Question

Use implicit differentiation to find the equations of the tangent and normal lines at the point (3, 1) for the curve (y2 + 4)y = 13.

Answer 1

To find the tangent and normal lines at (3, 1) for the curve (y² + 4)y = 13, we use implicit differentiation to get the slope -2/7, then use point-slope form for both lines, resulting in y - 1 = (-2/7)(x - 3) for the tangent and y - 1 = (7/2)(x - 3) for the normal.

Step-by-step explanation:

Using Implicit Differentiation to Find Tangent and Normal Lines

Finding the equations of the tangent and normal lines at the point (3, 1) for the curve defined by (y² + 4)y = 13 requires the use of implicit differentiation to determine the slope of the tangent line at that point. First, take the derivative of both sides of the equation with respect to x:

d/dx [(y² + 4)y] = d/dx [13]

This yields 3y²(dy/dx) + y(2dy/dx) = 0 after applying the product rule and the chain rule. Solving for dy/dx, we get:

dy/dx = -2y/(3y² + 4)

Plugging in y = 1, we find dy/dx at (3,1) is -2/7. This is the slope of the tangent line. To find the equation of the line, we use the point-slope form:

y - 1 = (-2/7)(x - 3)

For the normal line, which is perpendicular to the tangent, we take the negative reciprocal of the tangent's slope, which gives us 7/2. Using the point-slope form again, the equation of the normal line is:

y - 1 = (7/2)(x - 3)

These are the equations of the tangent and normal lines at the point (3, 1).

Answer 2

Given the equation below;

`(x^2+4)y=13`

To find the derivative of the equation above, we will defferentiate implicitly, this is shown below

`\begin{gathered} (x^2+4)y=13 \\ \text{ Remove bracket} \\ x^2(y)+4(y)=13 \\ x^2y+4y=13 \end{gathered}`
`\begin{gathered} 2xy+x^2\frac{\mathrm{d}y}{dx}+4\frac{\mathrm{d}y}{dx}=0 \\ x^2\frac{\mathrm{d}y}{dx}+4\frac{\mathrm{d}y}{dx}=-2xy \\ \text{factor }\frac{\mathrm{d}y}{dx} \\ \frac{\mathrm{d}y}{dx}(x^2+4)=-2xy \end{gathered}`
`\frac{\mathrm{d}y}{dx}=(-2xy)/((x^2+4))`

At point (3, 1) we can obtain the value of the slope ( dy/dx), by substituting for x and y.

`\begin{gathered} x=3,y=1 \\ \frac{\mathrm{d}y}{dx}=(-2xy)/((x^2+4)) \\ \frac{\mathrm{d}y}{dx}=(-2(3)(1))/(3^2+4)=(-6)/(9+4)=-(6)/(13) \end{gathered}`

Using the point-slope equation of a line formula as shown below

`y-y_1=m(x-x_1)`

We can use the above to find the tangent to the line equation. then we have:

`\begin{gathered} y-1=-(6)/(13)(x-3) \\ y-1=-(6)/(13)x+(18)/(13) \\ y=-(6)/(13)x+(18)/(13)+1 \\ y=-(6)/(13)x+(31)/(13) \end{gathered}`

To find the normal line equation, we have to use the condition for perpendicularity of two lines, that is

`\begin{gathered} m_1=-(1)/(m_2) \\ If_{} \\ m_1\Rightarrow\text{slope of the tangent line} \\ m_2\Rightarrow\text{slope of the normal line} \\ m_2=-(1)/(m_1) \\ \text{Thus, the slope of the normal line is} \\ =(13)/(6) \end{gathered}`

Using point-slope equation of a line we can also find the normal line equation.

`\begin{gathered} y-1=(13)/(6)(x-3) \\ y-1=(13)/(6)x-(13)/(2) \\ y=(13)/(6)x-(13)/(2)+1 \\ y=(13)/(6)x-(11)/(2) \end{gathered}`

Hence, the tangent line and normal line equation are

`\begin{gathered} \text{tangent line}\Rightarrow y=-(6)/(13)x+(31)/(13)\text{ } \\ \text{normal line}\Rightarrow y=(13)/(6)x-(11)/(2) \end{gathered}`

This can be simplified as;

`\begin{gathered} \text{tangent line}\Rightarrow y=-(6)/(13)x+2(5)/(13)\text{ } \\ \text{normal line}\Rightarrow y=2(1)/(6)x-5(1)/(2) \end{gathered}`