Find a tangent vector of unit length at the point with the given value of the parameter t. r(t) = 2 sin(t)i + 7 cos(t)j t = π/6

Applying derivatives, the tangent vector of unit length at the point given is:

The vector function is:

The tangent vector is it's derivative, which is given by:

At point , we have that:

The norm of the vector is:

The unit vector is given by each component divided by the norm, thus:

A similar problem is given at link

The tangent vector to r(t) at any t in the domain is

At t = π/6, the tanget vector is

To get the unit tangent, normalize this vector by dividing it by its magnitude:

So the unit tangent at the given point is

y=mx+b

step-by-step explanation:

m slope

b y-intercept

answer: 204

step-by-step explanation:

we are told that the revenue is

x⋅(4080−10x)=−10x^2+4080x

this is a quadratic function, so to maximize it we should find the x-value of its vertex. recall that the x-value of the vertex of the parabola ax^2+bx+c is given by the formula −b/2a.

so in this case, we have a=−10 and b=4080, so the x-value of the vertex is

−b/2a= −(4080)/2(−10)=204