A curve has parametric equations \( x=2t \), \( y=t^2 \). Find the Cartesian equation.
\( y=(x/2)^2=x^2/4 \).
Parametric Equations
10 exam-style questions on parametric curves, conversion to Cartesian and differentiation. Tap Show answer for full working.
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Find the coordinates of the point on the curve \( x=3\cos\theta \), \( y=2\sin\theta \) when \( \theta=\pi/2 \).
\( (0,2) \).
Find \( \dfrac{dy}{dx} \) in terms of \( t \) for \( x=t^2 \), \( y=t^3 \).
\( \dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{3t^2}{2t}=\dfrac{3t}{2} \) (\( t eq 0 \)).
The curve \( C \) has parametric equations \( x=1+t \), \( y=t^2-2 \). Find the equation of the tangent at the point where \( t=2 \).
Point \( (3,2) \). \( \dfrac{dy}{dx}=\dfrac{2t}{1}=2t \). At \( t=2 \), gradient 4. \( y-2=4(x-3) \) ⇒ \( y=4x-10 \).
Find the Cartesian equation of the curve \( x=2\cos\theta \), \( y=3\sin\theta \).
\( \dfrac{x^2}{4}+\dfrac{y^2}{9}=1 \) (ellipse).
Find the stationary points of the curve \( x=t^2 \), \( y=t^3-3t \).
\( \dfrac{dy}{dx}=\dfrac{3t^2-3}{2t} \). Set numerator 0 (and denom ≠0): \( t=\pm1 \). Points \( (1,-2) \) and \( (1,2) \).
A curve has parametric equations \( x=\dfrac{1}{t} \), \( y=t+\dfrac{1}{t} \). Find \( \dfrac{dy}{dx} \) in terms of \( t \) and the equation of the normal at \( t=1 \).
\( \dfrac{dy}{dt}=1-1/t^2 \), \( \dfrac{dx}{dt}=-1/t^2 \). \( \dfrac{dy}{dx}=-(t^2-1) \). At \( t=1 \), gradient 0 so normal is vertical: \( x=1 \).
Find the area under the curve \( x=t^2 \), \( y=2t \) from \( t=0 \) to \( t=2 \) (i.e. the area between the curve and the \( x \)-axis).
\( A=\int y\,dx=\int_0^2 2t\cdot 2t\, dt=\int_0^2 4t^2\, dt=\dfrac{32}{3} \).
The parametric equations of a curve are \( x=\cos^3 t \), \( y=\sin^3 t \). Find \( \dfrac{dy}{dx} \) in terms of \( t \) and the equation of the tangent at \( t=\pi/4 \).
\( \dfrac{dy}{dx}=\dfrac{3\sin^2 t\cos t}{-3\cos^2 t\sin t}=-\tan t \). At \( t=\pi/4 \), gradient \( -1 \), point \( (\dfrac{\sqrt{2}}{4},\dfrac{\sqrt{2}}{4}) \). \( y+\dfrac{\sqrt{2}}{4}=-(x-\dfrac{\sqrt{2}}{4}) \) ⇒ \( x+y=\dfrac{\sqrt{2}}{2} \).
A curve has parametric equations \( x=t-\sin t \), \( y=1-\cos t \). Show that \( \dfrac{dy}{dx}=\dfrac{\sin t}{1-\cos t} \) and find the Cartesian equation of the curve.
\( \dfrac{dy}{dt}=\sin t \), \( \dfrac{dx}{dt}=1-\cos t \) ⇒ ratio. From \( y=1-\cos t \), \( \cos t=1-y \), \( \sin^2 t=1-\cos^2 t=2y-y^2 \). Also \( x=t-\sin t \) is harder to eliminate fully; the curve is a cycloid. For Cartesian relation between \( x \) and \( y \) alone is non-elementary; often left parametric or \( x=\arccos(1-y)-\sqrt{2y-y^2} \).