If an arrow is shot straight up from the surface of the moon with an initial velocity of 100 ft/s, its height in feet after t second is given by s(t)=(100t)−(83/100)(t^2).

Use the limit definition of the derivative to find the answers to the following questions.

Find the velocity of the arrow when t=2.

Find the velocity of the arrow when t=a.

When will the arrow return to the moon's surface?

What will the velocity of the arrow be when it hits the surface?

s(2) = 100(2) - (83/100)(4) = 4917/25

s(2+h) = 100(2+h) - (83/100)(2+h)^2
= 200+100h - 332/100 - (332/100)h - (83/100)h^2


velocity = Limit ( s(2+h) - s(2) )/(2+h - 2) as h -->0
= Lim (200+100h - 332/100 - (332/100)h - (83/100)h^2 - 4917/25)/h
= Lim ( (2417/25)h - (83/100)h^2 )/h
= lim 2417/25 - (83/100)h , as h ---> 0
= 2417/25 ft/sec

carefully repeat the above steps using a instead of 2
you should get
v(a) = 100 - (83/50) a

the arrow will return to the surface when s(t) = 0

100t - (83/100)t^2 = 0
t (100 - (83/100)t ) = 0
t= 0 ----> the start of the shot
or
t = 100(100)/83 = appr 120.5 seconds

To find the velocity of the arrow when t = 2, we need to find the derivative of the height function with respect to time (t). By using the limit definition of the derivative, we can find the instantaneous rate of change of the height function at t = 2.

Step 1: Find the derivative of the height function, s(t), by using the limit definition of the derivative.

s'(t) = lim(h->0) [s(t+h) - s(t)] / h

Plugging in the values of the height function, we get:

s'(t) = lim(h->0) [(100(t+h) - (83/100)(t+h)^2) - (100t - (83/100)t^2)] / h

Step 2: Simplify the expression.

s'(t) = lim(h->0) [100t + 100h - (83/100)(t^2 + 2th + h^2) - 100t + (83/100)t^2] / h

s'(t) = lim(h->0) [100h - (83/100)(2th + h^2)] / h

s'(t) = lim(h->0) 100 - (83/100) * (2t + h)

Step 3: Take the limit as h approaches 0.

s'(t) = 100 - (83/100) * 2t

s'(t) = 100 - (166/100)t

Now we can find the velocity of the arrow at t = 2 by plugging in the value of t into the derivative equation:

s'(2) = 100 - (166/100)(2)

s'(2) = 100 - (332/100)

s'(2) = 100 - 3.32

s'(2) = 96.68 ft/s

Therefore, the velocity of the arrow when t = 2 is approximately 96.68 ft/s.

To find the velocity of the arrow when t = a, we use the same derivative equation:

s'(a) = 100 - (166/100)a

When will the arrow return to the moon's surface?

To find when the arrow returns to the moon's surface, we need to find the time when the height function, s(t), equals zero.

s(t) = 100t - (83/100)t^2

Setting s(t) equal to zero:

0 = 100t - (83/100)t^2

100t = (83/100)t^2

Dividing both sides by t:

100 = (83/100)t

t = (100 * 100) / 83

t ≈ 120.48 seconds

Therefore, the arrow will return to the moon's surface after approximately 120.48 seconds.

What will the velocity of the arrow be when it hits the surface?

To find the velocity of the arrow when it hits the surface, we use the derivative equation:

s'(t) = 100 - (166/100)t

Plugging in the value of t when the arrow hits the surface (t ≈ 120.48):

s'(120.48) = 100 - (166/100)(120.48)

s'(120.48) = 100 - 199.285

s'(120.48) ≈ -99.285 ft/s

Therefore, the velocity of the arrow when it hits the surface is approximately -99.285 ft/s (negative because the arrow is moving in the opposite direction when it hits the surface).

To find the velocity of the arrow at a certain time t using the limit definition of the derivative, we need to find the derivative of the height function s(t). The derivative of s(t) will give us the rate of change of height with respect to time, which is the velocity of the arrow.

1. Find the velocity of the arrow when t = 2:
To find the velocity at t = 2, we need to find the derivative of s(t) and evaluate it at t = 2.

First, let's find the derivative of s(t) using the limit definition of the derivative:
s'(t) = lim(h->0) [(s(t + h) - s(t))/h]

Substituting s(t) = (100t) - (83/100)(t^2):
s'(t) = lim(h->0) [((100(t + h) - (83/100)((t + h)^2)) - ((100t) - (83/100)(t^2)))/h]

Expanding and simplifying the expression, we get:
s'(t) = lim(h->0) [(100t + 100h - (83/100)(t^2 + 2th + h^2)) - (100t - (83/100)(t^2))]/h

Canceling out common terms, we get:
s'(t) = lim(h->0) [100h - (83/100)(2th + h^2)]/h

Expanding further and canceling out terms, we get:
s'(t) = lim(h->0) [100h - (166/100)th - (83/100)h^2]/h

Now, we can simplify the expression by factoring out h:
s'(t) = lim(h->0) h[100 - (166/100)t - (83/100)h]/h

Canceling out h and simplifying, we get:
s'(t) = lim(h->0) 100 - (166/100)t - (83/100)h

At this point, we can take the limit as h approaches 0 to find the derivative:
s'(t) = 100 - (166/100)t

Now, we can evaluate s'(t) at t = 2 to find the velocity at that time:
s'(2) = 100 - (166/100)(2)
= 100 - (332/100)
= 100 - 3.32
= 96.68 ft/s

Therefore, the velocity of the arrow when t = 2 is 96.68 ft/s.

2. Find the velocity of the arrow when t = a:
Using the derivative we obtained earlier, we can evaluate s'(t) at t = a:
s'(a) = 100 - (166/100)a

Therefore, the velocity of the arrow when t = a is 100 - (166/100)a ft/s.

3. When will the arrow return to the moon's surface?
The arrow will return to the moon's surface when its height is 0. We can set s(t) = 0 and solve for t:

0 = (100t) - (83/100)(t^2)

To solve this quadratic equation, we can factor it:
0 = (100t)(1 - (83/100)t)

Setting each factor equal to 0, we get two possible solutions:
100t = 0 -> t = 0
1 - (83/100)t = 0 -> t = 100/83

Since we're talking about time, the solution t = 0 is not very meaningful, as it implies the initial time when the arrow was shot. So, the arrow will return to the moon's surface at t = 100/83 seconds.

4. What will the velocity of the arrow be when it hits the surface?
Using the velocity equation derived earlier, we can substitute t = 100/83 to find the velocity at that time:
s'(100/83) = 100 - (166/100)(100/83)

Simplifying the expression, we get:
s'(100/83) = 100 - (16600/8300)

Calculating:
s'(100/83) = 100 - 2
= 98 ft/s

Therefore, the velocity of the arrow when it hits the surface will be 98 ft/s.