Secondary 3 Elementary Math (15 year olds)
Topic 5.4 - Gradient of a Curve
How do you estimate a gradient of a curve on a piece of graph paper? The gradients on each point on a curve are different, so this video shows you how to estimate the gradient of a point on a curve on a piece of graph paper.
(For 15 year olds)
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Okay let’s look at how you can find the gradient of a point on a curve that you draw. So for example if you have a curve that looks like this and you want to find a gradient of a point, say here… So what you need to do is to take a ruler and try to angle the ruler such that this angle here and this angle here is the same. now you can only do this visually and as I have mentioned before that curve sketching is essentially an estimate.
So as much as you can see if this angle you know, is a visual check. This and this angle is the same you can say that this line, the slope of this line or the gradient will be equivalent to the gradient of this point. Now if you remember the formula for gradient is y1 – y2 / x1 – x2.
So you need to select two points on your line, any two points, so say this point, and this point, and then find the coordinates so that you have y1 and y2, and then find the x coordinate: x1 and x2. This point if you minus y2 from y1 you will have the height here and if you minus x2 from x1, you will have the base so if you take this divided by this you have the gradient so see another example.
Say you have a coordinate system and you want to find the gradient of a point here so visually take a ruler and take an estimate such that your, your line is parallel to this point, to the curve here and make sure as best as you can the angle here and here are the same. So you find two points on your straight line.
Now this is coordinate x1, y1 and coordinate x2, y2, so put in your formula the gradient is equals to y1 – y2, over x1 – x. Lets take a look at a concrete example.
You are given this question: the variables x and y are connected by the equation y=1/2(5x-x^2) so the table below show some values of x and corresponding values of y. Now first you are asked to find the values of a and b. That shouldn’t be too difficult because you know the expression for y, so when y is a, you can put in the values for x to find y and the value for x is minus 0.5 -0.5 square, so if you work this out you will get minus one, three over eight.
It’s the same thing for b. To find b when y is b, x is 2.5. so just dump in the x values 5/2 minus 5/2 square if you work this out you will get b = 3 1/8. The main point of this question is to show how to find the tangent that occurs at point (1,2). In order to do that you have to draw out this curve and to figure out the curve, you will have to take out a graph paper, draw the x and y axis, use the correct scale, plot it out.
So after you plot out the curve you should have something like this. This is y-axis and this is your x-axis. On this graph, you can see that the scale for y. The y and x is 2cm represents one unit. so this scale is 1cm.. 1cm, 2cm represents 1 unit on the y axis, and 2 cm represents 1 unit on the x axis. So these points are the points on the table on the previous slide.
Once you mark out these points on the graph paper, you can sketch it using your ruler and your freehand to get something like this. For part (c), they ask you to find out the gradient.. at a point at point (1,2). So at point (1,2), it’s this point so what you should do is ready to take a ruler and try to angle it in a way
Now, your ruler you can angle it many many ways You can angle it like that, you can angle it like this you know but all these are incorrect. You should try to angle it in such a great way where the angle here and the angle here are the same. Or as similar as your naked eye can see. so once you have the green line something close to the green line you can arbitrarily choose two points say this point and this point and get the coordinate.
The coordinate on this point is 2 and the coordinate for the point here is 1 and 2 so, you can find the gradient to be equals to y1 minus y2 over x1 – x2 and if you work this out you will see that the gradient is one and a half. 1.5 so you have found the gradient at this point. Now the definition of a gradient is a slope, and this is a curve so obviously at every single point the gradient will change unlike a straight line.
On a straight line the gradient is the same because the slope of a straight line is always the same. But on a curve is you choose and other points here, the slope will be different. So you have to do the same thing to find the value of the gradient at another point. Alright so for part (d). The gradient of the curve at point (h,k) is 0. Draw the tangent at the point (h,k) hence find the value of h and k.
Now, this is easy because when the gradient is zero, m is 0, it means that there is no slope. So the line is horizontal. It’s a straight horizontal line. And if you look at the curve, there is only one point that the slope is flat. It is a the top point here… at the maximum point. Here, this point. So if you take the ruler and try to do the same thing, find the slope.. find the line such that this angle and this angle is the same.. is a straight line.. horizontal line. So you can see that there’s only this point of the graph that has gradient 0. This point is 2.5, 2.5 here the x value is 2.5 and the y value is 3.1 something but the accuracy of your graph is.. we can only get the closest possible value is 3.1.
So.. 2.5 and 3.1.
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Secondary 3 E-Maths (For 15 year olds)
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New Syllabus Mathematics is a series of textbooks specially designed to provide valuable learning experiences to engage the hearts and minds of students sitting for the GCE O level examination in Mathematics. Included in the textbooks are Investigation, Class Discussion, Thinking Time, Journal Writing, Performance Task and Problems in Real-world Contexts to support the teaching and learning of Mathematics.
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