for each function, determine the equation of the tangent line to the graph for the indicated value of the variable.
a) y = 3cosx + 4sinx + 1 , when x = pie
b) y = x - 3cosx , when x = (5pie)/3
π is PI not PIE!
a)
y'(x) = -3sinx + 4cosx
y'(π) = -3sin π + 4cos π = -4
y(π) = 3(-1) + 0 + 1 = -2
thus, the line through (π,-2) having slope -4 is
(y+2) = -4(x-pi)
b) y' = 1 + 3sinx
y'(5π/3) = 1 + 3(-√3/2) = 1 - 3√3/2
y(5π/3) = 5π/3 - 3(1/2) = 5π/3 - 3/2
(y-(5π/3 - 3/2) = (1 - 3√3/2)(x-5π/3)
thank you so much steve!
for b) (y-(5π/3 - 3/2) = (1 - 3√3/2)(x-5π/3)
is the square root for the whole 3/2 or is it just root 3 OVER 2.
I think by now you know the value of cos(pi/3)
(y-(5π/3 - 3/2) = (1 - (3√3)/2)(x-5π/3)
like that ? :(
To find the equation of the tangent line to the graph of a function at a specific value of the variable, you need to follow these steps:
1) Differentiate the function to find its derivative.
2) Evaluate the derivative at the given value of the variable to find the slope of the tangent line.
3) Use the point-slope form to write the equation of the tangent line using the slope and the given point.
Let's apply these steps to each function:
a) y = 3cos(x) + 4sin(x) + 1, when x = π
1) Differentiate the function to find its derivative:
dy/dx = -3sin(x) + 4cos(x)
2) Evaluate the derivative at x = π:
dy/dx = -3sin(π) + 4cos(π) = -3(0) + 4(-1) = -4
The slope of the tangent line at x = π is -4.
3) Use the point-slope form to write the equation of the tangent line:
y - y1 = m(x - x1), where x1 = π, y1 = 3cos(π) + 4sin(π) + 1
Substituting the values:
y - (3cosπ + 4sinπ + 1) = -4(x - π)
Simplifying:
y - (3 + 1) = -4(x - π)
y - 4 = -4x + 4π
y = -4x + 4π + 4
b) y = x - 3cos(x), when x = (5π)/3
1) Differentiate the function to find its derivative:
dy/dx = 1 + 3sin(x)
2) Evaluate the derivative at x = (5π)/3:
dy/dx = 1 + 3sin((5π)/3) = 1 + 3(-√3/2) = 1 - (3√3/2) = 1 - (3√3)/2
The slope of the tangent line at x = (5π)/3 is 1 - (3√3)/2.
3) Use the point-slope form to write the equation of the tangent line:
y - y1 = m(x - x1), where x1 = (5π)/3, y1 = (5π)/3 - 3cos((5π)/3)
Substituting the values:
y - ((5π)/3 - 3cos((5π)/3)) = (1 - (3√3)/2)(x - (5π)/3)
Simplifying:
y - ((5π)/3 - 3(-1/2)) = (1 - (3√3)/2)(x - (5π)/3)
y - ((5π)/3 + 3/2) = (1 - (3√3)/2)(x - (5π)/3)
These are the equations of the tangent lines to the graphs of the functions at the given values of the variable.