Question

Find the slope of the line having a parametric equation y = 4t + 6 and x = t + 1

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is D 4

Given,

$x=t+1$

$y=4t+6$

Slope of line is given by $\frac{dy}{dx}$

Now,

$\frac{dx}{dt}=\frac{d}{dt}(t+1)$

$=\frac{d}{dt}\left(t\right)+\frac{d}{dt}\left(1\right)$

$=1+0$

$=1$

And,

$\frac{dy}{dt}=\frac{d}{dt}(4t+6)$

$=\frac{d}{dt}\left(4t\right)-\frac{d}{dt}\left(6\right)$

$=4-0$

$=4$

$\therefore \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

$=\frac{4}{1}$

$=4$