Question

Please help differentiate this!!!!!!!!!

The height of a certain species 't' years after it was planted is given by
H=20ln(3t +2) +30cm
a) how tall was the shrub when it was planted
b) how long will it take for the shrub to reach a height of 1 metre
c)At what rate is the shrubs height changing
i) 3 years after being planted
ii) 10 years after being planted

Answer 1

H=20ln(3t +2) +30cm :

a) how tall was the shrub when it was planted. When it was planted t = 0, then

H = 20ln(0+2) +30 →→H = 20.ln(2) + 3 →→ and H = 43.86 cm

b) how long will it take for the shrub to reach a height of 1:

H = 1 or in cm, H=100→→ 100 = 20ln(3t+2) + 30
100-30 = 20ln(3t+2) →→ 70/20 = ln(3t+2)

ln(3t+2) = 7/2 →→ 3t+2 = e⁷/² →→ t = [(e⁷/²)-2]/3 →→t = 10.37 years

c)At what rate is the shrubs height changing 3 years after being planted?
H = 20ln(3t+2) +30
dH/dt = 20 . 3/(3t+2) (remember d(ln(u) = u'/u)

for t =3 years , the growth (dH/dt) will be (20).(3)(3.3 + 2) = 60/11 = 5.45 cm/year
for t = 10 , dH/dt = 1.81 cm/year

Answer 2

`\bf h=20ln(3t+2)+30\\\\ -------------------------------\\\\ \boxed{a}\\\\ \stackrel{0~years}{t=0}\qquad h=20ln[3(0)+2]+30\implies h=20ln(2)+30 \\\\\\ h\approx 43.86 \\\\\\ \boxed{b}\\\\ \stackrel{1~meter}{h=100}\qquad 100=20ln(3t+2)+30\implies 70=20ln(3t+2) \\\\\\ \cfrac{70}{20}=ln(3t+2)\implies \stackrel{\textit{log cancellation rule}}{e^{(7)/(2)}=e^(ln(3t+2))}\implies e^{(7)/(2)}=3t+2 \\\\\\ e^{(7)/(2)}-2=3t\implies \cfrac{e^{(7)/(2)}-2}{3}=t\implies 10.371817\approx t`


`\bf \boxed{c}\\\\ \cfrac{dh}{dt}=20\left(\cfrac{1}{3t+2}\cdot 3 \right)+0\implies \cfrac{dh}{dt}=20\left(\cfrac{3}{3t+2} \right)\\\\\\ \cfrac{dh}{dt}=\cfrac{60}{3t+2} \\\\\\ \left. \cfrac{dh}{dt} \right|_(3)\implies \cfrac{60}{3(3)+2}\implies \cfrac{60}{11} \\\\\\ \left. \cfrac{dh}{dt} \right|_(10)\implies \cfrac{60}{3(10)+2}\implies \cfrac{15}{8}`