Vector calculus — Question 273

ANSWER

(a) $\frac{(x-1)^2}{4}-(y-2)^2=1$, with $x\ge3$, $y\ge2$. (b) Upper right branch from $(3,2)$, moving up and right; asymptote $y-2=(x-1)/2$. (c) $dy/dx=\sqrt2/2$.

Worked solution

(a) $\sec t=(x-1)/2$ and $\tan t=y-2$. Use $\sec^2t-\tan^2t=1$. The time interval gives $\sec t\ge1$, $\tan t\ge0$, selecting the upper part of the right branch, including $(3,2)$.

(b) Both coordinates increase for $0<t<\pi/2$. The branch approaches the stated positive-slope asymptote. The rest of the hyperbola is not part of the particle path.

(c) $dx/dt=2\sec t\tan t$ and $dy/dt=\sec^2t$. Hence $dy/dx=\sec t/(2\tan t)$; at $\pi/4$ this is $\sqrt2/2$.

Equivalent justified methods accepted; respect any requested proof method.

Exact values unless specified. Sketches assessed by mathematical features, not artistic quality.

Part a: Express secant and tangent in terms of x and y.

[1 mark]

Part a: Use the Pythagorean identity to obtain the hyperbola.

[1 mark]

Part a: State both restrictions.

[1 mark]

Part b: Draw the restricted branch and included starting point.

[1 mark]

Part b: Indicate motion and the correct asymptote.

[1 mark]

Part c: Form the parametric derivative ratio.

[1 mark]

Part c: Evaluate at $\pi/4$.

[1 mark]

Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.