathrm{d}y}{\mathrm{d}x}&=-\frac{2x+y}{x+2y} && x+2y\ne0\\ m_P&=-\frac{2+2}{1+4}=-\frac45 && \text{substitute the point} \end{aligned}

Why this step? The product xyxy changes through both factors, so its derivative has two terms.

Answer. Tangent y−2=−45(x−1)y-2=-\frac45(x-1), or 4x+5y=144x+5y=14.

Check. PP satisfies the curve and tangent, and the line's gradient is −4/5-4/5.

Change only the coefficient of xyxy from 11 to 33. The constant is adjusted to keep the same point on the curve.

Worked example 2: a larger product coefficient

Question. Find the tangent to x2+3xy+y2=11x^2+3xy+y^2=11 at P(1,2)P(1,2).

Plan. Include the factor 33 on both terms from differentiating xyxy.

2x+3y+3xdydx+2ydydx=0differentiatedydx=−2x+3y3x+2ysolve for the derivativemP=−2+63+4=−87evaluate\begin{aligned} 2x+3y+3x\frac{\mathrm{d}y}{\mathrm{d}x}+2y\frac{\mathrm{d}y}{\mathrm{d}x}&=0 && \text{differentiate}\\ \frac{\mathrm{d}y}{\mathrm{d}x}&=-\frac{2x+3y}{3x+2y} && \text{solve for the derivative}\\ m_P&=-\frac{2+6}{3+4}=-\frac87 && \text{evaluate} \end{aligned}

Why this step? The constant on the right differentiates to zero, so the slope calculation depends on the changing terms on the left.

Answer. y−2=−87(x−1)y-2=-\frac87(x-1), or 8x+7y=228x+7y=22.

Check. The curve gives 1+6+4=111+6+4=11, and the tangent gives 8+14=228+14=22.

At (1,1)(1,1) on x2+3xy+y2=5x^2+3xy+y^2=5, the same derivative gives slope −1-1 and tangent x+y=2x+y=2. The point must always be checked against the particular curve.

Watch out

Wrong: (xy)′=x dy/dx(xy)'=x\,\mathrm{d}y/\mathrm{d}x. The missing yy is the contribution from differentiating xx. Write x dy/dx+yx\,\mathrm{d}y/\mathrm{d}x+y.

Change the product coefficient again while keeping PP on the relation.

Parametric Differentiation

A parameter describes both coordinates: x=x(t)x=x(t), y=y(t)y=y(t). To find the slope in the xx-yy plane, divide the vertical rate by the horizontal rate.

Parametric First Derivative

For differentiable x(t)x(t), y(t)y(t), where dx/dt≠0\mathrm{d}x/\mathrm{d}t\ne0,

dydx=dy/dtdx/dt\frac{\mathrm{d}y}{\mathrm{d}x} =\frac{\mathrm{d}y/\mathrm{d}t}{\mathrm{d}x/\mathrm{d}t}

Not in the formula booklet: learn it.

This follows from the chain rule dy/dt=(dy/dx)(dx/dt)\mathrm{d}y/\mathrm{d}t=(\mathrm{d}y/\mathrm{d}x)(\mathrm{d}x/\mathrm{d}t). It is a ratio of derivatives, not y/xy/x.

[DIAGRAM: asset_slug: p1_7_5_differentiate_simple_functions_and_relations_defined_implicitly_or_parametrically__diagram_02; asset_name: A parametric gradient; caption: At t=2, vertical rate 9 divided by horizontal rate 4 gives slope 9 over 4; description: The curve x=t squared plus 1, y=t cubed minus 3t has P(5,2) at t=2 with its tangent. A separate rate-vector panel shows horizontal component 4 and vertical component 9, explicitly labelled as rates.]
At t=2, vertical rate 9 divided by horizontal rate 4 gives slope 9 over 4

Worked example 3: a parametric tangent

Question. For x=t2+1x=t^2+1, y=t3−3ty=t^3-3t, find the tangent at t=2t=2.

Plan. Differentiate both coordinates, take the correct ratio, and find the curve point separately.

dxdt=2t,dydt=3t2−3coordinate ratesdydx=3t2−32tt≠0m=94,P=(5,2)substitute t=2\begin{aligned} \frac{\mathrm{d}x}{\mathrm{d}t}&=2t,\qquad \frac{\mathrm{d}y}{\mathrm{d}t}=3t^2-3 && \text{coordinate rates}\\ \frac{\mathrm{d}y}{\mathrm{d}x}&=\frac{3t^2-3}{2t} && t\ne0\\ m&=\frac94,\qquad P=(5,2) && \text{substitute }t=2 \end{aligned}

Why this step? Dividing vertical rate by horizontal rate gives rise per run.

Answer. y−2=94(x−5)y-2=\frac94(x-5).

Check. A small positive parameter change gives displacement approximately (4Δt,9Δt)(4\Delta t,9\Delta t), with slope 9/49/4.

Change only the parameter value to t=1t=1. This produces a horizontal tangent.

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