Is there an alternative method to find gradient, such as using tangents?

[PA3040D]

Dear Experts, the following question addresses finding the gradient of line B.. While my method works, I suspect there might be a shorter and less time-consuming approach. Is there an alternative method, such as using the tangent function (tan(theta)), to achieve the same result?

 

[khansaheb]

Dear Experts, the following question addresses finding the gradient of line B.. While my method works, I suspect there might be a shorter and less time-consuming approach. Is there an alternative method, such as using the tangent function (tan(theta)), to achieve the same result?

View attachment 38064

I cannot figure out your "method". In my opinion the shortest-way to calculate the gradient of a 'straight-line' joining points (X1,Y1) and (X2,Y2) is:

Gradient = (rise)/(run) = (Y2 - Y1) / (X2 - X1)

For the problem shown (X1 = 0, X2 = 2, Y1 = 0 and Y2 = 2)

Gradient = (X2 - X1) / (Y2 - Y1) = (2 - 0) / (2 - 0) = 2/2 = 1

 

[PA3040D]

I cannot figure out your "method". In my opinion the shortest-way to calculate the gradient of a 'straight-line' joining points (X1,Y1) and (X2,Y2) is:

Gradient = (rise)/(run) = (Y2 - Y1) / (X2 - X1)

For the problem shown (X1 = 0, X2 = 2, Y1 = 0 and Y2 = 2)

Gradient = (X2 - X1) / (Y2 - Y1) = (2 - 0) / (2 - 0) = 2/2 = 1

Thank you for your response to my post. I'm actually looking to determine the function of line "B" in my picture, with the intention of applying the Fourier series.

I employed the method of calculating the difference in Y-coordinates divided by the difference in X-coordinates.

 

[Dr.Peterson]

Dear Experts, the following question addresses finding the gradient of line B.. While my method works, I suspect there might be a shorter and less time-consuming approach. Is there an alternative method, such as using the tangent function (tan(theta)), to achieve the same result?

View attachment 38064

You wrote:

​

The RHS of this is the gradient (also called the slope). The LHS is nonsense; and on the third line, you seem to transmute it (illegally) to what it should have been if the goal was to derive the equation of the line:

​

This equation says that the gradient from [imath](x_1, y_1)[/imath] to [imath](x, y)[/imath] is the same as the gradient from [imath](x_1, y_1)[/imath] to [imath](x_2, y_2)[/imath].

But you don't need to find the equation of the line (in the form [imath]y = mx + b[/imath]) to find the slope; you already had the slope as soon as you wrote -1/2.

Thank you for your response to my post. I'm actually looking to determine the function of line "B" in my picture, with the intention of applying the Fourier series.

I employed the method of calculating the difference in Y-coordinates divided by the difference in X-coordinates.

Apparently you misstated your goal, and you did want the equation, not just the gradient.

For that, there is a somewhat easier way: First find the gradient (-1/2), and then use the point-slope form, [imath]y-y_1=m(x-x_1)[/imath], using the gradient just found, and either of the two known points.

You certainly don't need the tangent of an angle that you don't know.

 

[PA3040D]

Dear Sir, I greatly appreciate your support and advice. While it's much clearer now the mistakes that I have done. I'm still a bit confused about finding the line function easily without directly substituting values into y = mx + c, despite watching the tutorials provided on the webpage.

Thanks in Advanced

You wrote:

View attachment 38066​

The RHS of this is the gradient (also called the slope). The LHS is nonsense; and on the third line, you seem to transmute it (illegally) to what it should have been if the goal was to derive the equation of the line:

View attachment 38067​

This equation says that the gradient from [imath](x_1, y_1)[/imath] to [imath](x, y)[/imath] is the same as the gradient from [imath](x_1, y_1)[/imath] to [imath](x_2, y_2)[/imath].

But you don't need to find the equation of the line (in the form [imath]y = mx + b[/imath]) to find the slope; you already had the slope as soon as you wrote -1/2.


Apparently you misstated your goal, and you did want the equation, not just the gradient.

For that, there is a somewhat easier way: First find the gradient (-1/2), and then use the point-slope form, [imath]y-y_1=m(x-x_1)[/imath], using the gradient just found, and either of the two known points.

You certainly don't need the tangent of an angle that you don't know.

 

[Dr.Peterson]

Dear Sir, I greatly appreciate your support and advice. While it's much clearer now the mistakes that I have done. I'm still a bit confused about finding the line function easily without directly substituting values into y = mx + c, despite watching the tutorials provided on the webpage.

Thanks in Advanced

When you say "Thanks in advanced", I believe you mean "thanks in advance", meaning, before I actually do what you are thanking me for.

But you have not asked me a question, or implied one. What is it, specifically, that still confuses you? How can I make it clearer?

I gave you a link (not to a video tutorial to watch), and I told you what the point-slope form is. Plugging in values in y = mx + b and solving for b is one common way to find the equation, but this is a different way, and it does not appear that you have tried it yet.

Perhaps you need to show your work for whatever method confuses you, so I can see that nature of your confusion.

 

[Steven G]

Although the formula you write is correct I think that looking at the situation is better, especially with the numbers you were given.

To go from the top left of line B to the bottom right of line B you go down 2 and to the right 4. So the slope/gradient is simply -2/4 = -1/2.