Jack planted a mysterious bean just outside his kitchen wind

Jack planted a mysterious bean just outside his kitchen window. It immediately sprouted 2.56 cm above the ground. Jack kept a careful log of the plant’s growth. He measured the height of the plant each day at 8am and recorded these data.

Day 0 1 2 3 4

Height (cm) 2.56 6.4 16 40 100

a. Define variables and write an exponential equation for this pattern. If the pattern were to continue, what would be the heights on the fifth day

b. Jack’s younger brother measured the plant at 8pm on the evening of the third day and found it to be about 63.25 cm tall. Show how to find this value mathematically.

c. Find the height of the sprout at 12noon on the sixth day.

d. Find the time at which the sprout has doubled in height.

e. Find the day and time when the plant reaches a height of 1 km.

Solution

Let the exponential funtion that models the growth be : y = ab^x

Let y be the height of bean in cm and x be the no.of days

x =0 ; y = 2.56. So, a= 2,56

y = 2.56(b)^x .Lets find variable b

x =1 ; y = 6.4 use this point to find b:

6.4 = 2.56(b)^1

b = 2.5

a) So, Growth function: y = 2.56(2.5)^x

x = 5 ; y = 2.56(2.5)^5 = 250 cm

b) on third day at 8 pm , we find the no. of days elapsed = 3+0.5 = 3.5 =x

Plug x= 3.5 in y = 2.56(2.5)^3.5

c) sixth day 12 noon: x = 6+4/24 = 6.17

y = 2.56(2.5)^6.17 = 730.35 cm

d) height of sprout is doubled:

y = 2.56*2

So, 2*2,56 = 2.56(2.5)^x

2 = (2.5)^x

taking log on both sides: ln2 = x*ln(2.5)

x= 0.756 = 18.15 hrs = 2.00 am 9 min on day 1