how to find the slope of a tangent when the derivative involves x and y on both sides.

Question

I was given the function x²+y²=25  at (4,-3)

I took dy/dx of all the terms and got -x/y for the derivative.

Using that i assumed the line would be y(x)= -(x²/y) -(4x/y) -3 but I got that wrong. How am I supposed to find hte tangent line?

Answer 1

Hi Olivia,

 

I'm thinking of two different ways to solve this problem.

 

In one method, you solve the original equation for y: y = ±√(25 - x²), then choose the appropriate sign. You'd choose the negative sign because the point you're working with is (4, -3). Then, take the derivative and solve. The problem with this method is that you'll end up taking the derivative of a function inside of a square root, and that just gets messy. So, let's try something else...

 

Instead, let's take the x-derivative of both sides:

 

     [d/dx]{x² + y²} = [d/dx]{25}

 

We'll split up the left-hand side:

 

     [d/dx]{x²} + [d/dx]{y²} = [d/dx]{25}

 

We'll use the power rule to find the derivative of x² with respect to x, and the derivative of 25 is just zero:

 

     2x + [d/dx]{y²} = 0

 

Keep in mind that y changes with x. This means that we need to use the chain rule to take the x-derivative of y²:

 

     2x + 2y · (dy/dx) = 0

 

Solve for dy/dx:

 

     dy/dx = -x/y.

 

This is the same thing you got, so it looks like your calculus skills are still great. :)

 

Remember that we can interpret dy/dx as the slope m of the tangent line at the point (x, y). Since we're working with the point (4, -3), we can calculate the numerical value of the slope of the tangent line at that point:

 

     dy/dx|(x = 4, y = -3) = -4/(-3) = 4/3.

     m = 4/3 at point (4, -3).

 

Now we can use the point-slope form of a line to obtain the equation of the tangent line:

 

     (y - y₀) = m(x - x₀)

     y - (-3) = (4/3)(x - 4)

     y = (4/3)x - (16/3) - 3

     y = (4/3)x - (25/3)

 

You were very, very close to the answer. It looks like you simply forgot to evaluate the actual numerical value of the slope (dy/dx) before continuing.  :)

Answer 2

x² + y² = 25

y² = 25 - x²

y = √(25 - x²)

 

dy/dx = (1/2) [1 / √(25 - x²)] (-2x) = -x / √(25 - x²) = -x/y

 

At (4,-3), dy/dx = -4 / (-3) = 4/3

 

Slope of tangent line at (4,-3) is m = 4/3

 

Equation of tangent line is

 

y = (4/3)x + b

 

Use (4.-3) to find b.

 

-3 = (4/3)(4) + b

 

b = -25/3

 

Final equation of tangent line:

 

y = (4/3)x - 25/3