The derivative of a function $f(x)$ is given by $$f'(x) = ext{sin} \, x.$$ Find $f(0)$, given that $f(0) = 2$ - HSC - SSCE Mathematics Extension 1 - Question 2 - 2010 - Paper 1

Starting with the mass equation:

$M = 36 - 35.5e^{-kt},$
we differentiate both sides with respect to $t$:

$\frac{dM}{dt} = 0 - 35.5 \cdot (-k)e^{-kt} = 35.5ke^{-kt}.$
Substituting $M = 36 - 35.5e^{-kt}$ into this gives:

$\frac{dM}{dt} = k(36 - M).$

Using the mass equation at $t = 10$ years:

$M = 36 - 35.5e^{-10k} = 20.$
Solving for $k$:

$36 - 20 = 35.5 e^{-10k} \implies 16 = 35.5 e^{-10k} \implies e^{-10k} = \frac{16}{35.5}.$
Taking the natural logarithm:

$-10k = \ln\left(\frac{16}{35.5}\right) \implies k = -\frac{1}{10} \ln\left(\frac{16}{35.5}\right) \approx 0.0797.$

The limiting mass is found when $t \to \infty$. At this limit, $e^{-kt} \to 0$:

$\lim_{t \to \infty} M = 36 - 35.5 \cdot 0 = 36.$ Thus, the limiting mass of the whale is 36 tonnes.

Since $P(3) = 0$, substituting $x = 3$ into the polynomial gives:

$P(3) = (3 + 1)(3 - 3)Q(3) + 3a + b = 0.$
Thus, we have:

$3a + b = 0.$
For the remainder when divided by $x + 1$, $P(-1) = 8$:

$(1)(-4)Q(-1) - a + b = 8.$
Substituting $b = -3a$ gives:

$-4Q(-1) - a - 3a = 8\implies -4Q(-1) - 4a = 8.$
This allows us to solve for $a$ and $b$.

Using the polynomial remainder theorem, we substitute $x = -1$ and $x = 3$ into $P(x)$:

$R = P(-1)$ and $P(3)$. The remainder when divided by $(x + 1)(x - 3)$ is a linear expression of the form $Ax + B$.

Applying the Pythagorean theorem in the context of the situation:

$r^2 = x^2 + 6^2.$
Differentiate both sides with respect to $t$:

$2r \frac{dr}{dt} = 2x \frac{dx}{dt}.$ Replacing rac{dx}{dt} = 100 km/h gives:

$\frac{dr}{dt} = \frac{x}{r} \cdot 100.$
This provides the required expression.