d The equation for A can be re-expressed in the form A = b × e^−kt, where b and k are constants and e is the natural base.
i Write down the value of b.
ii Find the value of k, correct to 4 decimal places.
thank you
Hello,Originally Posted by rachael
to be honest, I've got a few problems giving you a correct value when there isn't any number available.
But in general, you have to transform the given equation.
to i) $\displaystyle A=b \cdot e^{-k \cdot t} \ \Leftrightarrow \ b=\frac{A}{e^{-k \cdot t}}=A \cdot e^{k \cdot t}$
to ii) $\displaystyle A=b \cdot e^{-k \cdot t}\ \Leftrightarrow \ ln \left( \frac{A}{b} \right)=-k\cdot t\ \Leftrightarrow \ k= -\frac{1}{t} \cdot (ln(A)-ln(b))$
Maybe this was of some help.
Greetings
EB
4 Joshua fell over and has a large wound on his shin. As his wound heals, the surface area of
the scab decreases. The surface area of the scab, A square centimetres, t days after the
wound was received, can be modelled by the equation A = 10 × 2−0.1t.
a What was the original surface area of Joshua’s wound?
b What is the surface area of the wound after 3 days?
c How many days after the wound was received has its surface area reduced by
i 50%? ii 95%?
d The equation for A can be re-expressed in the form A = b × e−kt, where b and k are
constants and e is the natural base.
i Write down the value of b.
ii Find the value of k, correct to 4 decimal places.
oops... this is the question
could you help me with part d?
Hello,Originally Posted by rachael
to d) i) from the given text of your problem, you can get the value of $\displaystyle A(0)=10 \cdot 2^{-.1 \cdot 0}=b \cdot e^{k \cdot 0}=10=b$
to d) ii) You know one equation for A and now you know the 2nd equation for A where only the value for k is missing. Both equation must be the same:
$\displaystyle A = 10 \cdot 2^{-0.1 \cdot t}=10 \cdot e^{k \cdot t}$
Divide both sides by 10 and you'll get
$\displaystyle 2^{-0.1 \cdot t}= e^{k \cdot t}$ Now you have to logarithmize(?) this equation to
$\displaystyle ln(2) \cdot (-0.1 \cdot t)= k \cdot t$
And this will give: $\displaystyle k=-.1 \cdot ln(2)\approx -0.069314718...$
Greetings
EB