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

To derive the differential equation, we differentiate the mass model:

$M = 36 - 35.5e^{-kt}$

Differentiating with respect to ( t ):

$\frac{dM}{dt} = 0 - (35.5)(-k)e^{-kt} = k(35.5)e^{-kt}.$

Now recognizing that ( M ) approaches 36 as ( t ) increases, we rewrite the above as:

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

At ( t = 10 ), ( M = 20 ) tonnes:

Using the mass model:

$20 = 36 - 35.5e^{-10k} \Rightarrow 35.5e^{-10k} = 16 \Rightarrow e^{-10k} = \frac{16}{35.5}.$

Taking the natural logarithm:

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

Calculating this gives:

$k \approx 0.0706943,$ which rounds to 0.071.

The limiting mass occurs when ( t \to \infty ). As ( t \to \infty ):

$M \to 36 - 35.5e^{-kt} \to 36 - 35.5 \cdot 0 = 36.$

Thus, the limiting mass of the whale is 36 tonnes.

Since ( x - 3 ) is a factor, ( P(3) = 0 ):

$P(3) = (3+1)(3-3)Q(3) + 3a + b = 0 \Rightarrow 3a + b = 0.\label{1}$

Using the remainder, when dividing by ( x + 1 ):

$P(-1) = 0 + (-1 - 3)Q(-1) + (-a + b) = 8 \Rightarrow -4Q(-1) -a + b = 8.\label{2}$

Solving equations ({1}, {2} ), we find the values of ( a ) and ( b ).

Use polynomial long division or synthetic division. When ( P(x) ) is divided by ( (x + 1)(x - 3) ), find the resulting remainder, which can be expressed in the form:

$R(x) = Ax + B$

Use the previously obtained values for ( a ) and ( b ) to determine the specific remainder.

By applying the Pythagorean theorem:

$M^2 = S^2 + r^2 \Rightarrow 36 = x^2 + r^2.$

Differentiate with respect to t:

$0 = 2x\frac{dx}{dt} + 2r\frac{dr}{dt} \Rightarrow \frac{dr}{dt} = -\frac{x}{r}\frac{dx}{dt}.$

Given ( \frac{dx}{dt} = 100 ) km/h, substitute to find ( \frac{dr}{dt}. $$