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Tangent to a Curve on mousemove with d3.js and math.js

January 26, 2016

Following on from my last post on axes positioning, I have added the functionality to add a tangent to the curve on mousemove. You can see a working example here by dragging the mouse over the svg document.

Below is a screenshot of the end result:


The first steps are to create the elements that I will use to display the tangent indicator and also to hook up an event listener for mousemove on the svg document:

  .attr('class', 'diff')
  .attr('cx', 0)
  .attr('cy', 0)
  .attr('r', 7)
  .style('fill', 'red')

g.append('text').attr('class', 'difflabel')

  .style('stroke', 'red')
  .attr('class', 'tangent')
  .attr('x1', xScale(0))
  .attr('y1', yScale(0))
  .attr('x2', xScale(0))
  .attr('y2', yScale(0))'svg').on('mousemove', mouseMove)

The above creates a circle to indicate where on the curve the mouse is relative to the x axis. A label is created to display the coordinate that the mouse is currently on with respect to the curve and lastly I create a line that will display the tangent.

Line 19 adds a mousemove handler to the svg element that has been previously created with the code below:

componentDidMount() {
  const el = this.refs.curve;

  const dimensions = this.getDimensions();

  this.svg ="svg")
        .attr("width", dimensions.width + dimensions.margin.left + dimensions.margin.right)
        .attr("height", dimensions.height + + dimensions.margin.bottom)
        .attr("transform", "translate(" + dimensions.margin.left + "," + + ")");

The goal of the mousemove handler is to draw the tangent of the curve with respect to the x axis as the mouse moves over the svg document.

In geometry, the tangent line to a plane curve at a given point is the straight line that just touches the curve at that point.

I can get the x for the tangent line from the mouse coordinates of the mousemove function:

const mouseMove = function() {
  const m = d3.mouse('.curve').node());

  let x = m[0];

With this, I can use mathematical differential calculus to work out the tangent line. If I was to perform these steps with pen and paper, I would take the following steps:

  • Find the derivative of the curve
  • Substitute the x retreived from the mousemove event into the derivative to calculate the gradient (or slope for the US listeners) of the line.
  • Substitute the gradient of the tangent and the coordinates of the given point into the equation of the line in the format y = mx + c.
  • Solve the equation of the line for y.

What was surprising and enjoyable for me was that the steps on paper transferred into machine instructions quite well which is not always the case.

Before I plot the line, I want to position my circle and label onto the curve. I am already using the excellent mathjs library to get the coordinates to draw the curve:

getDataFromProps(expr) {
  const expression = math.parse(expr);

  const fn = (x) => {
    return expression.eval({x: x});

  return d3.range(-10, 11).map( (d) => {
    return {x:d, y:fn(d)};

drawCurve(data) {
  const xScale = this.xScale;
  const yScale = this.yScale;
  const svg = this.svg;

  const line = d3.svg.line()
          .x( (d) => {return xScale(d.x);})
          .y( (d) => {return yScale(d.y);});

Line 2 of the above code uses mathjs’s parse function to create an expression from the string input of the form:


Once I have the expression, I can evaluate it with different values. Lines 5 and line 9 evaluates the expression for each x value in a predetermined range of values. Line 19 plots the line.

As I know what x is from the mouse event, I can use mathjs to parse my expression with respect to x and get the y coordinate to position my label on the curve:

const mouseMove = function() {
  const m = d3.mouse('.curve').node());

  let x = m[0];

  let y = yScale(math.parse(me.props.expression).eval({
    x: xScale.invert(x)

  const point = {
    x: xScale.invert(x),
    y: yScale.invert(y)

  if(point.x > maxX) {
    point.x = maxX;
    point.y = maxY;

    x = xScale(maxX);
    y = yScale(maxY);
    .attr('cx', x)
    .attr('cy', y);'.difflabel')
    .text( function() {
      const xLabel = Math.round(point.x);
      const yLabel = Math.round(point.y);

      return `(${xLabel}, ${yLabel})`;
    .attr('dx', x + 10)
    .attr('dy', y + 8);

Lines 2 and 4 retrive x from the mouse event and line 6 evaluates y by parsing and evaluating the equation of the curve with respect to x. I then use the x and y coordinates to position my label elements.

Mathjs does not come with its own differentiation module to work out the derivative of the user entered expression but I found this plugin that seems to work out well for this task.

Armed with this module, it was plain sailing to create an equation for the tangent line that I could use to find out y values for the tangent.

const derivative = math.diff(math.parse(me.props.expression), 'x')

const gradient = derivative.eval({ x: point.x })

const yIntercept = getYIntercept(point, gradient)

const lineEquation = math.parse('m * x + c')

const getTangentPoint = delta => {
  const deltaX = xScale.invert(x + delta)

  const tangentPoint = {
    x: deltaX,
    y: lineEquation.eval({
      m: gradient,
      x: deltaX,
      c: yIntercept,

  return tangentPoint

const length = xScale(200)

const tangentPoint1 = getTangentPoint(+length)
const tangentPoint2 = getTangentPoint(-length)'.tangent')
  .attr('x1', xScale(tangentPoint1.x))
  .attr('y1', yScale(tangentPoint1.y))
  .attr('x2', xScale(tangentPoint2.x))
  .attr('y2', yScale(tangentPoint2.y))
  • Line 1 creates a derivative function from the equation of the curve.
  • The derivative function is evaluted to get the gradient or slope for the US viewers on line 3.
  • The c constant or y-intercept of the equation of the line y = mx + c is retrieved on line 5.
  • Mathjs is employed to create an expression in the format of y = mx + c.
  • A getTangentPoint function is created that will be used to get the points at either end of the function.
  • Lines 24 - 26 create 2 points for x that will be far off the length and height of the svg document to give the impression of stretching off to infinity.
  • Each point gets its y value by calling the getTangentPoint funtion on line 9 that in turn will solve for x for the equation of the line function previously created on line 7.
  • Once we have the pair of points, the line can be plotted on lines 29 - 33.

You can see the end result here.

The following util function will return the y-intercept for a point and gradient:

export function getYIntercept(vertex, slope) {
  return vertex.y - slope * vertex.x

Paul Cowan

Nomadic cattle rustler and inventor of the electric lasso.
Company Website
Follow me on twitter
Contact me for frontend answers.