y sin 12x = x cos 2y, (pi/2, pi/4)
Find the tangent line to the curve at the given point.
y=???
Can anyone else elaborate?
To find the tangent line to the curve at the given point (π/2, π/4), we need to find the slope of the curve at that point.
First, let's differentiate the given equation implicitly with respect to x.
Differentiating both sides of the equation y sin(12x) = x cos(2y) with respect to x:
(d/dx) [y sin(12x)] = (d/dx) [x cos(2y)]
Using the product rule, we get:
sin(12x) * dy/dx + 12y * cos(12x) = cos(2y) - x * sin(2y) * dy/dx
Now, let's rearrange the equation to solve for dy/dx:
sin(12x) * dy/dx + x * sin(2y) * dy/dx = cos(2y) - 12y * cos(12x)
Factoring out dy/dx:
dy/dx * (sin(12x) + x * sin(2y)) = cos(2y) - 12y * cos(12x)
Dividing both sides by (sin(12x) + x * sin(2y)):
dy/dx = (cos(2y) - 12y * cos(12x)) / (sin(12x) + x * sin(2y))
Now, we can substitute the given values π/2 for x and π/4 for y:
dy/dx = (cos(2(π/4)) - 12(π/4) * cos(12(π/2))) / (sin(12(π/2)) + (π/2) * sin(2(π/4)))
Simplifying the expression:
dy/dx = (cos(π/2) - 12(π/4) * cos(6π)) / (sin(6π) + (π/2) * sin(π/2))
dy/dx = (0 - 12(π/4) * cos(6π)) / (0 + (π/2) * 1)
dy/dx = (-12(π/4) * cos(6π)) / (π/2)
Simplifying further, we have:
dy/dx = -6π
Therefore, the slope of the curve at the point (π/2, π/4) is -6π.
Now, we can use the point-slope form to find the equation of the tangent line.
The point-slope form of a line is given by y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope of the line.
Using the point (π/2, π/4) and the slope -6π, the equation of the tangent line is:
y - π/4 = -6π(x - π/2)
Multiplying out:
y - π/4 = -6πx + 6π²
Adding π/4 to both sides:
y = -6πx + 6π² + π/4
Simplifying:
y = -6πx + 6π² + π/4
Therefore, the equation of the tangent line to the curve at the point (π/2, π/4) is y = -6πx + 6π² + π/4.
To find the tangent line to the curve at a given point, we need to calculate the derivative of the curve and substitute the given point into the derivative formula.
Given the equation y sin(12x) = x cos(2y), we can implicitly differentiate both sides with respect to x.
Differentiating y sin(12x) with respect to x gives us:
(d/dx)(y sin(12x)) = (d/dx)(x cos(2y))
Using the product rule, the left side becomes:
y' sin(12x) + y(12cos(12x))
Using the chain rule, the right side becomes:
cos(2y) - x (-sin(2y))(2y')
Next, we can rearrange the equation and solve for y' (the derivative of y with respect to x):
y' sin(12x) + y(12cos(12x)) = cos(2y) + 2xy' sin(2y)
Now we can substitute the given point (pi/2, pi/4) into the equation. This will give us the value of y' at that point.
Let's plug in x = pi/2 and y = pi/4:
(pi/4)sin(12(pi/2)) + (pi/4)(12cos(12(pi/2))) = cos(2(pi/4)) + 2(pi/2)(pi/4)sin(2(pi/4))
Simplifying the equation, we get:
(pi/4)(-1) + (pi/4)(0) = 1 + pi/2(1)
Therefore, the equation simplifies to:
(-pi/4) = 1 + pi/2
Now, we can find the value of y' by rearranging the equation and solving for y':
y' = (1 + pi/2 - pi/4) / (2 pi/2)
Simplifying further, we get:
y' = (4 + 2pi - pi) / (4 pi)
y' = (3 + pi) / (4 pi)
Finally, we have found the value of y' at the given point. Hence, the equation of the tangent line to the curve at the point (pi/2, pi/4) is:
y = ((3 + pi) / (4 pi))(x - pi/2) + pi/4
dy/dx * sin12x+ 12 y cos12x= cos2y-xsin2y dy/dx
dy/dx(sin12x+xsin2y)=cos2y-12ycos12x
first, solve for dy/dx, that is the slope.at x,y.
y= mx+b
Now put in the x,y in that equation, and solve for b.