Tangents with Parametric Equations — Question 4

Question 4

Problem

For $x=e^t$, $y=te^t$, find the tangent line at the point whose $x$-coordinate is $1$.

Question 4 – Solution

Solution

1. Locate the required parameter value by solving $$x=e^t=1.$$ Since the exponential function equals $1$ only at $t=0$, the required parameter value is $t=0$.

2. Evaluate the $y$-coordinate: $$y(0)=0\cdot e^0=0.$$ Thus the point of tangency is $(1,0)$.

3. Differentiate both coordinates. The product rule is needed for $y$: $$\frac{dx}{dt}=e^t, \qquad \frac{dy}{dt}=e^t+te^t=e^t(1+t).$$

4. Because $dx/dt=e^t\ne0$, divide the derivatives: $$\frac{dy}{dx} =\frac{e^t(1+t)}{e^t} =1+t.$$ At $t=0$, the slope is $m=1$.

5. Use point–slope form through $(1,0)$: $$y-0=1(x-1).$$ Therefore, the tangent line is $$\boxed{y=x-1}.$$