Suppose r⃗ (t)=cos(πt)i+sin(πt)j+5tkr→(t)=cos( πt)i+sin(πt)j+5tk represents the position of a particle on a helix, where zz is the height of the particle. (a) What is tt when the particle has height 2020? t=t= (b) What is the velocity of the particle when its height is 2020? v⃗ =v→= (c) When the particle has height 2020, it leaves the helix and moves along the tangent line at the constant velocity found in part (b). Find a vector parametric equation for the position of the particle (in terms of the original parameter tt) as it moves along this tangent line.

a) t = 4

b) v = pi j + 5 k

c) rt = 1i + (pi t) j + (20 +5t )k

Step-by-step explanation:

You have the following vector equation for the position of a particle:

   (1)

(a) The height of the helix is given by the value of the third component of the position vector r, that is, the z-component.

For a height of 20 you have:

(b) The velocity of the particle is the derivative, in time, of the vector position:

   (2)

and for t=4 (height = 20):

(c) The vector parametric equation of the tangent line is given by:

     (3)

ro: position of the particle for t=4

Then you replace ro and v in the equation (3):

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!

the answer is 15.50 for one mile

b) is slope: 0   y- intercept 15.5