Q5P - Mathematical Methods in Physical Sciences

Question

If $\mathit{x}{\mathit{y}}^{\mathbf{3}}\mathbf{-}\mathit{y}{\mathit{x}}^{\mathbf{3}}\mathbf{=}\mathbf{6}$ is the equation of a curve, find the slope and the equation of the tangent line at the point $\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{)}$ . Computer plot the curve and the tangent line on the same axes.

Step-by-Step Solution

Answer

The tangent has a slope of $\frac{-2}{11}$ .

The equation for the tangent is.....$2x+11y-24=0$.

Step1: Explanation of Solution

The provide equation or curve is $x{y}^3-y{x}^3=6.$

Step 2: Implicit Differentiation

To find the slopes of tangents to curves that aren't clearly functions, we can utilize implicit differentiation (they fail the vertical line test). Parts of y are assumed to be functions that satisfy the supplied equation, but y is not a function of x .

Step 3: Calculation

Consider the equation of curve below:

$x{y}^3-y{x}^3=6$

Differentiate the provided above equation of curve,

$$\begin{gathered} x.3y^3\frac{dy}{dx}+y^3-\left(y.3x^2+x^3.\frac{dy}{dx}\right)=0 \\ 3x{y}^3\frac{dy}{dx}+y^3-3x^{2}y-x^3.\frac{dy}{dx}=0 \\ {y}^3-3x^{2}y+\left(3x{y}^2-x^3\right)\frac{dy}{dx}=0 \end{gathered}$$

Hence,

$\frac{dy}{dx}=\frac{3x^{2}y-y^3}{3x{y}^2-x^3}$

Put the  $(1,2)$ for $(x,y)$

$$\begin{gathered} {\left(\frac{dy}{dx}\right)}_{(1,2)}=\frac{3{\left(1\right)}^2\left(2\right)-{\left(2\right)}^3}{3\left(1\right){\left(2\right)}^2-{\left(1\right)}^3} \\ =\frac{6-8}{12-1} \\ =\frac{-2}{11} \end{gathered}$$

Then, the equation for tangent line, 

$y-2=\frac{-2}{11}(x-1)$

Implies that,

$2x+11y-24=0$