Question

If $x=\sin t-\cos t$ and $y=\sin t\cos t$, find $\dfrac{dy}{dx}$ at $t=\dfrac{\pi}{4}$.

Hint:

For parametric equations \(x=f(t)\) and \(y=g(t)\), always use \[ \frac{dy}{dx}=\frac{dy/dt}{dx/dt} \] to find the derivative.

Answer 1

Step 1: Use the parametric differentiation formula.
Since both \(x\) and \(y\) are given in terms of the parameter \(t\), the derivative \(\dfrac{dy}{dx}\) is calculated using \[ \frac{dy}{dx}=\frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}} \] Thus we first compute \(\dfrac{dx}{dt}\) and \(\dfrac{dy}{dt}\).
Step 2: Differentiate $x=\sin t-\cos t$.
\[ \frac{dx}{dt}=\cos t+\sin t \] because \[ \frac{d}{dt}(\sin t)=\cos t \] and \[ \frac{d}{dt}(-\cos t)=\sin t \] Thus \[ \frac{dx}{dt}=\cos t+\sin t \] Step 3: Differentiate $y=\sin t\cos t$.
Use the product rule: \[ \frac{dy}{dt}=(\sin t)'(\cos t)+(\sin t)(\cos t)' \] \[ =\cos t\cos t+\sin t(-\sin t) \] \[ =\cos^2 t-\sin^2 t \] Using the trigonometric identity \[ \cos^2 t-\sin^2 t=\cos 2t \] Thus \[ \frac{dy}{dt}=\cos 2t \] Step 4: Form the derivative $\dfrac{dy{dx}$.}
\[ \frac{dy}{dx}=\frac{\cos 2t}{\cos t+\sin t} \] Step 5: Substitute $t=\dfrac{\pi{4}$.}
First compute \[ \cos 2t=\cos\left(\frac{\pi}{2}\right)=0 \] Next, \[ \cos\frac{\pi}{4}+\sin\frac{\pi}{4} =\frac{\sqrt2}{2}+\frac{\sqrt2}{2} =\sqrt2 \] Thus \[ \frac{dy}{dx}=\frac{0}{\sqrt2}=0 \] Step 6: Final conclusion.
Therefore the value of the derivative at \(t=\dfrac{\pi}{4}\) is \[ 0 \] Final Answer: \[ \boxed{0} \]