(a)
x=2sin(3t)....(1)
y=9cos(3t)....(2)
square each equation
x²=4sin²(3t)...(1a)
y²=81cos²(3t)...(2a)
divide by coefficient of RHS and add:
(x²/4+y²/81)=sin²(3t)+cos²(3t)=1
(x²/4+y²/81)-1=0
Multiply by 4*81 to have common denominator of 1
P(x,y)=81x²+4y²-324=0
b.
To find tangent, calculate dy/dx=y'(x,y) using implicit differentiation.
dy/dx at t0=pi/9:
evaluate m=y'(2sin(3t0), 9cos(3t0))
Equation of tangent line:
L : (y-9cos(3t0))=m(x-2sin(3t0))
a)
Find a Cartesian equation relating and corresponding to the parametric equations: x=2sin(3t), y=9cos(3t).
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 4.
b)Find the equation of the tangent line to the curve at the point corresponding to t=pi/9
2 answers
x=2sin(3t), y=9cos(3t)
Start by squaring each equation to achieve the "y^2" and polynomial
x^2=4sin^2(3t)
y^2=81cos^2(3t)
Move the coefficients to the left and add these polynomials:
(x^2/4+y^2/81)=sin^2(3t)+cos^2(3t)
From our trig formulas, we know that sin^2(3t)=cos^2(3t) = 1
SO, (x^2/4+y^2/81)-1=0
Multiply the two squares by their denominators to achieve a common denominator of 1 and the coefficient of "4"
Our final answer is: P(x,y)=81x^2+4y^2-324=0
Start by squaring each equation to achieve the "y^2" and polynomial
x^2=4sin^2(3t)
y^2=81cos^2(3t)
Move the coefficients to the left and add these polynomials:
(x^2/4+y^2/81)=sin^2(3t)+cos^2(3t)
From our trig formulas, we know that sin^2(3t)=cos^2(3t) = 1
SO, (x^2/4+y^2/81)-1=0
Multiply the two squares by their denominators to achieve a common denominator of 1 and the coefficient of "4"
Our final answer is: P(x,y)=81x^2+4y^2-324=0
1 answer