Find the slope of the tangent line to the curve 2(x^2+y^2)^2=25(x^2−y^2) at the point (−3,−1)?
Here's what I did:
2(x^4 + y^4) = 25(x^2-y^2)
2x^4 + 2y^4 = 25x^2 - 25y^2
8x^3 + 8y^3(dy/dx) = 50x - 50y(dy/dx)
d/dx(8y^3 + 50y) = 50x - 8x^3
d/dx = (50x-8x^3)/(8y^3 + 50y)
and got m= -1.13793
What did I do wrong?
in algebra I you learned that (a+b)^2 ≠ a^2 + b^2
That's still true, even if you're taking pre-cal! :-)
2(x^2+y^2)^2=25(x^2−y^2)
2 * 2(x^2+y^2) (2x+2yy') = 25(2x-2yy')
y' =
-x(4x^2+4y^2-25)
--------------------------
y(4x^2+4y^2+25)
So, at (-3,1), y' = -9/13
Also, don't lose the y in your dy/dx
thanks...though I asked for (-3,-1).
To find the slope of the tangent line to the curve at a specific point, you need to take the derivative of the equation with respect to x, and then substitute the x and y values of the given point into the derivative. It looks like you made a small mistake in your calculation.
Let's go through the steps together:
Step 1: Start with the given equation.
2(x^2 + y^2)^2 = 25(x^2 - y^2)
Step 2: Take the derivative of the equation with respect to x.
[2(x^2 + y^2)^2]' = [25(x^2 - y^2)]'
4(x^2 + y^2)(2x + 2y(dy/dx)) = 25(2x - 2y(dy/dx))
Step 3: Substitute the x and y values of the given point (-3,-1).
4((-3)^2 + (-1)^2)(2(-3) + 2(-1)(dy/dx)) = 25(2(-3) - 2(-1)(dy/dx))
Step 4: Simplify the equation.
4(9 + 1)(-6 - 2(dy/dx)) = 25(-6 + 2(dy/dx))
40(-8 - 2(dy/dx)) = 25(-12 + 2(dy/dx))
-320 - 80(dy/dx) = -300 + 50(dy/dx)
Step 5: Collect like terms.
30(dy/dx) = 20
dy/dx = 20/30
dy/dx = 2/3
So, the slope of the tangent line to the curve at the point (-3,-1) is 2/3.
It seems like you missed a negative sign in the equation when simplifying. Always double-check your calculations to avoid small errors.