find an equation to the curve at the point corresponding to the given value of the parameter.
x = tcost y = tsint when t = π
i know I am supposed to find dy and dx which is:
dy = (product form) t*-sint + 1*cost
simplifying = -tsint+cost
dx = tcost+sint
now, to find the I realize it is simply just dy/dx
which equals -tsint+cost/tcost+sint
however slope intercept form is y-y1=m(x-x1) right?
no, that is point-slope form.
To me, it's just as good as the slop-intercept form, since it contains all the necessary info.
For this kind of problem, you have a point on the curve, and the slope of the tangent. The point-slope form is ideal.
x = tcost
dx/dt = cost - tsint
y = tsint
dy/dt = sint - tcost
dy/dx = -π
Yes, you are correct in finding the derivative of y with respect to x, which is dy/dx.
To find the slope-intercept form for the equation of the tangent at the point corresponding to t=π, you first need to substitute t=π into your expressions for dy and dx:
dy = -tsin(t) + cos(t)
dx = tcos(t) + sin(t)
Next, evaluate these expressions at t=π:
dy = -πsin(π) + cos(π) = 0 + (-1) = -1
dx = πcos(π) + sin(π) = (-π) + 0 = -π
Now you have the slope of the tangent at the point corresponding to t=π, which is -1. To find the equation of the tangent line, you also need the coordinates of the point on the curve.
Substitute t=π into the given parametric equations:
x = tcos(t) = πcos(π) = -π
y = tsin(t) = πsin(π) = 0
So the point corresponding to t=π is (-π, 0).
Now, use the point-slope form of a line, y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line. Plug in the values:
y - 0 = (-1)(x - (-π))
y = -x - π
Therefore, the equation of the tangent line to the curve at the point corresponding to t=π is y = -x - π.