Tangents with Parametric Equations — Question 2

Question 2 – Solution

Solution

1. Differentiate the coordinate functions: $$\frac{dx}{dt}=1+\cos t, \qquad \frac{dy}{dt}=\sin t.$$

2. At $t=\pi/2$, $$\frac{dx}{dt}=1+\cos\frac{\pi}{2}=1, \qquad \frac{dy}{dt}=\sin\frac{\pi}{2}=1.$$ Since $dx/dt\ne0$, the tangent slope is $$\frac{dy}{dx} =\frac{dy/dt}{dx/dt} =\frac{1}{1} =\boxed{1}.$$

3. The point on the path is $$\left(x\left(\frac\pi2\right),y\left(\frac\pi2\right)\right) =\left(\frac\pi2+1,1\right).$$ Therefore, the tangent line is $$y-1=x-\left(\frac\pi2+1\right).$$

4. Both velocity components are positive at this instant. Thus the marker is moving to the right and upward, which is consistent with the positive slope.