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

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- Mar 24th 2006, 11:29 PMrachaelalgebraic technique
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 - Mar 25th 2006, 12:54 AMearbothQuote:

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 - Mar 25th 2006, 01:10 AMrachael
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? - Mar 25th 2006, 01:44 AMearbothQuote:

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