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Equation of the tangent to the curve at a point

Calculus is one of those branches of math that I must say I love it. It help us to find solutions to problems where the other branches fail.

The following is just an example where calculus allows us to find elegantly the equation of the tangent to the curve at a point.

Let's suppose that we have a curve given by f(x)=\frac{sin(x)}{x-1}. Now let's assume that I would choose an arbitrary point on the graph at x=1.62.

What is the equation of the tangent to the curve at the point [1.62,f(1.62)] ?

Equation of the tangent to the curve

This is the area where calculus comes in rescue.

We know that the concept of derivative is related to the rate of change of y-value with respect to x-values, also \frac{\Delta_x}{\Delta_y}. The slope of a line (eg. y=kx+m) is also \frac{\Delta_x}{\Delta_y}. If we want to determine the tangent to the curve we should determine the slope of that tangent. If we know a point that belongs to that tangent (and we know it, it's [1.62,f(1.62)]) and we know also the slope of that tangent then we can determine the equation of the tangent to the curve at that point.

So how fast is changing the y-value with respect to the change of the x-value? This question/statement can be written (in Leibnitz notation) as \frac{d}{dx}f or f'(x) and it is what we know it as the derivative of the function f. The derivative is just another function, as the name says, derived from the original function (like the speed which is just a function derived from a function that shows the change of the distance with respect to the time-axe).

Let's derivate our original function:

\frac{d}{dx}f(x)=\frac{cos(x)\cdot (x-1)-sin(x)}{(x-1)^2}

This is the that can answer us to the question underlined earlier. If we plug in a value the function will tell us what is the rate of the change of y-value for that x-value (also the slope of the original function at that location). So "what's the slope of tangent line to the curve f at x=1.62?" is the same like asking "what is f'(1.62)?".

f'(1.62)=\frac{cos(1.62)\cdot (1.62-1)-sin(1.62)}{(1.62-1)^2}\approx -2.68

The equation of the tangent (like the equation of any straight line) is y=kx+m where k is the slope of the line and m is the y-intercept (also the y-value when x=0).

So the equation that we are looking for looks like y=-2.68\cdot x+m

What is the value of m and how to determine it? Well, just because we already know a point that belongs to the tangent line we could use it to determine the m-value, right?

We know that the point [1.62,f(1.62)] belongs to the curve but it is also the point where the tangent touches the curve. With other words the point [1.62,f(1.62)] belongs to the tangent line too. It means that when x=1.62 the tangent should have a value of 1.61 (which is f(1.62); don't get confused by the fact that 1.62 just happened to be very close to 1.61, it's just the hazard).

Let's check what this means when we plug it in our y-line:

y(1.62)=1.61 <=> y(1.62)=-2.68\cdot 1.62+m=1.61

If we solve this equation for m we get that m=5.9516. So our tangent has the equation y=-2.68\cdot x +5.95.

This is our final answer and as you can see it corresponds to the green line in the image above.

If you like it the read another post about the derivative application. You will not regret it!

Now, if you think that this article was interesting don't forget to rate it. It shows me that you care and thus I will continue write about these things.

 
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Equation of the tangent to the curve at a point

Eugen Mihailescu

Founder/programmer/one-man-show at Cubique Software
Always looking to learn more about *nix world, about the fundamental concepts of math, physics, electronics. I am also passionate about programming, database and systems administration. 16+ yrs experience in software development, designing enterprise systems, IT support and troubleshooting.
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