An equation of the line tangent to y=sinx+2cosx at (pi/2, 1) is
A. 2x-y=pi-1
B. 2x+y=pi+1
C. 2x-2y=2-pi
D. 4x+2y=2-pi
I got A
y' = cosx - 2sinx
y'(pi/2) = 0 - 2
y-1 = -2(x-pi/2)
or
2x+y = 1+pi
I get B.
http://www.wolframalpha.com/input/?i=plot+y%3Dsinx%2B2cosx%2C+y+%3D+1%2Bpi-2x+for+x+%3D+0+to+3
To find the equation of the tangent line to the curve represented by the equation y = sin(x) + 2cos(x), you can follow these steps:
Step 1: Determine the derivative of the given equation.
The derivative of y with respect to x will give us the slope of the tangent line at any given point. Taking the derivative of y with respect to x, we have:
dy/dx = d/dx (sin(x) + 2cos(x))
The derivative of sin(x) is cos(x), and the derivative of 2cos(x) is -2sin(x) (using the chain rule). So, the derivative dy/dx is:
dy/dx = cos(x) - 2sin(x)
Step 2: Substitute the x-coordinate of the given point into dy/dx.
To find the slope of the tangent line at the point (pi/2, 1), we need to substitute x = pi/2 into dy/dx:
dy/dx (at x = pi/2) = cos(pi/2) - 2sin(pi/2)
= 0 - 2(1)
= -2
So, the slope of the tangent line at (pi/2, 1) is -2.
Step 3: Use the point-slope form of a linear equation to write the equation of the tangent line.
The point-slope form of a linear equation is y - y1 = m(x - x1), where m is the slope of the line and (x1, y1) is a point on the line. In this case, we have the slope m = -2 and the point (x1, y1) = (pi/2, 1).
Plugging these values into the equation, we get:
y - 1 = -2(x - pi/2)
To simplify, distribute the -2:
y - 1 = -2x + pi
Finally, rearrange the equation to get it in the standard form Ax + By = C:
2x + y = pi - 1
Comparing the equation 2x + y = pi - 1 to the given options:
A. 2x - y = pi - 1
B. 2x + y = pi + 1
C. 2x - 2y = 2 - pi
D. 4x + 2y = 2 - pi
We can see that option A (2x - y = pi - 1) matches the equation we derived.
Therefore, the correct answer is A. 2x - y = pi - 1.
To find the equation of the line tangent to the curve y = sin(x) + 2cos(x) at the point (π/2, 1), we need to find the derivative of the function at that point and use it to write the equation of the tangent line.
First, let's find the derivative of y = sin(x) + 2cos(x).
dy/dx = cos(x) - 2sin(x)
Now, substitute the x-coordinate of the given point, which is π/2, into the derivative to find the slope (m) of the tangent line at that point.
m = cos(π/2) - 2sin(π/2)
= 0 - 2(1)
= -2
Using the point-slope form of the equation of a line, we can write the equation of the tangent line as:
y - y1 = m(x - x1)
Substituting the values of (x1, y1) = (π/2, 1) and m = -2:
y - 1 = -2(x - π/2)
y - 1 = -2x + π
2x + y = π + 1
Comparing this equation with the given options, we see that the correct answer is B. 2x + y = π + 1.