How do you use implicit differentiation to find the points on the curve #x^2+y^2=5x+4y# where tangent line is horizontal and where tangent line is vertical?

Question

Charles Arocha · Accepted Answer

Horizontal tagts. at #(5/2,(4+sqrt41)/2), &, (5/2,(4-sqrt41)/2).#

Eqn. of the Curve C is given to be, # C :x^2+y^2-5x-4y=0.#

Diff.ing w.r.t. #x,# using the Chain Rule, we get,

# 2x+2ydy/dx-5-4dy/dx=0.#

# :. dy/dx=(5-2x)/{2(y-2)}.#

We know that, #dy/dx# denotes the Slope of a tangent to the

Curve at the pt.#(x,y).#

Now, the slope of Horizontal Tangent is #0.#

#:.," for horizontal tgt., "dy/dx=0 rArr (5-2x)/{2(y-2)}=0.#

# :. x=5/2," &, for the corresponding y, we appeal to "C," to get,"#

# x=5/2, C : 25/4+y^2-25/2-4y=0, i.e., y^2-4y-25/4=0.#

#:. y={4+-sqrt(16+25)}/2=(4+-sqrt41)/2.#

Thus, there are two horizontal tagts. at

#(5/2,(4+sqrt41)/2), &, (5/2,(4-sqrt41)/2).#

Similarly, for the Vertical Tangent, we must have undefined

#dy/dx, i.e., y=2,# &, the corresponding #x# can be obtained from #C.#

Luke Alvarez · Answer

undefined To find the points on the curve $x^2 + y^2 = 5x + 4y$ where the tangent line is horizontal and vertical using implicit differentiation, follow these steps:

1. Differentiate both sides of the equation with respect to $x$.
2. Solve for $dy/dx$, which represents the slope of the tangent line.
3. For horizontal tangent lines, set $dy/dx = 0$ and solve for $x$ and $y$.
4. For vertical tangent lines, find the points where the derivative $dy/dx$ is undefined.

Let's go through these steps:

1. Differentiating both sides with respect to $x$, we get:
$$2x + 2y \frac{dy}{dx} = 5 + 4 \frac{dy}{dx}$$

2. Rearrange to solve for $dy/dx$:
$$2y \frac{dy}{dx} - 4 \frac{dy}{dx} = 5 - 2x$$
$$\frac{dy}{dx}(2y - 4) = 5 - 2x$$
$$\frac{dy}{dx} = \frac{5 - 2x}{2(y - 2)}$$

3. For horizontal tangent lines, set $dy/dx = 0$:
$$\frac{5 - 2x}{2(y - 2)} = 0$$
$$5 - 2x = 0$$
$$x = \frac{5}{2}$$

Substitute $x = \frac{5}{2}$ into the original equation to find $y$:
$$\left(\frac{5}{2}\right)^2 + y^2 = 5\left(\frac{5}{2}\right) + 4y$$
$$y^2 = \frac{25}{4} - \frac{25}{2} + 4y$$
$$y^2 - 4y + \frac{25}{4} = 0$$

Solve this quadratic equation for $y$ to find the $y$-coordinate(s) of the point(s) where the tangent line is horizontal.

4. For vertical tangent lines, the derivative $dy/dx$ is undefined where the denominator becomes zero:
$$2(y - 2) = 0$$
$$y = 2$$

Substitute $y = 2$ into the original equation to find $x$. This will give the $x$-coordinate(s) of the point(s) where the tangent line is vertical.