Question

Find the Slope of the normal to the functiony = -5t^2+ 9t at t = -3

Answer 1

Given

$y=-5t^2+9t,t=-3$

Find

The slope of the normal

Explanation

to find the slope we need to differentiate the function with respect to t.

$\begin{gathered} y=-5t^2+9t \\ \frac{dy}{dt}=-10t+9 \end{gathered}$

Slope = dy/dt

now, slope at t = -3

$\begin{gathered} \frac{dy}{dt}_{t=-3} \\ -10\times(-3)+9 \\ 30+9 \\ 39 \end{gathered}$

as we know that the slope of normal is perpendicular to tangent. hence the slope of normal is given by

slope of tangent * slope of normal = -1

so, slope of normal =

$-\frac{1}{39}$

Final Answer

slope of normal =

$-\frac{1}{39}$