Chapter 3.1, Problem 46E

To determine

To find : the equation of line tangent to the curve at the point $t=-1$ .

Answer to Problem 46E

The equation of line tangent to the curve at point $t=-1$ is $y-1=1\left(x-5\right)$ .

Explanation of Solution

Given information :

  $x=2t^2+3,y=t^4$ .

Formula needed :

The equation of line tangent to the curve at point $t=a$ is $y-y_1=m\left(x-x_1\right)$ where $x_1,y_1$ are the value of $x$ and $y$ at $t=a$ respectively. Slope of line is $m={\left.\frac{dy}{dx}\right|}_{t=a}$ .

  $\frac{dy}{dx}$ partially expressed as $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ .

Partially differentiate $x$ and $y$ with respect to $t$ .

  $\begin{array}{c}\frac{dx}{dt}=\frac{d}{dt}\left(2t^2+3\right)\\ =4t\\ \frac{dy}{dt}=\frac{d}{dt}\left(t^4\right)\\ =4t^3\end{array}$

So,

  $\begin{array}{c}\frac{dy}{dx}=\frac{dy/dt}{dx/dt}\\ =\frac{4t^3}{4t}\\ =t^2\end{array}$

Slope is given by $m={\left.\frac{dy}{dx}\right|}_{t=-1}$ .

  $\begin{array}{c}m={\left.\frac{dy}{dx}\right|}_{t=-1}\\ ={\left(-1\right)}^2\\ =1\end{array}$

The value of $x$ and $y$ at $t=-1$ is given by $x_1=2{\left(-1\right)}^2+3,y_1={\left(-1\right)}^4$ .

  $\begin{array}{c}{x}_1=2+3\\ =5\end{array}$

And

  $y_1=1$

Thus, the equation of line tangent to the curve at point $t=-1$ is $y-1=1\left(x-5\right)$ .