Find the slope of the tangent line to the curve
2sin(x) + 6cos(y) - 6sin(x)cos(y) + x = 3pi
at the point (3pi, 7pi/2)
Thank you very much for your help.
2cosx - 6siny y' - 6cos^2(y) + 6sin^2(x) y' + 1 = 0
at (3pi,7pi/2) we have
2(-1) - 6(-1)y' - 6(0) + 6(0)y' + 1 = 0
-2 + 6y' + 1 = 0
6y' = 1
y' = 1/6
Oops. Make that
2cosx - 6siny y' - 6cosx cosy + 6sinx siny y' + 1 = 0
2(-1) - 6(-1)y' - 6(-1)(0) + 6(0)y' + 1 = 0
-2 + 6y' + 1 = 0
y' = 1/6
got the right answer before, but only because all the wrong stuff evaluated to 0!
To find the slope of the tangent line to the curve, we need to find the derivative of the curve and then evaluate it at the given point.
The equation of the curve is 2sin(x) + 6cos(y) - 6sin(x)cos(y) + x = 3π.
To find the derivative with respect to x, we differentiate each term with respect to x:
d/dx (2sin(x)) = 2cos(x),
d/dx (6cos(y)) = 0 (since y is a constant),
d/dx (-6sin(x)cos(y)) = -6cos(x)cos(y),
d/dx (x) = 1.
Summing up all the derivatives, we have:
2cos(x) - 6cos(x)cos(y) + 1 = 0.
Now, we can find the derivative with respect to y, by differentiating each term with respect to y:
d/dy (2sin(x)) = 0 (since x is a constant),
d/dy (6cos(y)) = -6sin(y),
d/dy (-6sin(x)cos(y)) = -6sin(x)sin(y),
d/dy (x) = 0 (since x is a constant).
Summing up all the derivatives, we have:
-6sin(y) - 6sin(x)sin(y) = 0.
Now we have a system of two equations:
2cos(x) - 6cos(x)cos(y) + 1 = 0,
-6sin(y) - 6sin(x)sin(y) = 0.
To find the values of x and y at the point (3π, 7π/2), we substitute these values into the equations:
2cos(3π) - 6cos(3π)cos(7π/2) + 1 = 0,
-6sin(7π/2) - 6sin(3π)sin(7π/2) = 0.
Simplifying the equations, we have:
-2 - 6(-1)(0.5) + 1 = 0,
0 + 6(1)(0) = 0.
Simplifying further, we get:
-2 + 3 + 1 = 0,
0 = 0.
Since both equations are satisfied, the point (3π, 7π/2) lies on the curve.
Now, let's find the slope of the tangent line at this point.
To find the slope, we can differentiate the equation of the curve with respect to x:
2cos(x) - 6cos(x)cos(y) + 1 = 0.
Differentiating, we get:
-2sin(x) + 6sin(x)cos(y) = 0.
Substituting x = 3π and y = 7π/2 into this equation, we have:
-2sin(3π) + 6sin(3π)cos(7π/2) = 0.
Since sin(3π) = 0, the equation simplifies to:
6sin(3π)cos(7π/2) = 0.
Simplifying further, we get:
0 = 0.
Therefore, the slope of the tangent line to the curve at the point (3π, 7π/2) is 0.
To find the slope of the tangent line to the given curve at the point (3pi, 7pi/2), we will differentiate the curve with respect to x and y separately and then substitute the given coordinates into the resulting equations.
Let's start by differentiating the equation with respect to x:
d/dx (2sin(x) + 6cos(y) - 6sin(x)cos(y) + x) = d/dx (3pi)
Differentiating sin(x) with respect to x gives us cos(x), and differentiating x with respect to x gives us 1. The partial derivative of cos(y) with respect to x is 0 since y is not a function of x. Thus, the equation becomes:
cos(x) + 1 = 0
At this point, we can solve this equation for x to find the x-coordinate of the point where the slope is required. In this case, we are given that x = 3pi, so we substitute 3pi into the equation:
cos(3pi) + 1 = 0
Since cos(3pi) is equal to -1, we have:
-1 + 1 = 0
Which is true. Therefore, x = 3pi is a valid x-coordinate.
Now, let's differentiate the equation with respect to y:
d/dy (2sin(x) + 6cos(y) - 6sin(x)cos(y) + x) = d/dy (3pi)
Differentiating cos(y) with respect to y gives us -sin(y), and since -sin(y) is a constant with respect to y, the partial derivative of any constant with respect to y is 0. Therefore, our equation becomes:
-6sin(x)cos(y) = 0
Since sin(x) and cos(y) are trigonometric functions, this equation will be satisfied if either sin(x) = 0 or cos(y) = 0. However, we cannot directly substitute y = 7pi/2 into this equation since it is not valid for this specific equation. Instead, we need to solve the equation for y to find the y-coordinate of the required point.
Let's solve the equation cos(y) = 0:
cos(y) = 0
To find the values of y that satisfy this equation, we need to look for the x-intercepts of the cosine function. The x-intercepts occur at integer multiples of pi/2. Therefore, we have:
y = (2n + 1)pi/2
where n is an integer. Now, let's substitute the given x-coordinate, x=3pi, into the equation to find y:
y = (2n + 1)pi/2 = (2(0) + 1)pi/2 = pi/2
Therefore, the y-coordinate of the point is y = pi/2.
Now that we have both the x and y coordinates of the point (3pi, pi/2), we can substitute them back into the equation:
cos(x) + 1 = 0
cos(3pi) + 1 = 0
-1 + 1 = 0
Which is true. Therefore, the point indeed lies on the curve.
To find the slope of the tangent line at this point, we need to evaluate the derivative of y with respect to x at this point. The derivative dy/dx gives the rate of change of y with respect to x, which is the slope of the tangent line.
dy/dx = -6sin(x)cos(y) / (cos(x) + 1)
Substituting x=3pi and y=pi/2 into the equation:
dy/dx = -6sin(3pi)cos(pi/2) / (cos(3pi) + 1)
sin(3pi) = 0 and cos(pi/2) = 0, therefore:
dy/dx = -6(0)(0) / (cos(3pi) + 1)
dy/dx = 0 / (cos(-pi) + 1)
cos(-pi) = -1, therefore:
dy/dx = 0 / (-1 + 1)
dy/dx = 0 / 0
The expression 0/0 is an indeterminate form, which means we can't determine the slope directly from this equation. To handle this, we need to simplify further or use an alternate method such as taking the limit of the function as it approaches the point. However, this might vary depending on the context and application you are working with.