The value of an investment in Canada Savings Bonds is modelled by $A(t) = A_0 e^{0.0255t}$... Rest of question below.

Question

The value of an investment in Canada Savings Bonds is modeled by $$A(t) = A_0 e^{0.0255t}$$, where A is the amount the investment is worth after $t$ years, and $A_0$ is the initial amount invested. At what rate, correct to 3 decimal places, is the investment growing at the time when its value has doubled?

I understand that we have to find the derivative but how can we find it if we don't know an initial amount?

Answer 1

The question asks for rate when the investments value has been doubled or when

$$A(t)=2A_0 = A_0 e^{0.0255t}$$ So the initial value cancels and you have: $$ 2=e^{0.0255t}\Rightarrow \ln(2)=0.0255t\Rightarrow t=\frac{\ln(2)}{0.0255} $$ Now, taking the derivative: $$A'(t)=0.0255A_0e^{0.0255t}\Rightarrow A'(\frac{\ln(2)}{0.0255})=0.0255A_0e^{0.0255(\frac{\ln(2)}{0.0255})}\\ =.0255A_0e^{\ln(2)}=2(.0255)A_0=.051A_0$$ Which does depend on the principal, as it should.

Answer 2

You have $A = 2A_0 \implies 2A_0 = A_0e^{0.0255t} \implies e^{0.0255t} = 2\implies 0.0255t = \ln 2 \implies t = \dfrac{\ln 2}{0.0255}$. To find the rate, we have: $A'(t) = 0.0255A_0e^{0.0255t} \implies A'\left(\frac{\ln 2}{0.0255}\right) = 0.0255A_0e^{0.0255\cdot \left(\frac{\ln 2}{0.0255}\right)} = 0.0255A_0e^{\ln 2} = 0.051A_0$.

Answer 3

Just set $A_0e^{(0.0255t)}=2A_0$, divide by $A_0$ and you get $2=e^{(0.0255t)}$. Use the natural log to find what the value of t is at this time (it's 27.182). Take the derivative of $A(t)$, plug in 27.182 you get $A'(27.182)=A_0(.0255)e^{(.0255)(27.182)}$. Compute that and it's your answer. It'll be some number multiplied by $A_0$.