Question

A child wanders slowly down a circular staircase from the top of a tower. With x,y,zx,y,z in feet and the origin at the base of the tower, her position tt minutes from the start is given by x = 10cos t, y = 10sin t, z = 90 - 5t.


(a) How tall is the tower?


(b) when does the child reach the bottom?


(c) What is her speed at tme t?


(d) What is her acceleration at time t?

Answer 1

Answer:

a) The tower is 90 feet tall

b) She reaches the bottom at t = 18 minutes.

c) Her speed at time t is 5 \sqrt[]{5} ft/minute

d) Her acceleration at time t is 10 ft/minute^2

Step-by-step explanation:

Consider the path described by the child as going down the tower to have the following parametrization $\gamma(t) = (10\cos t, 10 \sin t, 90-5t)$

a) Assuming that the child is at the top of the tower when she starts going down, we have that at the initial time (t=0) we will have the value of the height of the tower. That is z = 90-5*0 = 90 ft.

b) The child reaches the bottom as soon as z =0. We want to find the value of t that does that. Then we have 0 = 90-5t, which gives us t = 18 minutes.

c) Given the parametrization we are given, the velocity of the child at time t is given by $\frac{d\gamma}{dt}= (\frac{d}{dt}(10\cos t), \frac{d}{dt} (10 \sin t ), \frac{d}{dt}(90-5t)) = (-10 \sin t, 10 \cos t, -5)$. The speed is defined as the norm of the velocity vector,

so, the speed at time t is given by $v = \sqrt[]{(-10 \sin t)^2+(10 \cos t)^2+(-5)^2} = \sqrt[]{100(\sin^2 t + \cos^2 t)+25} = \sqrt[]{125}= 5 \sqrt[]{5}$

d) ON the same fashion we want to know the norm of the second derivative of $\gamma$.

We have that $\gamma ^{''}(t) =(-10\cost t, -10 \sin t , 0)$ so the acceleration is given by $\sqrt[]{100(\cos^2 t+ \sin^2 t )} = 10$ 

Answer 2

a) The height of the tower corresponds to the value of $z$ evaluated at $t = 0$, that is to say:

$z = 90-5\cdot (0)$

$z = 90$

The tower is 90 feet tall.

b) The time taken by the child to reach the bottom is the value of $t$ so that $z = 0$.

$90 - 5\cdot t = 0$

$t = 18$

The child takes 18 minutes to reach the bottom of the tower.

c) The speed is the magnitude of the velocity, which is determined by concepts of first derivative and Pythagorean theorem:

$v = \sqrt{\left(\frac{dx}{dt} \right)^{2}+\left(\frac{dy}{dt} \right)^{2}+\left(\frac{dz}{dt} \right)^{2}}$ (1)

Where the first derivatives are, respectively:

$\frac{dx}{dt} = -10\cdot \sin t$, $\frac{dy}{dt} = 10\cdot \cos t$, $\frac{dz}{dt} = -5$ (2, 3, 4)

By applying (2), (3) and (4) in (1):

$v = \sqrt{100 + 25}\,\left[\frac{ft}{min} \right]$

$v = \sqrt{125}\,\left[\frac{ft}{min} \right]$

$v = 5\sqrt{5}\,\left[\frac{ft}{min} \right]$

Her speed at time $t$ is $5\sqrt{5}$ feet per minute.

d) The acceleration is the magnitude of the acceleration, which is determined by concepts of second derivative and Pythagorean theorem:

$a = \sqrt{\left(\frac{d^{2}x}{dt^{2}} \right)^{2}+\left(\frac{d^{2}y}{dt^{2}} \right)^{2}+\left(\frac{d^{2}z}{dt^{2}} \right)^{2}}$ (5)

Where the second derivatives are, respectively:

$\frac{d^{2}x}{dt^{2}} = -10\cdot \cos t$, $\frac{d^{2}y}{dt^{2}} = -10\cdot \sin t$, $\frac{d^{2}z}{dt^{2}} = 0$ (6, 7, 8)

By applying (6), (7) and (8) in (5):

$a = 10\,\left[\frac{ft}{s^{2}} \right]$

Her acceleration at time $t$ is 10 feet per square second.

We kindly invite to check this question on parametric equations: https://brainly.com/question/23070611