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: Find $\dfrac{dx}{dt}$
To find $\dfrac{dy}{dx}$, we first need to calculate $\dfrac{dx}{dt}$. Since $x = \sin t - \cos t$, we differentiate with respect to $t$: 
$\dfrac{dx}{dt} = \cos t + \sin t$.

Step 2: Find $\dfrac{dy}{dt}$
Now, differentiate $y = \sin t \cos t$ with respect to $t$. Using the product rule: 
$\dfrac{dy}{dt} = \cos^2 t - \sin^2 t$.

Step 3: Use the formula for $\dfrac{dy}{dx}$
The formula for $\dfrac{dy}{dx}$ is: 
$\dfrac{dy}{dx} = \dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}$. 
Substituting the values we found: \[ \dfrac{dy}{dx} = \dfrac{\cos^2 t - \sin^2 t}{\cos t + \sin t}. \]

Step 4: Evaluate at $t = \dfrac{\pi}{4}$
At $t = \dfrac{\pi}{4}$, we have: \[ \cos \dfrac{\pi}{4} = \sin \dfrac{\pi}{4} = \dfrac{\sqrt{2}}{2}. \] Substituting these values: \[ \dfrac{dy}{dx} = \dfrac{\left(\dfrac{\sqrt{2}}{2}\right)^2 - \left(\dfrac{\sqrt{2}}{2}\right)^2}{\dfrac{\sqrt{2}}{2} + \dfrac{\sqrt{2}}{2}} = \dfrac{0}{\sqrt{2}} = 0. \]

Final Answer:
$\dfrac{dy}{dx} = 0$ at $t = \dfrac{\pi}{4}$.