y = 2ᵗ , x = eᵗ²

So y/x = 2ᵗ/eᵗ²

①Using quotient rule,

d(y/x) / dt
= (eᵗ² · 2ᵗ ln2 - 2t eᵗ² · 2ᵗ) / (eᵗ²)²
= (2ᵗ ln2 - 2t · 2ᵗ)/eᵗ²
= 2ᵗ(ln2 - 2t)/eᵗ²

②Next, an increase of 0.1% means an increase of 0.1/100 times (or 1/1000 times or 0.001 times)

So from t = 1, ∆t = 1/1000 × 1 = 1/1000

(∆ is the Greek capital letter delta , stands for 'change' or 'change in')


③For this approximate change question,

∆(y/x) = d(y/x)/dt · ∆t

It is basically : change in y/x = (rate of change of y/x with respect to t) × (change in t)

∆(y/x) = 2ᵗ(ln2 - 2t)/eᵗ² · ∆t

Lastly,

Δt = 1/1000 so we are going from t = 1 to t = 1.001.

Since we know ∆t = 0.001, to get the approximate change in y/x (Δ(y/x)), we use t = 1 to approximate it.


∆(y/x) = 2¹(ln2 - 2(1))/e^(1²) · 1/1000

= 2(ln2 - 2) / 1000e
= (ln2 - 2) / 500e
≈ -0.000961528569818
= -0.000962 (3s.f)



This is very close to the actual change of -0.000961635352875
(when you actually do f(1.001) - f(1) , whereby y/x = f(t) = 2ᵗ/eᵗ²)

Approximate change was last tested in the old A Math syllabus (last year of examination was 2007) so this is probably an old A Level question from that time.

(Knowledge of A Maths is assumed in A level/H2 Math so students in that era would be expected to know approximate change)

okay thanks !

I learnt it and now I forgot about it already. Sigh.

Quite easy to pick up anyway. Wasn't in my syllabus

Thank u!!