Please help me to find an equation for the tangent to the curve xsin2y=ycos2x at the point (phi/4, phi/2)
first, it's PI, not PHI
pi(π) = 3.14, the ratio of circumference to diameter
phi(φ) = 1.62 (1+√5)/2, the golden ratio
So, we use implicit differentiation to get
xsin2y = ycos2x
sin2y + 2xcos2y y' = cos2x y' - 2ysin2x
y' = -(sin2y + 2y sin2x)/(2x cos2y - cos2x)
at (π/4,π/2), that gives a slope of
y' = -(0 + π*(1))/(π/2*(-1) - 0)
= -π/(-π/2)
= 2
So, bow you want the line with slope=2 through (π/4,π/2):
y - π/2 = 2(x - π/4)
y = 2x
xsin2y=ycos2x
To find the equation of the tangent to the curve at the point (φ/4, φ/2), we need to find the slope of the tangent line at that point and then use the point-slope form to write the equation of the line.
Step 1: Find the slope of the tangent line
To find the slope of the tangent line, we can use implicit differentiation. Implicitly differentiate the equation xsin(2y) = ycos(2x) with respect to x.
d/dx(xsin(2y)) = d/dx(ycos(2x))
Applying the product rule, we get:
sin(2y) + x * d/dx(sin(2y)) = y * d/dx(cos(2x)) + cos(2x) * d/dx(y)
The derivative of sin(2y) with respect to x is 2y' * cos(2y), where y' represents dy/dx (the derivative of y with respect to x). Similarly, the derivative of cos(2x) with respect to x is -2x' * sin(2x).
This yields:
sin(2y) + x * 2y' * cos(2y) = y * (-2x' * sin(2x)) + cos(2x) * y'
Simplifying and solving for y', we have:
2xy' * cos(2y) + 2xy * sin(2x) = y * cos(2x) - sin(2y)
y' * cos(2y) = (y * cos(2x) - sin(2y) - 2xy * sin(2x)) / (2x)
Step 2: Substitute the coordinates of the point (φ/4, φ/2)
Substitute x = φ/4 and y = φ/2 into the equation derived in step 1:
y' * cos(φ) = (φ/2 * cos(φ) - sin(φ) - φ/8 * sin(2φ)) / (φ/2)
Step 3: Simplify the equation
Divide both sides of the equation by cos(φ):
y' = (φ/2 * cos(φ) - sin(φ) - φ/8 * sin(2φ)) / (φ/2 * cos(φ))
Simplify further:
y' = (2 * φ * cos(φ) - 4 * sin(φ) - φ * sin(2φ)) / (4 * φ * cos(φ))
Step 4: Write the equation of the tangent line
Using the point-slope form of a line, the equation of the tangent line can be written as:
y - y1 = m(x - x1)
where (x1, y1) are the coordinates of the given point, and m is the slope of the tangent line. Substituting the values, we get:
y - φ/2 = ((2 * φ * cos(φ) - 4 * sin(φ) - φ * sin(2φ)) / (4 * φ * cos(φ)))(x - φ/4)
Simplifying further, we have:
y - φ/2 = (2 * cos(φ) - 4 * φ^(-1) * sin(φ) - sin(2φ)/2) / (4 * cos(φ))(x - φ/4)
This is the equation of the tangent line to the curve xsin(2y) = ycos(2x) at the point (φ/4, φ/2).
To find the equation of the tangent to the curve xsin(2y) = ycos(2x) at the point (φ/4, φ/2), we will need to find the derivative of the function and then use it to find the slope of the tangent line.
Step 1: Find the derivative of the function
To find the derivative, we need to apply the chain rule and differentiate both sides of the equation with respect to x.
Differentiating xsin(2y) with respect to x:
d/dx [xsin(2y)] = d/dx [ycos(2x)]
Differentiating the left side using the product rule, we get:
sin(2y) + x * d/dx [sin(2y)]
Differentiating the right side with respect to x, we get:
-d/dx [ysin(2x)]
Step 2: Evaluate the derivative at the given point
Substitute the given point (φ/4, φ/2) into the derivative expression we obtained in Step 1 to find the slope of the tangent line at that point.
Let's evaluate the derivative at the point (φ/4, φ/2).
sin(2 * φ/2) + (φ/4) * d/dx [sin(2 * φ/2)] = -d/dx [((φ/2)sin(2 * φ/4))]
Using the values φ/2 = 1.5708 and φ/4 = 0.7854, we have:
sin(1.5708) + (0.7854) * d/dx [sin(1.5708)] = -d/dx [(1.5708 * sin(0.7854))]
Simplifying, we get:
1 + (0.7854) * d/dx [1] = -d/dx [(1.5708 * 0.7071)]
Since d/dx [1] is 0 (the derivative of a constant is zero), we have:
1 + 0.7854 * 0 = -0.7071 * d/dx [1.1115]
Simplifying further, we find:
1 = -0.7071 * d/dx [1.1115]
Therefore, the derivative at the point (φ/4, φ/2) is d/dx [1.1115] = -1/√2.
Step 3: Use the slope and point to find the equation of the tangent
We have the slope of the tangent line from Step 2. Now, we can use the point-slope form of a line to find the equation of the tangent line. The point-slope form is given by:
y - y₁ = m(x - x₁),
where (x₁, y₁) is the given point and m is the slope we found.
Substituting the values, we have:
y - φ/2 = (-1/√2)(x - φ/4)
Simplifying, we get:
y - φ/2 = (-1/√2)x + φ/(4√2) + φ/2
Consolidating terms, we have:
y = (-1/√2)x + (φ + 2φ√2)/(4√2)
Therefore, the equation of the tangent to the curve xsin(2y) = ycos(2x) at the point (φ/4, φ/2) is y = (-1/√2)x + (φ + 2φ√2)/(4√2).