Conceptual question

[Nazariy]

Hello,

I want to know how you would think about the following, as my method was not very time efficient, nor was it neat.

You have a graph of sinx. Now you draw a tangent at the origin of sin x as well. Now how do you go about deciding how to sketch it? You cannot use any wolframs or any software packages, this is purely conceptual. I can tell you how I did it, but without getting you biased I can only do that once I have received a couple neat answers. Thanks

the domain is 0 to pi/2

 

[Ishuda]

Hello,

I want to know how you would think about the following, as my method was not very time efficient, nor was it neat.

You have a graph of sinx. Now you draw a tangent at the origin of sin x as well. Now how do you go about deciding how to sketch it? You cannot use any wolframs or any software packages, this is purely conceptual. I can tell you how I did it, but without getting you biased I can only do that once I have received a couple neat answers. Thanks

the domain is 0 to pi/2

Draw an x-y axis. Lightly sketch a slope 1 line slope 1 up and to right. Go up line 'a ways', drop down about 1/3 of the distance to the x axis & put a dot [represents point (\(\displaystyle \frac{\pi}{2},\, 1\))]. Lightly sketch a horizontal line through the dot. Fair a curved line tangent to the slope one line and slope zero line going through the origin and dot.

 

[Steven G]

Hello,

I want to know how you would think about the following, as my method was not very time efficient, nor was it neat.

You have a graph of sinx. Now you draw a tangent at the origin of sin x as well. Now how do you go about deciding how to sketch it? You cannot use any wolframs or any software packages, this is purely conceptual. I can tell you how I did it, but without getting you biased I can only do that once I have received a couple neat answers. Thanks

the domain is 0 to pi/2

As you stated you are given the graph of sin x. So you just need to draw the tangent line. This line crosses through the origin and has a slope of 1. If the x and y scale are the same then you just draw a line that is a 45 degree angle above the positive x axis. If the scales are different you need to adjust the angle.

 

[Nazariy]

As you stated you are given the graph of sin x. So you just need to draw the tangent line. This line crosses through the origin and has a slope of 1. If the x and y scale are the same then you just draw a line that is a 45 degree angle above the positive x axis. If the scales are different you need to adjust the angle.

Sure, there is no problem with that. However, how do you know whether the tangent line crosses or does not cross the sin x graph before reaching pi/2? Now I know that the sine graph goes below the tangent line, but how would I have decided that? I realized that slope of tangent is 1 because I differentiated sin x and then used x = 0, so cos 0 = 1, hence 45 degree line. I then realized that at pi/2 the tangent line is above sin(pi/2) because pi/2 > 1, around 1.57; that is as far as I got.

However it appears that sin(x)<x in the domain 0 to pi/2 .. Well I should have arrived at that conclusion without wolfram. And I wanted to know how I could have realized that on my own, without just plugging in numbers and comparing, because if u use something like pi/3, sine if which is sqrt3/2.. It becomes less and less obvious whether tangent line is above sine graph (i.e. U are comparing sin(pi/3) and pi/3 I.e. Sqrt3/2 and pi/3 ...)

 

[Deleted member 4993]

Sure, there is no problem with that. However, how do you know whether the tangent line crosses or does not cross the sin x graph before reaching pi/2? Now I know that the sine graph goes below the tangent line, but how would I have decided that? I realized that slope of tangent is 1 because I differentiated sin x and then used x = 0, so cos 0 = 1, hence 45 degree line. I then realized that at pi/2 the tangent line is above sin(pi/2) because pi/2 > 1, around 1.57; that is as far as I got.

However it appears that sin(x)<x in the domain 0 to pi/2 .. Well I should have arrived at that conclusion without wolfram. And I wanted to know how I could have realized that on my own, without just plugging in numbers and comparing, because if u use something like pi/3, sine if which is sqrt3/2.. It becomes less and less obvious whether tangent line is above sine graph (i.e. U are comparing sin(pi/3) and pi/3 I.e. Sqrt3/2 and pi/3 ...)

Have taken any calculus?

Have you worked through the proof of:

\(\displaystyle \lim_{x \to 0}\dfrac{sin(x)}{x} \to 1\)?

 

[Nazariy]

Have taken any calculus?

Have you worked through the proof of:

\(\displaystyle \lim_{x \to 0}\dfrac{sin(x)}{x} \to 1\)?

I'm learning everything by myself. I have done some calculus so far, however I have not come across the above. But I can see that as x approaches 0 from the positive side, then sin(x) becomes very small and is positive and x becomes very small and is positive. When you divide by infinitely small number you obtain a large number, however the numerator is also small, so the outcome depends on whether the numerator is larger or smaller or same when compared to denominator. It is clear from what you have written that two numbers converge, i.e. become the same. But why that is, I wouldn't be able to say, and how would this help me in the mentioned problem? I can see that sinx is multiplied by 1/x but how does that relate to y=x.