Yes. It all goes back to h = angle = arc length, when you use radians, in the derivation of the proofs of the derivatives of sine and cosine. In degrees the arc length corresponding to angle h is s = Pi*x/180. I always just mumbled something about "radians and arc length..." when students asked. *sigh*
Nice demonstration. I had one question though. At 17:57, you're dividing through by h*cos(h). And you mention that as h is small and positive, this is fine and the inequality signs aren't affected. But the actual limit we're interested in is h->0, not h->0+. I.e. we want the whole limit, not just approaching zero from the positive side. So, it's important that we also look at h->0- isn't it? My impression is that since you're using a geometric argument in the proof, and a "negative angle" doesn't really have meaning in that context, maybe this isn't actually an issue. Is that right? If you allow for "negative angles", I think you might have to start over with the areas and have the relevant areas of the sectors including the absolute value of h instead of just the naked h. I'm sure the end result is the same, but a mention of why we can just ignore the small, negative h case or how it is dealt with would be better than no mention.
Good question: from the odd property of the sine function, we see that the ratio sin(h)/h is an even function. Thus we can look at the limit for h +ve and know we'd get the same result for h -ve. Hence the geometric argument suffices.
Same reason really. If you want the derivative of arctan(x) to be 1/(1+x^2) you need the angle defined via arctan to be measured in radians. (Think about using implicit differentiation to obtain the result.)
Changing the units to measure angles changes the scale on the x-axis. It thus stretches (or compresses) the graph horizontally. Limits depend upon the angle measure used. Eg, the limit as x->0 for sin(x)/x is one in radians and is pi/180 if you use degrees. Hope this helps.
I have a question in calculus and that i never the got answer and searched online also but couldnt find satisfactory result. What does limit at infinity actually mean please help me when can find limit at a certain point this make sense but when limit at infinity what does this actually please tell.
Dear Ankur, Limits are treated rigorously in analysis but here is a quick response. You can't be at infinity - which I think you recognise. Instead ask what happens to a function as the dependent variable becomes larger and larger. For example, 1/x will tend to zero. It is not zero for any finite value of x but gets smaller and smaller as x increases. We say that 1/x tends to zero as x tends to infinity. On a graph the x axis is a horizontal asymptote. Similarly x/(1+x) will tend to 1 as x tends to infinity. (You can rewrite it as 1/(1+1/x) and, as x tends to infinity, this approaches 1/(1+0)=1.) Hope this helps.
It would mean more if you can demonstrate why using "degrees" would go bonkers compared to using "rad" in a real world application eg Sniper Bullet leaving the barrel.
It'd only mess you up if you're taking derivatives, and then the conversion is relatively simple: (d/dx)[ sin(x) ] = (Pi/180)cos(x) if you want to use degrees, by the Chain Rule, and the conversion of x to radians inside the sine, via the factor Pi/180, giving you the Pi/180*cos(x*Pi/180). Convert back to degrees inside the cosine: Pi/180*cos(x), when x is in degrees.
Yes. It all goes back to h = angle = arc length, when you use radians, in the derivation of the proofs of the derivatives of sine and cosine. In degrees the arc length corresponding to angle h is s = Pi*x/180.
I always just mumbled something about "radians and arc length..." when students asked. *sigh*
Nice demonstration. I had one question though. At 17:57, you're dividing through by h*cos(h). And you mention that as h is small and positive, this is fine and the inequality signs aren't affected. But the actual limit we're interested in is h->0, not h->0+. I.e. we want the whole limit, not just approaching zero from the positive side. So, it's important that we also look at h->0- isn't it? My impression is that since you're using a geometric argument in the proof, and a "negative angle" doesn't really have meaning in that context, maybe this isn't actually an issue. Is that right? If you allow for "negative angles", I think you might have to start over with the areas and have the relevant areas of the sectors including the absolute value of h instead of just the naked h. I'm sure the end result is the same, but a mention of why we can just ignore the small, negative h case or how it is dealt with would be better than no mention.
Good question: from the odd property of the sine function, we see that the ratio sin(h)/h is an even function. Thus we can look at the limit for h +ve and know we'd get the same result for h -ve. Hence the geometric argument suffices.
@@PlymUniMaths Thanks. That's a very straightforward way to deal with it.
Excellent,sir why we use radian in inverse circular function
Same reason really. If you want the derivative of arctan(x) to be 1/(1+x^2) you need the angle defined via arctan to be measured in radians. (Think about using implicit differentiation to obtain the result.)
Can i ask? Do limits must be in radian mode? like sin 0.6?
It is different when i used degree mode
Changing the units to measure angles changes the scale on the x-axis. It thus stretches (or compresses) the graph horizontally. Limits depend upon the angle measure used. Eg, the limit as x->0 for sin(x)/x is one in radians and is pi/180 if you use degrees. Hope this helps.
I have a question in calculus and that i never the got answer and searched online also but couldnt find satisfactory result. What does limit at infinity actually mean please help me when can find limit at a certain point this make sense but when limit at infinity what does this actually please tell.
Dear Ankur, Limits are treated rigorously in analysis but here is a quick response. You can't be at infinity - which I think you recognise. Instead ask what happens to a function as the dependent variable becomes larger and larger. For example, 1/x will tend to zero. It is not zero for any finite value of x but gets smaller and smaller as x increases. We say that 1/x tends to zero as x tends to infinity. On a graph the x axis is a horizontal asymptote.
Similarly x/(1+x) will tend to 1 as x tends to infinity. (You can rewrite it as 1/(1+1/x) and, as x tends to infinity, this approaches 1/(1+0)=1.)
Hope this helps.
It would mean more if you can demonstrate why using "degrees" would go bonkers compared to using "rad" in a real world application eg Sniper Bullet leaving the barrel.
It'd only mess you up if you're taking derivatives, and then the conversion is relatively simple:
(d/dx)[ sin(x) ] = (Pi/180)cos(x) if you want to use degrees, by the Chain Rule, and the conversion of x to radians inside the sine, via the factor Pi/180, giving you the Pi/180*cos(x*Pi/180). Convert back to degrees inside the cosine: Pi/180*cos(x), when x is in degrees.