(a) Find the unit vectors that are parallel to the tangent...

(a) Find the unit vectors that are parallel to the tangent line to the curve $ y = 2 \sin x $ at the point $ (\frac{\pi}{6}, 1) $. (b) Find the unit vectors that are perpendicular to the tangent line. (c) Sketch the curve $ y = 2 \sin x $ and the vectors in parts (a) and (b), all starting at $ (\frac{\pi}{6}, 1) $.

(a) Find the unit vectors that are parallel to the tangent line to the curve $ y = 2 \sin x $ at the point $ (\frac{\pi}{6}, 1) $.
(b) Find the unit vectors that are perpendicular to the tangent line.
(c) Sketch the curve $ y = 2 \sin x $ and the vectors in parts (a) and (b), all starting at $ (\frac{\pi}{6}, 1) $.

Transcript

00:01 Okay, for this specific function, i mean, which is given by x is equal to two times sine x, we're required to first find the tangent line, that's called l, i mean, at this specific point, pi over six and one.

00:21 So to find the tangent line for a very nice function, a very useful method for us is to use a derivative to first find its slope, right? that's called, suppose kl is the slope of our tangent line, then kl will be just equal to the derivative of f at this point, i mean, at pi over six, right? so the first thing for us is to compute its derivative or f prime is just equal to y prime because f and y are the same thing.

00:55 Prime x, by definition, it's just equal to two times cosine x.

00:59 So kl will be just equal to two times cosine pi over six, which is square root of three.

01:09 So we know as we have the slope of the straight line, the equation of the straight line can be written as y is equal to square root of three times x plus some b.

01:22 This straight line also passes through this point, plug this number into expression, we know our straight line will be actually like that.

01:35 Y is equal to square root of three times x plus one minus pi times square root of three over six.

01:47 Okay, this is the expression of our straight line.

01:50 For the second one, we are required to find the unit vector b that is perpendicular to our tangent line, l.

02:00 Okay, suppose, okay, let's call it two.

02:05 Let's add an arrow to make it clear.

02:09 Suppose this vector has a coordinate, let's call it x naught and y naught.

02:15 Then as it is a unit vector, we know the length of b is one, and the length will be just equal to square root of x naught square plus y naught square.

02:30 It is equivalent to say, okay, now x naught square plus y naught square is one.

02:37 And now use this property as we know the slope of kl.

02:44 Kl is the slope of the tangent line will be square root of three...

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