Differentiation — practice questions

First, we differentiate the equation implicitly with respect to $x$. $d/dx(x^{3}) + d/dx(y^{3}) - d/dx(3xy) = d/dx(0)$

$3x^{2} + 3y^{2}(dy/dx) - [3x(dy/dx) + 3y(1)] = 0$ Note the use of the product rule on the 3xy term.

$3x^{2} + 3y^{2}(dy/dx) - 3x(dy/dx) - 3y = 0$

Now, we gather the dy/dx terms: $(3y^{2} - 3x)(dy/dx) = 3y - 3x^{2}$

Solve for dy/dx: $dy/dx = (3y - 3x^{2}) / (3y^{2} - 3x) = (y - x^{2}) / (y^{2} - x)$

Next, find the gradient at the point P(3/2, 3/2): $m = dy/dx |_(3/2, 3/2) = ((3/2) - (3/2)^{2}) / ((3/2)^{2} - (3/2))$ $m = (3/2 - 9/4) / (9/4 - 3/2) = (-3/4) / (3/4) = -1$

The gradient of the tangent at P is -1.

Finally, use the point-gradient formula $y - y_{1} = m(x - x_{1})$: $y - 3/2 = -1(x - 3/2)$ $y - 3/2 = -x + 3/2$ $y + x = 3$

The equation of the tangent is $x + y = 3$.

(i) First, find the derivatives with respect to $t$: $dx/dt = -2sin(t)$ $dy/dt = 2cos(2t)$

Now, apply the formula for dy/dx: $dy/dx = (dy/dt) / (dx/dt) = (2cos(2t)) / (-2sin(t)) = -cos(2t) / sin(t)$ Using the double angle identity $cos(2t) = 1 - 2sin^{2}(t)$: $dy/dx = -(1 - 2sin^{2}(t)) / sin(t) = (2sin^{2}(t) - 1) / sin(t) = 2sin(t) - csc(t)$

(ii) Stationary points occur when $dy/dx = 0$. $2sin(t) - csc(t) = 0$ $2sin(t) = 1/sin(t)$ $2sin^{2}(t) = 1$ $sin^{2}(t) = 1/2$ $sin(t) = \pm 1/\sqrt2$

Since $0 < t < \pi$, sin(t) is positive. So, $sin(t) = 1/\sqrt2$. This gives $t = \pi /4$ and $t = 3\pi /4$.

For $t = \pi /4$: $x = 2cos(\pi /4) = 2(1/\sqrt2) = \sqrt2$ $y = sin(2 * \pi /4) = sin(\pi /2) = 1$ Point is (√2, 1).

For $t = 3\pi /4$: $x = 2cos(3\pi /4) = 2(-1/\sqrt2) = -\sqrt2$ $y = sin(2 * 3\pi /4) = sin(3\pi /2) = -1$ Point is (-√2, -1).

The stationary points are (√2, 1) and (-√2, -1).

(iii) To find d²y/dx², we use the formula [d/dt(dy/dx)] / (dx/dt). We use the expression $dy/dx = 2sin(t) - csc(t)$. $d/dt(dy/dx) = 2cos(t) - (-csc(t)cot(t)) = 2cos(t) + csc(t)cot(t)$

We already know $dx/dt = -2sin(t)$. $d^{2}y/dx^{2} = (2cos(t) + csc(t)cot(t)) / (-2sin(t))$ $d^{2}y/dx^{2} = - (2cos(t) + (1/sin(t))(cos(t)/sin(t))) / (2sin(t))$ $d^{2}y/dx^{2} = - (2cos(t) + cos(t)/sin^{2}(t)) / (2sin(t))$ $d^{2}y/dx^{2} = - (2cos(t)sin^{2}(t) + cos(t)) / (2sin^{3}(t))$ $d^{2}y/dx^{2} = - cos(t)(2sin^{2}(t) + 1) / (2sin^{3}(t))$