Question

How do you find the slope of the line tangent to the curve $3x^2-2xy+y=11$ at the point $\left(1,-2\right)$?

Answer 1

Finding the slope of the tangent:

Given, that the equation of the curve is $\begin{array}{rcl}3x^2-2xy+y& =& 11\end{array}$.
$\begin{array}{rcl}& \Rightarrow & 3x^2-2xy+y-11=0\end{array}$

A point on the curve is $\left(1,-2\right)$.

Step- 1: Differentiating the equation of the curve.

$\begin{array}{rcl}\frac{d\left(3x^2-2xy+y-11\right)}{dx}& =& 0\\ & \Rightarrow & 6x-2y-2x\frac{dy}{dx}+\frac{dy}{dx}=0\end{array}$

Step 2. Substituting the value of $x=1$ and $y=-2$ in order to get the slope of the tangent.

$\begin{array}{rcl}6\left(1\right)-2\left(-2\right)-2\left(1\right)\frac{dy}{dx}+\frac{dy}{dx}& =& 0\\ & \Rightarrow & 6+4-2\frac{dy}{dx}+\frac{dy}{dx}=0\\ & \Rightarrow & 10-\frac{dy}{dx}=0\\ & \Rightarrow & \frac{dy}{dx}=10\end{array}$

As, $\frac{dy}{dx}$ is the slope of the tangent.

Thus the slope of the tangent is $10$.

Hence, the slope of the tangent to the curve at point $\left(1,-2\right)$ is $\mathbf{10}$.