The question
According to a psychologist, the ability of a person to understand spatial concepts is given by
A = 1 3 t , A = \frac13\sqrt t, A = 3 1 t , where t t t t ∈ [ 5 , 18 ] t \in [5, 18] t ∈ [ 5 , 18 ]
Exam answer Show all explanations
A = 1 3 t A = \frac13\sqrt t A = 3 1 t ∴ d A d t = 1 6 t − 1 2 = 1 6 t \therefore\ \frac{dA}{dt} = \frac16t^{-\frac12} = \frac{1}{6\sqrt t} ∴ d t d A = 6 1 t − 2 1 = 6 t 1 Why?
Write t = t 1 2 \sqrt t = t^{\frac12} t = t 2 1
d d t ( t n ) = n t n − 1 \frac{d}{dt}(t^n) = nt^{n - 1} d t d ( t n ) = n t n − 1 1 3 × 1 2 = 1 6 , 1 2 − 1 = − 1 2 \frac13 \times \frac12 = \frac16,\quad \frac12 - 1 = -\frac12 3 1 × 2 1 = 6 1 , 2 1 − 1 = − 2 1 d A d t = 1 3 ⋅ d d t ( t 1 2 ) = 1 3 ⋅ 1 2 t 1 2 − 1 = 1 6 t − 1 2 \begin{aligned} \frac{dA}{dt} &= \frac13 \cdot \frac{d}{dt}\big(t^{\frac12}\big) \\ &= \frac13 \cdot \frac12\,t^{\frac12 - 1} \\ &= \frac16\,t^{-\frac12} \end{aligned} d t d A = 3 1 ⋅ d t d ( t 2 1 ) = 3 1 ⋅ 2 1 t 2 1 − 1 = 6 1 t − 2 1 A negative power means a reciprocal: t − 1 2 = 1 t t^{-\frac12} = \frac{1}{\sqrt t} t − 2 1 = t 1
for all t ∈ [ 5 , 18 ] t \in [5, 18] t ∈ [ 5 , 18 ]
Why?
d A d t \frac{dA}{dt} d t d A
The question is about how this rate changes. So we differentiate d A d t \frac{dA}{dt} d t d A
d 2 A d t 2 = 1 6 ( − 1 2 ) t − 3 2 \frac{d^2A}{dt^2} = \frac16\left(-\frac12\right)t^{-\frac32} d t 2 d 2 A = 6 1 ( − 2 1 ) t − 2 3 Why?
Power rule again on 1 6 t − 1 2 \frac16t^{-\frac12} 6 1 t − 2 1 − 1 2 -\frac12 − 2 1
− 1 2 − 1 = − 3 2 -\frac12 - 1 = -\frac32 − 2 1 − 1 = − 2 3 ∴ d 2 A d t 2 = − 1 12 t t \therefore\ \frac{d^2A}{dt^2} = -\frac{1}{12t\sqrt t} ∴ d t 2 d 2 A = − 12 t t 1 Why?
1 6 × ( − 1 2 ) = − 1 12 \frac16 \times \left(-\frac12\right) = -\frac{1}{12} 6 1 × ( − 2 1 ) = − 12 1
t 3 2 = t 1 ⋅ t 1 2 = t t t^{\frac32} = t^1 \cdot t^{\frac12} = t\sqrt t t 2 3 = t 1 ⋅ t 2 1 = t t Put these together:
1 6 ( − 1 2 ) = − 1 × 1 6 × 2 = − 1 12 t − 3 2 = 1 t 3 2 = 1 t t − 1 12 ⋅ 1 t t = − 1 12 t t \begin{aligned} \frac16\left(-\frac12\right) &= -\frac{1 \times 1}{6 \times 2} = -\frac{1}{12} \\[4pt] t^{-\frac32} &= \frac{1}{t^{\frac32}} = \frac{1}{t\sqrt t} \\[4pt] -\frac{1}{12} \cdot \frac{1}{t\sqrt t} &= -\frac{1}{12t\sqrt t} \end{aligned} 6 1 ( − 2 1 ) t − 2 3 − 12 1 ⋅ t t 1 = − 6 × 2 1 × 1 = − 12 1 = t 2 3 1 = t t 1 = − 12 t t 1 So, d 2 A d t 2 < 0 \dfrac{d^2A}{dt^2} < 0 d t 2 d 2 A < 0 t ∈ [ 5 , 18 ] t \in [5, 18] t ∈ [ 5 , 18 ]
Why?
For t t t [ 5 , 18 ] [5, 18] [ 5 , 18 ] t > 0 t > 0 t > 0 12 t t > 0 12t\sqrt t > 0 12 t t > 0
Hence d A d t \dfrac{dA}{dt} d t d A A A A t t t [ 5 , 18 ] [5, 18] [ 5 , 18 ]
Why?
A function whose derivative is negative on an interval is decreasing there. Here the function is d A d t \frac{dA}{dt} d t d A d 2 A d t 2 \frac{d^2A}{dt^2} d t 2 d 2 A
Watch out: A A A
This means that the rate of increase of the ability to understand spatial concepts decreases (slows down) with age in between 5 and 18. Hence proved.
Where this comes from Chapter 6, Application of Derivatives : Rate of change: sign of the second derivative.
Question type: Show that a function is increasing everywhere
How to spot it: click to see Type 4. Show that a function is increasing everywhere
Spot it: "Show that (or prove that) f f f R \R R
Here: The function to test is the rate d A d t = 1 6 t \frac{dA}{dt} = \frac{1}{6\sqrt t} d t d A = 6 t 1 d 2 A d t 2 = − 1 12 t t \frac{d^2A}{dt^2} = -\frac{1}{12t\sqrt t} d t 2 d 2 A = − 12 t t 1 t t t [ 5 , 18 ] [5, 18] [ 5 , 18 ]
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