Practical lesson 1. The geometric meaning of the derivative
The practical lesson is devoted to the geometric meaning of the derivative. We will talk about tangents. Let’s solve a simple problem. So, a quadratic function is given, remember, the graph is a parabola. We need to solve several problems for one and the same function f(x)=x2-2x-3. First, we make an equation of the tangent and the normal to this parabola at the point (3; 0). Second, we determine the angle at which the graph intersects the Ox axis at this point (point (3; 0) lies on the Ox axis). And also we answer the question at which point the graph has a horizontal tangent. We guess that the tangent passes through the vertex, but we have to get it analytically. And make the equation of the tangent and the normal at this point.
Let’s get it started. So, make the equation of the tangent and the normal to the graph of the function at the point (3; 0). Let’s check whether the point belongs to the graph of the function: х0=3, indeed, f(х0)=0, everything is correct, it is always worth checking. Let’s see how we will use the tangent equation. The equation of the tangent: y-y0=f’(х0)(x-х0).
Let’s start with this. We need to find the value of the derivative at the point х0: f’(х)=2х-2, f’(3)=4.. Substitute the found value into the equation: y-0=4(x-3), y=4x-12. The equation found is the equation of the tangent.
To write the normal equation, you can only look at the tangent equation, which we wrote down as an equation with an angular coefficient. The point through which the normal passes remains the same, but the angular coefficient of the normal is kn=-1/4. We keep in mind that two straight lines are perpendicular if the product of their angular coefficients is -1. So, the normal is a straight line passing through the point of contact perpendicular to the tangent: y-0=-1/4(x-3), y=-1/4x+3/4. We got the answer – the normal equation. The first point of the problem is solved. Let’s move on to the next point.
At what angle at point (3; 0) does the graph intersect the Ox axis? Previously, we determined that f’(3)=4. The geometric meaning of the derivative is that the tangent of the angle at which the tangent intersects the Ox axis is 4, which means that the angle is the arctangent 4: tgα =4, α=arctg 4. All we need to know is just the value of the derivative. You don’t need to look at the tangent equation at all, everything turns out at once.
Furthermore, at which point does the graph have a horizontal tangent? The horizontal tangent has an angular coefficient equal to zero k=0. A straight line y=kx + b is horizontal if k=0. For a tangent, the angular coefficient is a derivative, so we must solve the equation and determine where the derivative is zero: f’(х)=0 => 2х-2=0 => х=1. At point x, the value of the derivative is zero, which means that the tangent is horizontal at the corresponding point on the graph. Let’s see what this point is: х0=1, у0=1-2-3=-4. The horizontal tangent passes through the point with coordinates (1; -4), so y=-4 is the equation of the tangent, and the line x=1 is the normal.
We didn’t even need to know the tangent equation and the normal equation (where, by the way, the derivative doesn’t have to be zero). This problem has also been solved. Let’s look at the figure, what we have found out (see the video): the parabola with the vertex at (1; -4), the initial tangent and the normal, the angle α=arctg 4, the tangent and the normal at (1, -4).
The next task is very interesting and is worth solving. At what angle is the parabola у=х2 visible from some point A(0, -2) (note that the point does not belong to the parabola)?
What is meant here? So, look at the figure (see the video), we can draw two tangents from point A to the parabola, the resulting solution (two rays) forms the angle we are talking about. At this angle, the parabola is visible from the point A. The problem is that we don’t know the points where the tangents are drawn. Let’s solve the problem. We will assume that х0 is the desired point, there are even two of them. This means that the coordinates of the tangency point are (х0; х02).
In general, the tangent equation is: y-y0=f’(х0)(x- х0), y0= х0^2. Let’s look at the value of the derivative at the point х0: f’(х0)=2х0. Despite the fact that х0 is unknown, we can make the tangent equation: y-х02=2х0(x- х0). This is the equation of the tangent passing through the point (х0, у0). Everything would be fine, but х0 is unknown to us. We have not yet used the condition that A belongs to this tangent, that is, the coordinates of point A satisfy this equation y-х02=2х0(x- х0). We substitute the coordinates of the point A in the equation (y and x are variables, and х0 is a fixed point whose value we need to find): -2-х02= -2х02. We get: х02=2 => х01=21/2, х02=-21/2 (the points can be seen in the figure). The coordinates of the tangency point become known to us. To avoid starting all over again from scratch, we can use the equation y-х02=2х0(x-х0). We substitute х0=√2 into this equation and get y=√2x-2 – this is the equation of the first tangent.
In the second case, we substitute -√2 instead of х0, the sign will change in the equation, y= - √2x-2. We have two tangents. Our task is to find the angle at which the parabola is visible from the point A. The task is to find the angle between the lines y=√2x-2 and y=-√2x-2. You need to use the tangent of the angle between the straight lines. If the straight lines are given by angular coefficients, the formula is as follows: tgα=|(k1-k2)/(1+k1*k2)|. Note that k1=√2, k2=-√2. After substituting k1, k2 into the formula and simplifying the expression, we get: 4*√2/7. We write out the answer: α=arctg(4*√2/7). The problem is solved.
We remembered the condition of perpendicular lines, learned how to calculate the angle between straight lines, and now we will consider another problem where the parallelism of straight lines is considered.
So, again, the same parabola у=х2, and the problem is to answer at which point the tangent to this parabola is parallel to the secant passing through the points of the parabola A(-1, 1) and B(2, 4). Our task is to find the point х0, belonging to the parabola through which the tangent passes. We remember that two straight lines defined by the equations with angular coefficients k1, k2 are parallel if the values of the angular coefficients coincide, k1=k2. We need to find the angular coefficients of the straight line AB and the tangent at the point х0.
To find the angular coefficient of a straight line passing through points A and B, we divide the ordinate difference by the abscissa difference: (4-1)/(2-(-1))=1.
The angular coefficient of the tangent of the parabola at the point х0, which is unknown to us, will be equal to 2х0. So, 2х0=1 => х0=1/2. We have found at which point the tangent is parallel to the straight line AB. The first coordinate of the point is equal to 1/2, and the second is easy to calculate, it is equal to 1/4. The problem is solved.