Tangents with Parametric Equations — Question 6

Question 6 – Solution

Solution

1. Because $x=\ln t$, the parameter domain is $t>0$. Differentiate the coordinates: $$\frac{dx}{dt}=\frac1t, \qquad \frac{dy}{dt}=1-\frac1{t^2}.$$

2. Compute the first derivative with respect to $x$: $$\frac{dy}{dx} =\frac{dy/dt}{dx/dt} =\frac{1-t^{-2}}{t^{-1}} =t-\frac1t.$$

3. For a parametric curve, differentiate $dy/dx$ with respect to $t$ and divide once more by $dx/dt$: $$\frac{d^2y}{dx^2} =\frac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)}{dx/dt}.$$ Since $$\frac{d}{dt}\left(t-\frac1t\right)=1+\frac1{t^2},$$ we obtain $$\frac{d^2y}{dx^2} =\frac{1+t^{-2}}{t^{-1}} =t+\frac1t.$$

4. Evaluate at $t=1$: $$\left.\frac{d^2y}{dx^2}\right|_{t=1} =1+1 =\boxed{2}.$$

5. Because the second derivative is positive, the curve is concave up at this point. Also, $(x(1),y(1))=(0,2)$ and $dy/dx=0$, so $(0,2)$ is a local minimum of the curve when viewed as $y$ versus $x$.