Chapter 4.1, Problem 45E

To determine

To find:the equation of tangent line to the curve at a point.

Answer to Problem 45E

Required equation of the tangent to the given curve at $t=\frac{1}{4}$ is $y=x+\frac{1}{4}$ .

Explanation of Solution

Given information:

1. The given equation of the curve is as follows

  $x=t,y=\sqrt{t}$ .

2. Need to find the equation of tangent at point $t=\frac{1}{4}$ .

Formula used:

1. The slope of the tangent of the curve is $m=\frac{dy}{dx}$ .
2. equation of the tangent line is as follows

  $\left(y-y_0\right)=m\left(x-x_0\right)$ .

3. $\frac{dy}{dx}=\frac{\left(\frac{dy}{dt}\right)}{\left(\frac{dx}{dt}\right)}$

Calculation:

Give equation of curve is

  $x=t,y=\sqrt{t}$

As $x=t$ , $\frac{dx}{dt}=\frac{d}{dt}t=1$

And $y=\sqrt{t}$ , $\frac{dy}{dt}=\frac{d}{dt}\sqrt{t}=\frac{1}{2\sqrt{t}}$

Calculate slope of the tangent to the curve that is $m$ .

  $m=\frac{dy}{dx}$

  $=\frac{\left(\frac{dy}{dt}\right)}{\left(\frac{dx}{dt}\right)}$

  $=\frac{\left(\frac{1}{2\sqrt{t}}\right)}{1}$

  $m$ $=\frac{1}{2\sqrt{t}}$

Calculate value of $x_0,y_0\text{and}m$ at $t=\frac{1}{4}$ .

  $x_0=t=\frac{1}{4}$

  $y_0=\sqrt{t}=\sqrt{\frac{1}{4}}=\frac{1}{2}$

  $m$ $\frac{1}{2\sqrt{t}}=\frac{1}{2\sqrt{\frac{1}{4}}}=1$

Substitute calculated value of $x_0,y_0\text{and}m$ in equation of tangent line.

  $\left(y-y_0\right)=m\left(x-x_0\right)$

  $\left(y-\frac{1}{2}\right)=1\left(x-\frac{1}{4}\right)$

  $y=x-\frac{1}{4}+\frac{1}{2}$

  $y=x+\frac{1}{4}$

Hence required equation of the tangent to the given curve at $t=\frac{1}{4}$ is $\underset{\_}{y=x+\frac{1}{4}}$ .