The parametric curve
x(t)=3cos(t)
y(t)=7sin(2t)
intersects itself at a point P=(x0,y0) where x0= y0=
The slopes of the 2 tangents at this point are?
THANK YOU!
To find the point of intersection on the curve, we need to find the values of t at which the x and y coordinates are equal. So, setting x(t) = y(t), we have:
3cos(t) = 7sin(2t)
Now we need to solve this equation for t. To do this, we can use trigonometric identities and simplify the equation:
3cos(t) = 7(2sin(t)cos(t))
Dividing both sides by cos(t), we get:
3 = 14sin(t)
Now solving this equation for sin(t), we have:
sin(t) = 3/14
We can take the inverse sine (sin^(-1)) of both sides to find t:
t = sin^(-1)(3/14)
Now that we have found the value of t, we can substitute it back into the parametric equations to find the x and y coordinates at the point of intersection:
x0 = 3cos(t) = 3cos(sin^(-1)(3/14)) ≈ 2.91
y0 = 7sin(2t) = 7sin(2sin^(-1)(3/14)) ≈ 1.21
So, the point of intersection on the curve is P = (x0, y0) ≈ (2.91, 1.21).
To find the slopes of the two tangents at this point, we need to find the derivative of the parametric equations with respect to t. Taking the derivatives gives us:
dx/dt = -3sin(t)
dy/dt = 14cos(2t)
Now, we can find the slopes of the tangents by substituting the value of t at the point of intersection:
Slope of the first tangent = dy/dt / dx/dt = (14cos(2t)) / (-3sin(t)) = (-14cos(2t)) / (3sin(t))
Slope of the second tangent = dy/dt / dx/dt = (14cos(2t)) / (-3sin(t)) = (-14cos(2t)) / (3sin(t))
So, the slopes of the two tangents at point P are (-14cos(2t)) / (3sin(t)) and (-14cos(2t)) / (3sin(t)) respectively.