Question

18.01SC Single Variable Calculus.

My question is from Session 8 on the geometric proof of d/dx sinx = cosx

I will try my best to be clear on my questions.

1) Why does it matter that angle OPQ is a right angle?

2) We know that PR is vertical, we know that θ
is the angle OP makes with the horizontal, and we can combine these facts to prove that angle RPQ and θ are (nearly) congruent angles.

Why is this true?

If I just accept that angle PQR is equal to theta then everything makes sense. I'm just missing why angle PQR equals to theta.

Thanks for any help!

Answer 1

Thank you for the drawing :)

Answer 2

It's beautiful!!!

Answer 3

I think it's easier to see the answer from this version of the drawing, which extends line RP down to the x-axis.
Notice that triangle OSP is a right triangle, so angle OPS is the complement to theta (in other words, 90 degrees minus theta). The total number of degrees in a straight angle is 180, so if OPS is complementary to theta, and OPQ is "approximately" 90 degrees, then the remaining piece of the straight angle, which is QPR, is "approximately" equal to theta. Prof. Jerison's way of explaining this is valid also but harder to see (for me at least).

Answer 4

Hey Creeksider, I just understood it the way you said it as well before. I just wanted to understand prof. Jerison's way. Can you explain why angle OPQ is almost a right angle? Is the tangent line at point P?

Answer 5

While perhaps harder to visualize, I can briefly state what Prof. Jerison was explaining. Line RP is a 90 degree rotation of line OS, which forms one side of angle theta. If angle OPQ is "approximately" 90 degrees, then line PQ is "approximately" a 90 degree rotation of line OP, which is the other side of angle theta. When you rotate both sides of an angle by the same amount (90 degrees in this case) you preserve the size of the angle, so angle QPR has to be equal to theta (approximately). The fact that angle OPQ approaches 90 degrees as delta theta approaches zero is essential to either of these two ways of arriving at the conclusion that angle QPR equals theta at the limit.

Answer 6

The reason OPQ is almost a right angle is that we are looking at the limit as delta theta goes to zero. For example, when delta theta is 1 degree, OPQ is 89.5 degrees.

Answer 7

sorry to bother you but if delta theta goes to 0, wouldn't it be approximately line segment OP? I'm just curious why delta theta tends to 0 would mean OPQ is near 90 degrees.

Answer 8

ah wait a moment, when delta theta goes to 0, the arc QP will essentially be a line segment QP right?

Answer 9

but why does that mean line segment QP is perpendicular to line segment OP?

Answer 10

That's a reasonable question. You're right, if delta theta goes all the way to zero, we no longer have an angle and all the reasoning goes out the window. But here we're working with the concept of a limit, which allows us to escape that contradiction. We haven't gone deep into the theory in this course, but a key element of the concept of a limit is to treat something as being equal if we can get absolutely as close as we want, even if we can't get all the way there. Recall that this is how derivatives work, because the basic definition is the limit as delta x goes to zero of the difference quotient, which has delta x in the denominator. We can't take it all the way to zero because we can't divide by zero, but we can see what it does as it approaches zero.

In this case we're applying the same concept to a geometric argument. We can't take delta theta all the way to zero, but we can observe what happens as it gets closer and closer to zero. Two important things happen. One is that line segment PQ gets closer and closer to arc length PQ, and the other is that angle OPQ approaches 90 degrees. Strictly speaking, it isn't enough to show that angle OPQ is approximately 90 degrees, we have to be able to make it as close to 90 degrees as we want. No matter what nonzero number you choose (say, 0.00001 degrees) I can find a figure for delta theta close enough to zero so that angle OPQ is even closer to 90 degrees than that. This establishes that the limit of OPQ as delta theta approaches zero is 90 degrees, even though we can never take delta theta all the way to zero.

Answer 11

As to your question about why line segment QP is perpendicular to line segment OP, let's just backtrack a moment. There are 180 degrees in a triangle, and triangle OPQ is an isosceles triangle with angles OPQ and OQP being equal. When angle delta theta (which is angle POQ) approaches zero, the other two angles approach 90 degrees. Although OPQ never gets all the way to 90 degrees, it approaches 90 degrees at the limit, and line segments that differ by 90 degrees are perpendicular.

Answer 12

By the way, don't be discouraged if this particular argument is hard to grasp. Prof. Jerison seems to have seen confusion in the faces of the students in his classroom, and toward the end, after laying it out as well as he could, he said something like, "Well, I never said all arguments were supposed to be easy."

Answer 13

yea im just trying to wrap my head around it. THanks for the help man!!