Mc9 - Tangents and normals to curves

Mc9 - Tangents and normals to curves

Construct tangent and normal equations from a curve point and its derivative, as required by Mc9. You will deal with horizontal and vertical directions, find contacts from a specified gradient, and read a short mathematical passage which develops a new relation from familiar normals.

A tangent needs a point and a gradient

Can a line capture a curve's direction at one point and still be useful somewhere else? At x=1x=1, the curve y=2x3−3x+1y=2x^3-3x+1 passes through the horizontal axis with tangent gradient 33. A point and direction are enough to identify that line exactly.

You already know

A line through (a,b)(a,b) with finite gradient mm has equation y−b=m(x−a)y-b=m(x-a). Differentiate first and evaluate afterwards; the derivative gives the tangent gradient, while the function gives the contact height.

Keep the contact input fixed when writing the tangent. Its slope is a number, even though the gradient function varies along the curve.

Tangent at a curve point

For y=f(x)y=\mathrm f(x), at P=(a,f(a))P=(a,\mathrm f(a)) with a finite derivative,

y−f(a)=f′(a)(x−a).y-\mathrm f(a)=\mathrm f'(a)(x-a).

Not in the formula sheet: learn it.

The point-gradient form makes both ingredients visible and works without first finding an intercept.

Worked example 1: constructing the tangent

Question. Find the tangent to y=2x3−3x+1y=2x^3-3x+1 at x=1x=1.

Plan. Find the contact point and the derivative value, then combine them.

y(1)=2−3+1=0contact heightdydx=6x2−3differentiatem=6(1)2−3=3evaluate at contacty=3(x−1)=3x−3line through (1,0)\begin{aligned} y(1)&=2-3+1=0 && \text{contact height}\\ \frac{\mathrm dy}{\mathrm dx}&=6x^2-3 && \text{differentiate}\\ m&=6(1)^2-3=3 && \text{evaluate at contact}\\ y&=3(x-1)=3x-3 && \text{line through }(1,0) \end{aligned}

Why this step? Using the derivative function itself as a changing slope would produce a curve, rather than this straight tangent.

Answer. y=3x−3y=3x-3.

Check. The line contains (1,0)(1,0) and has the required gradient 33.

Change only the sign of the contact input. The height changes, although this derivative happens to have the same value.

Worked example 2: a parallel tangent elsewhere

Question. Find the tangent to the same curve at x=−1x=-1.

Plan. Evaluate the new point and gradient separately before writing the line.

y(−1)=−2+3+1=2new heightm=6(−1)2−3=3new slopey−2=3(x+1)use the new pointy=3x+5simplify\begin{aligned} y(-1)&=-2+3+1=2 && \text{new height}\\ m&=6(-1)^2-3=3 && \text{new slope}\\ y-2&=3(x+1) && \text{use the new point}\\ y&=3x+5 && \text{simplify} \end{aligned}

Why this step? Equal gradients give parallel lines, but different contact points can give different intercepts.

Answer. y=3x+5y=3x+5.

Check. At x=−1x=-1, the line gives 22. It is 88 units above the previous tangent at every input.

A tangent can meet the curve again away from contact. For the first line,

2x3−3x+1=3x−3⟺2(x−1)2(x+2)=0.2x^3-3x+1=3x-3 \quad\Longleftrightarrow\quad 2(x-1)^2(x+2)=0.

So it also meets the cubic at (−2,−9)(-2,-9). That is the useful surprise: tangent means the local direction, not “a line with only one intersection”.

Watch out

Wrong: use a supplied point without checking it lies on the curve. Right: confirm its height and evaluate the derivative at the same input. The right slope through the wrong point gives a parallel line.

Choose a fresh contact input on the same cubic.

The normal is perpendicular to the tangent

A normal keeps the contact point but turns the tangent direction through a right angle. Think about the two directions before doing any algebra with gradients.

Normal

The normal at a curve point is the line through that point perpendicular to its tangent. The derivative gives the tangent's gradient; it does not directly give the normal's gradient.

In this equal-scale diagram, the tangent and normal to y=x2y=x^2 at (1,1)(1,1) have gradients 22 and −1/2-1/2. Both pass through the same curve point.

The perpendicular tangent and normal at a parabola point

Normal gradient and equation

If the tangent gradient m=f′(a)m=\mathrm f'(a) is finite and nonzero, then

n=−1m,y−f(a)=n(x−a).n=-\frac1m,\qquad y-\mathrm f(a)=n(x-a).

The negative-reciprocal step requires m≠0m\ne0.

Not in the formula sheet: learn it.

For a fractional gradient, invert the fraction as well as changing its sign. Tangent gradient −3/4-3/4 gives normal gradient 4/34/3, since their product is −1-1.

Worked example 3: finding a normal

Question. Find the normal to y=2x2y=2x^2 at x=1x=1.

Plan. Calculate the contact height and tangent gradient, then use the perpendicular direction.

P=(1,2)evaluate the curvedydx=4x,m=4differentiate; substituten=−14negative reciprocaly−2=−14(x−1)line through Px+4y=9rearrange\begin{aligned} P&=(1,2) && \text{evaluate the curve}\\ \frac{\mathrm dy}{\mathrm dx}&=4x,\qquad m=4 && \text{differentiate; substitute}\\ n&=-\frac14 && \text{negative reciprocal}\\ y-2&=-\frac14(x-1) && \text{line through }P\\ x+4y&=9 && \text{rearrange} \end{aligned}

Why this step? The line must be perpendicular at the curve point, so keep the point while replacing the gradient.

Answer. x+4y=9x+4y=9.

Check. 1+4(2)=91+4(2)=9, and 4(−1/4)=−14(-1/4)=-1.

Change only the contact input's sign. This checks both the reciprocal and the signs in the point-gradient form.

Worked example 4: a reflected normal

Question. Find the normal to y=2x2y=2x^2 at x=−1x=-1.

Plan. Use the new point and derivative value before taking the negative reciprocal.

P=(−1,2)new curve pointm=4(−1)=−4,n=14perpendicular gradienty−2=14(x+1)point-gradient formx−4y=−9rearrange\begin{aligned} P&=(-1,2) && \text{new curve point}\\ m&=4(-1)=-4,\qquad n=\frac14 && \text{perpendicular gradient}\\ y-2&=\frac14(x+1) && \text{point-gradient form}\\ x-4y&=-9 && \text{rearrange} \end{aligned}

Why this step? The tangent's negative gradient produces a positive normal gradient, not another negative gradient.

Answer. x−4y=−9x-4y=-9.

Check. The line contains (−1,2)(-1,2), and (−4)(1/4)=−1(-4)(1/4)=-1.

Writing separate symbols mm and nn is a small habit that prevents answering a tangent question when a normal was requested.

Exam tip

Check both conditions: the line contains the curve point, and its gradient is perpendicular to the tangent's. Passing through the correct point alone does not establish that a line is the normal.

Move the contact input to a fresh value.

Horizontal and vertical directions

A horizontal tangent has gradient zero. Its perpendicular normal is vertical and has no finite gradient, so this case needs geometry rather than division by zero.

Horizontal and vertical pair

At P=(a,f(a))P=(a,\mathrm f(a)), if f′(a)=0\mathrm f'(a)=0,

tangent: y=f(a),normal: x=a.\text{tangent: }y=\mathrm f(a),\qquad \text{normal: }x=a.

Not in the formula sheet: learn it.

One equation keeps height constant; the other keeps the horizontal coordinate constant. Check which direction the question asks for.

Worked example 5: a normal with no finite gradient

Question. Find the tangent and normal to y=x2+6x+2y=x^2+6x+2 at x=−3x=-3.

Plan. Find the contact point and verify that the tangent gradient is zero.

y(−3)=9−18+2=−7contact heightdydx=2x+6differentiatem=2(−3)+6=0horizontal tangent\begin{aligned} y(-3)&=9-18+2=-7 && \text{contact height}\\ \frac{\mathrm dy}{\mathrm dx}&=2x+6 && \text{differentiate}\\ m&=2(-3)+6=0 && \text{horizontal tangent} \end{aligned}

Why this step? A zero tangent gradient makes the normal vertical; assigning it gradient zero would make it parallel instead.

Answer. Tangent y=−7y=-7; normal x=−3x=-3.

Check. Both contain (−3,−7)(-3,-7) and have perpendicular directions.

Change only the constant term. The derivative remains unchanged, but the contact height moves.

Worked example 6: the same vertical normal

Question. Find the tangent and normal to y=x2+6x+5y=x^2+6x+5 at x=−3x=-3.

Plan. Recalculate the height and use the same derivative.

y(−3)=9−18+5=−4new heightm=2(−3)+6=0same gradient\begin{aligned} y(-3)&=9-18+5=-4 && \text{new height}\\ m&=2(-3)+6=0 && \text{same gradient} \end{aligned}

Why this step? Moving a curve upwards changes its horizontal tangent's height, but not its vertical normal's horizontal position.

Answer. Tangent y=−4y=-4; normal x=−3x=-3.

Check. The tangent has moved up 33, matching the curve's translation.

Vertical tangents are possible too. For the real cube-root curve y=x1/3y=x^{1/3}, the derivative 13x−2/3\tfrac13x^{-2/3} exists for x≠0x\ne0 and grows positively without bound towards zero from both sides. At the origin, write x=y3x=y^3: the horizontal change divided by vertical change is y3/y=y2→0y^3/y=y^2\to0. This establishes a vertical tangent x=0x=0 and horizontal normal y=0y=0. Their equations come from the geometry and are not in the formula sheet: learn the reasoning.

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