StudentAidUnit 2: Functions and Graphs — Exercise 2 2
11th Class Mathematics · Unit 2: Functions and Graphs
4.1.A building's height over time is modeled by H(t) = 100 + 20t which is in meters and t is the time in months. The height of a growing tree nearby is given by T(t) = 50 + 10t + t². At what time will the building and tree have the same height?
Given
Building height
H(t) = 100 + 20t
Tree height
T(t) = 50 + 10t + t2
Formula
Condition for intersection
T(t) = H(t)
Working
Substitute
50 + 10t + t2 = 100 + 20t
Rearrange to standard form
t2 - 10t - 50 = 0
Formula
Quadratic formula
t = frac{-b ± sqrt{b2 - 4ac}{2a}
With a = 1, b = -10, c = -50
t = frac{10 ± sqrt{100 + 200}{2} = frac{10 ± sqrt{300}{2}
Simplify
t = frac{10 ± 10sqrt{3}{2} = 5 ± 5sqrt{3}
Result
Taking positive root
t = 5 + 5sqrt{3} ≈ 13.66 months
4.2.What will that height be?
Substitute t = 13.66 into H(t)
H(13.66) = 100 + 20(13.66)
Result
Height at intersection
H(13.66) = 373.2 m
5.A radioactive substance has a half-life of 2 years. If the initial quantity is 200 grams and the exponential decay function is Q(t) = Q₀(1/2)^(t/2), then find the remaining quantity after 6 years graphically?
Given
Decay function
Q(t) = Q0left(12right)t/2
Working
Initial quantity Q₀ = 200 grams
Q(t) = 200left(12right)t/2
Evaluate at t = 6 years
Q(6) = 200left(12right)6/2 = 200left(12right)3 = 200 × 18
Result
Remaining quantity
Q(6) = 25 grams