Find a Cartesian equation relating and corresponding to the parametric equations: x=4sin(2t), y=3cos(2t).
Write your answer in the form P(x,y)=0, where P(x,y) is a polynomial in x and y, such that the coefficient of y^2 is 16.
b)Find the equation of the tangent line to the curve at the point corresponding to t=pi/6
well, we have
sin(2t) = x/4
cos(2t) = y/3
so
(x/4)^2 + (y/3)^2 = 1
x^2/16 + y^2/9 = 1
9x^2 + 16y^2 = 144
dy/dx = (dy/dt)/(dx/dt) = -6sin2t / 8cos2t = -3/4 tan2t
so at t = π/6, we have dy/dx = -3/4 * √3 at the point (2√3,3/2)
so the tangent line is
y - 3/2 = -3√3/4 (x - 2√3)
see the graphs at (eliding the extra spaces)
www.w olframalpha.c om/input?i=plot+9x%5E2+%2B+16y%5E2+%3D+144%2C+y+-+3%2F2+%3D+-3%E2%88%9A3%2F4+%28x+-+2%E2%88%9A3%29
To find the Cartesian equation corresponding to the given parametric equations, we can eliminate the parameter t by using the substitution method.
Given:
x = 4sin(2t)
y = 3cos(2t)
Step 1: Square both equations to eliminate the trigonometric functions:
x^2 = (4sin(2t))^2
= 16sin^2(2t)
y^2 = (3cos(2t))^2
= 9cos^2(2t)
Step 2: Use the Pythagorean identity to replace sin^2(2t) in terms of cos^2(2t):
1 - cos^2(2t) = sin^2(2t)
Step 3: Substitute the expressions from Step 2 into the equation obtained in Step 1:
x^2 = 16(1 - cos^2(2t))
Simplifying the equation:
x^2 = 16 - 16cos^2(2t)
Step 4: Rearrange the equation to match the required form P(x, y) = 0:
16cos^2(2t) + x^2 - 16 = 0
Since the coefficient of y^2 needs to be 16, we can multiply the equation by 16:
256cos^2(2t) + 16x^2 - 256 = 0
Thus, the Cartesian equation relating to the given parametric equations is:
256cos^2(2t) + 16x^2 - 256 = 0
b) To find the equation of the tangent line at t = pi/6, we need to find the corresponding point on the curve.
Given:
t = pi/6
Substitute the value of t into the parametric equations:
x = 4sin(2(pi/6))
= 4sin(pi/3)
= 4(√3/2)
= 2√3
y = 3cos(2(pi/6))
= 3cos(pi/3)
= 3(1/2)
= 3/2
The point corresponding to t = pi/6 is (2√3, 3/2).
Now, let's find the derivative of the parametric equations:
dx/dt = 4cos(2t)
dy/dt = -6sin(2t)
Substituting t = pi/6:
dx/dt = 4cos(2(pi/6))
= 4cos(pi/3)
= 4(1/2)
= 2
dy/dt = -6sin(2(pi/6))
= -6sin(pi/3)
= -6(√3/2)
= -3√3
The derivative of y with respect to x is dy/dx, which is given by:
dy/dx = (dy/dt)/(dx/dt)
= (-3√3)/2
Now we can use the point-slope form of a line to find the equation of the tangent line:
y - y1 = m(x - x1)
Substituting the values into the equation:
y - (3/2) = (-3√3/2)(x - 2√3)
Simplifying:
2y - 3 = -3√3x + 6
Rearranging the equation:
-3√3x + 2y = 9
Therefore, the equation of the tangent line to the curve at the point corresponding to t = pi/6 is:
-3√3x + 2y = 9