The diagram shows the curve with parametric equations
\(x = \\sin t + \\cos t, \quad y = \\sin^3 t + \\cos^3 t,\)
for \(\frac{1}{4}\pi < t < \frac{5}{4}\pi.\)
(i) Show that \(\frac{dy}{dx} = -3 \sin t \cos t.\)
(ii) Find the gradient of the curve at the origin.
(iii) Find the values of \(t\) for which the gradient of the curve is 1, giving your answers correct to 2 significant figures.
Solution
(i) Differentiate \(y\) to obtain \(3\sin^2 t \cos t - 3\cos^2 t \sin t\).
Use \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\).
Obtain given result \(-3\sin t \cos t\).
(ii) Identify parameter at origin as \(t = \frac{3}{4}\pi\).
Use \(t = \frac{3}{4}\pi\) to obtain \(\frac{3}{2}\).
(iii) Rewrite equation as equation in one trig variable, e.g. \(\sin 2t = -\frac{2}{3}\).
Find at least one value of \(t\) from equation of form \(\sin 2t = k\).
Obtain \(t = 1.9\).
Obtain \(t = 2.8\) and no others.
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