Question

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.

Answer 1

Answer:

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:

$r(t)=cos(\pi t)\hat{i}+sin(\pi t)\hat{j}+5t\hat{k}$    (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:

$5t=20\\\\t=\frac{20}{5}=4$

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

$v(t)=\frac{dr(t)}{dt}=-\pi sin(\pi t)\hat{i}+\pi cos(\pi t)\hat{j}+5\hat{k}$    (2)

and for t=4 (height = 20):

$v(t=4)=-\pi sin(\pi (4))\hat{i}+\pi cos(\pi (4))\hat{j}+5\hat{k}\\\\v(t=4)=-0\hat{i}+\pi\hat{j}+5\hat{k}$

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

$r_t(t)=r_o+vt$      (3)

ro: position of the particle for t=4

$r_o=cos(\pi (4))\hat{i}+sin(\pi (4))\hat{j}+20\hat{k}\\\\r_o=\hat{i}+0\hat{j}+20\hat{k}$

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

$r_t=(1\hat{i}+20\hat{k})+(\pi \hat{j}+5\hat{k})t\\\\r_t=1\hat{i}+\pi t \hat{j}+(20+5t)\hat{k}$

Answer 2

Part(a): The value of $t=4$

Part(b): Required vector $L(t)=(1\widehat{i}+0\widehat{j}+10\widehat{k})+(t-4)(0\widehat{i}+\pi \widehat{j}+5\widehat{k})$

Given vector equation is,

$r(t)=cos(\pi t)\widehat{i}+sin(\pi t)\widehat{j}+5t\widehat{j}$

Vector equation:

A vector is an object that has both a magnitude and a direction. Geometrically, we can picture a vector as a directed line segment, whose length is the magnitude of the vector, and with an arrow indicating the direction.

Part(a):

When the particle has a height of 20

$5t=20\\t=4$

Part(b):

The point on the curve is $(cos(4\pi),sin(4\pi),20) =(1,0,20)$

Differentiating the given equation with respect to $t$.

$r'(t)=- \pi sin(\pi t)\widehat{i}+\pi cos(\pi t)\widehat{j}+5\widehat{k}\\r'(t)=- \pi sin(4\pi t)\widehat{i}+\pi cos(4\pi t)\widehat{j}+5\widehat{k}\\r'(4)=0\widehat{i}+\pi \widehat{j}+5\widehat{k}\\L(t)=r(4)+(t-4)r'(4)\\L(t)=(1\widehat{i}+0\widehat{j}+10\widehat{k})+(t-4)(0\widehat{i}+\pi \widehat{j}+5\widehat{k})$

Learn more about the topic Vector equation:

https://brainly.com/question/24906745