Question

Consider the tangent line to the curve y = 6 sin(x) at the point (π/6, 3). (a) Find a unit vector that is parallel to the tangent line to the curve at the given point.

Answer 1

Answer:

$y=\frac{\pi }{6}+5.196(x-3)$

Step-by-step explanation:

We have given the equation y = 6 sin (x)

On differentiating both side $\frac{dy}{dx}=m=6cosx$

As it passes through the point $(\frac{\pi }{6},3)$

So $\frac{dy}{dx}=6cos\frac{\pi }{6}=5.196$

Now the unit vector is parallel to the tangent so m will be 5.196

This passes through the point $(\frac{\pi }{6},3)$

So unit vector will be $y-\frac{\pi }{6}=5.196(x-3)$

$y=\frac{\pi }{6}+5.196(x-3)$