#### Topic: tangents to graph

Is it possible to show more than one tangent on the same graph? This doesn't appear to be possible but would be a useful improvement to graph.

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Graph Forums → Support → tangents to graph

Is it possible to show more than one tangent on the same graph? This doesn't appear to be possible but would be a useful improvement to graph.

Yes, you can add all the tangents you want. You add the second tangent similar to the first by selecting the function in the list and selecting *Function|Insert tangent* in the menu.

reply to administrator Ivan Johanson, that doesn't work. Once you have 1 tangent the "insert tangent" function has faded out and no amount of clicking on it will put a second tangent on the graph. The box for where to put the tangent does not come up.

Please advise. thanks.

Remember to select the function in the list to the left before adding the tangent. I have attached a grf file showing a function with 4 tangents.

graph of 4x^0.6. In this graph at x=1, y=4 you get a point on the graph which measured along the curve from the origin is 4.1449. At x=2, y=6.0629 the length of the curve is 6.4385.

How would you express x and y as a function of the distance along the curve from the origin? Can this function be found and if so how do you do it? My maths book "calculus with analytic geometry" does not have anything in this regard.

Thanks for your help so far Ivan.

You can find the math for the distance along the curve here: http://en.wikipedia.org/wiki/Arc_length

Ivan, I know the integration methods to get the length of a curve. These can be very hard to solve. For instance just look at the curve y=4x^0.6 and the integration you need to solve.

A computer would not be able to solve these integration problems. It must be using a series to get the arc length by adding a lot of small increments of the arc length. Can you tell me the series it uses in the calculation?

I have found a way of expressing x and y in terms of the arc length from Graph. You plot points for arc length against the corresponding x and y values. I plotted only 4 points. Then I found a curve of best fit which turned out to be a power series. For the function y=4x^0.6 I obtained x=0.1109L^1.5492, y= 1.0694L^0.9295 where L is the curve length. y is almost linear but the power series fits better.

Graph is using the GNU Scientific Library to calculate the definite integral. Your method seems to be a good way to find it as a function.

Graph can actually plot the the length of the curve from origin to x using this:

`f(x) = integrate(sqrt(1+5.76a^-0.8), a, 0, x)`

However the inverse function where you plot x from a given length seems to be a lot more tricky.

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