X2 + Y4 = 16
find the slopes of the 2 tangent lines to the curve from (5,1)
Are the 2 and 4 exponents?
Yes, sorry about that.
Require that
Y^4 = 16 - X^2
and also require a tangency condition that
(Y-1)/(X-5) = dY/dX
Implict differentiation results in
4Y^2 dY/dX = -2X
dY/dX = -X/(2Y^2)
Therefore
(Y-1)/(X-5) = -X/(2Y^2)
You now have two equations in two unknowns, X and Y. That should allow to solve for any (X,Y) solutions
Thanks.
To find the slopes of the tangent lines to the curve, we need to find the derivative of the equation and then substitute the value (5,1) to calculate the slopes.
Step 1: Differentiate the equation with respect to x.
Let's differentiate x^2 + y^4 = 16 with respect to x.
The derivative of x^2 with respect to x is 2x.
The derivative of y^4 with respect to x will involve the chain rule. Since y is also a function of x, we need to use the chain rule to differentiate it.
The chain rule states that if we have a composite function y(u) and we differentiate with respect to x, we get dy/du * du/dx. In this case, y(u) = u^4 and u = y(x), so we need to differentiate y^4 with respect to y, then multiply it with dy/dx.
The derivative of y^4 with respect to y is 4y^3.
Therefore, the derivative of y^4 with respect to x is 4y^3 * dy/dx.
So, the overall derivative of x^2 + y^4 with respect to x is:
2x + 4y^3 * dy/dx = 0
Step 2: Solve the derivative equation for dy/dx.
Let's isolate dy/dx on one side:
4y^3 * dy/dx = -2x
dy/dx = -2x / 4y^3
dy/dx = -x / (2y^3)
Step 3: Substitute the values (5,1) to find the slopes of the tangent lines.
dy/dx = -x / (2y^3)
Substituting x=5 and y=1:
dy/dx = -(5) / (2(1)^3)
dy/dx = -5/2
The slope of the tangent line to the curve from (5,1) is -5/2.
Since there are two tangent lines, we can obtain the slope of the other tangent line by finding the negative reciprocal of -5/2. The negative reciprocal of a/b is -b/a.
Therefore, the slope of the other tangent line is the negative reciprocal of -5/2, which is 2/5.