A curve CC has parametric equations
x=etcos⁡t, y=e−tsin⁡t.x = \mathrm{e}^t \cos t,\ y = \mathrm{e}^{-t} \sin t.

  1. (i)

    Find the equation of the tangent to CC at the point with parameter pp. Give your answer in the form y=mx+cy = mx + c.

  2. (ii)

    The tangent meets the xx-axis at point AA and the yy-axis at point BB. Find, in terms of pp, the area of the triangle OABOAB, where OO is the origin.

Tangents normalsImplicit parametric中文
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Worked solution

4(i)

1. Find the derivatives of $x$ and $y$ with respect to $t$ using the product rule.

dxdt=etcos⁡t−etsin⁡t=et(cos⁡t−sin⁡t)\frac{dx}{dt} = \mathrm{e}^t \cos t - \mathrm{e}^t \sin t = \mathrm{e}^t(\cos t - \sin t)
dydt=−e−tsin⁡t+e−tcos⁡t=e−t(cos⁡t−sin⁡t)\frac{dy}{dt} = -\mathrm{e}^{-t} \sin t + \mathrm{e}^{-t} \cos t = \mathrm{e}^{-t}(\cos t - \sin t)

2. Apply the chain rule to find the gradient function $\frac{dy}{dx}$.

dydx=dydtdxdt=e−t(cos⁡t−sin⁡t)et(cos⁡t−sin⁡t)=e−tet=e−2t\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\mathrm{e}^{-t}(\cos t - \sin t)}{\mathrm{e}^t(\cos t - \sin t)} = \frac{\mathrm{e}^{-t}}{\mathrm{e}^t} = \mathrm{e}^{-2t}

3. Determine the gradient and the coordinates of the point on the curve at parameter $t = p$.

Substitute t=pt = p:
Gradient m=e−2pm = \mathrm{e}^{-2p}
Coordinates: xp=epcos⁡px_p = \mathrm{e}^p \cos p and yp=e−psin⁡py_p = \mathrm{e}^{-p} \sin p

4. Form the equation of the tangent and rearrange it into the form $y = mx + c$.

Using y−yp=m(x−xp)y - y_p = m(x - x_p):
y−e−psin⁡p=e−2p(x−epcos⁡p)y - \mathrm{e}^{-p} \sin p = \mathrm{e}^{-2p}(x - \mathrm{e}^p \cos p)
y−e−psin⁡p=e−2px−e−pcos⁡py - \mathrm{e}^{-p} \sin p = \mathrm{e}^{-2p}x - \mathrm{e}^{-p} \cos p
y=e−2px−e−pcos⁡p+e−psin⁡py = \mathrm{e}^{-2p}x - \mathrm{e}^{-p} \cos p + \mathrm{e}^{-p} \sin p
y=e−2px+e−p(sin⁡p−cos⁡p)y = \mathrm{e}^{-2p}x + \mathrm{e}^{-p}(\sin p - \cos p)

Answer

y=e−2px+e−p(sin⁡p−cos⁡p)y = \mathrm{e}^{-2p} x + \mathrm{e}^{-p} (\sin p - \cos p)

4(ii)

1. Find the equation of the tangent to the curve at $t = p$ (result required from part i).

dxdt=et(cos⁡t−sin⁡t),dydt=e−t(cos⁡t−sin⁡t)\frac{dx}{dt} = \mathrm{e}^t(\cos t - \sin t), \quad \frac{dy}{dt} = \mathrm{e}^{-t}(\cos t - \sin t)
dydx=e−t(cos⁡t−sin⁡t)et(cos⁡t−sin⁡t)=e−2t\frac{dy}{dx} = \frac{\mathrm{e}^{-t}(\cos t - \sin t)}{\mathrm{e}^t(\cos t - \sin t)} = \mathrm{e}^{-2t}
At t=pt=p, the gradient is m=e−2pm = \mathrm{e}^{-2p}.
The equation of the tangent is:
y−e−psin⁡p=e−2p(x−epcos⁡p)y - \mathrm{e}^{-p} \sin p = \mathrm{e}^{-2p}(x - \mathrm{e}^p \cos p)
y=e−2px+e−p(sin⁡p−cos⁡p)y = \mathrm{e}^{-2p}x + \mathrm{e}^{-p}(\sin p - \cos p)

2. Find the coordinates of $A$ and $B$ by evaluating the $x$ and $y$ intercepts of the tangent.

At the yy-axis (point BB), x=0x=0:
yB=e−p(sin⁡p−cos⁡p)y_B = \mathrm{e}^{-p}(\sin p - \cos p)
At the xx-axis (point AA), y=0y=0:
0=e−2pxA+e−p(sin⁡p−cos⁡p)0 = \mathrm{e}^{-2p}x_A + \mathrm{e}^{-p}(\sin p - \cos p)
e−2pxA=e−p(cos⁡p−sin⁡p)\mathrm{e}^{-2p}x_A = \mathrm{e}^{-p}(\cos p - \sin p)
xA=ep(cos⁡p−sin⁡p)x_A = \mathrm{e}^p(\cos p - \sin p)

3. Formulate the area of triangle $OAB$, taking the absolute value to ensure a positive geometric area.

Area=12∣xA∣∣yB∣\text{Area} = \frac{1}{2} |x_A| |y_B|
Area=12∣ep(cos⁡p−sin⁡p)∣∣e−p(sin⁡p−cos⁡p)∣\text{Area} = \frac{1}{2} \left| \mathrm{e}^p(\cos p - \sin p) \right| \left| \mathrm{e}^{-p}(\sin p - \cos p) \right|
Area=12epe−p∣−(cos⁡p−sin⁡p)2∣\text{Area} = \frac{1}{2} \mathrm{e}^p \mathrm{e}^{-p} \left| -(\cos p - \sin p)^2 \right|
Area=12(cos⁡p−sin⁡p)2\text{Area} = \frac{1}{2} (\cos p - \sin p)^2

4. Simplify the area expression using trigonometric identities.

Area=12(cos⁡2p−2sin⁡pcos⁡p+sin⁡2p)\text{Area} = \frac{1}{2} (\cos^2 p - 2\sin p \cos p + \sin^2 p)
Area=12(1−sin⁡2p)\text{Area} = \frac{1}{2} (1 - \sin 2p)

Answer

\frac{1}{2}(1 - \sin 2p)

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