Question

1. -/1 points let $ell$ be the tangent line to the curve $y = 4x^{2}$ at the point $(1,4)$. the angle of inclination of $ell$ is the angle $varphi$ that $ell$ makes with the positive direction of the $x$-axis. find $y$. $y=square$ find the slope $m$ of the tangent line at the point $(1,4)$. $m = square$ calculate $varphi$ to the nearest degree. $varphi=square^{circ}$ resources read it watch it

Explanation:

Step1: Differentiate the function

Using the power - rule $\frac{d}{dx}(ax^n)=nax^{n - 1}$, for $y = 4x^2$, we have $y'=\frac{d}{dx}(4x^2)=8x$.

Step2: Find the slope at the given point

Substitute $x = 1$ into $y'$. When $x = 1$, $m=y'(1)=8\times1 = 8$.

Step3: Calculate the angle of inclination

We know that the slope $m=\tan\varphi$. So, $\varphi=\arctan(m)$. Substitute $m = 8$ into the formula, $\varphi=\arctan(8)\approx83^{\circ}$.

Answer:

$y' = 8x$
$m = 8$
$\varphi\approx83$