Welcome :: Homework Help and Answers :: Mathskey.com

Recent Visits

  
Welcome to Mathskey.com Question & Answers Community. Ask any math/science homework question and receive answers from other members of the community.

13,435 questions

17,804 answers

1,438 comments

771,167 users

I need help with a parametrics question.?

0 votes

Write down the cartesian equation of the tangent and normal to the parabola indicated. 

x= cos2t and y= cost at the point where t= pi/4 radians

asked Nov 12, 2014 in PRECALCULUS by anonymous

2 Answers

0 votes

Equation for tangent :

The functions are x= cos2t and y= cost .

x= cos 2t

Derivative with respect to t .

dx/dt = -sin 2t (2)                                 [ derivative of cos at = -a sin at ]

dx/dt = -2sin 2t 

y= cost

Derivative with respect to t .

dy/dt = -sin t 

Slope of the tangent dy/dx = (dy/dt)*(dt/dx)

dy/dx = (-sin t )*(1/-2sin 2t)

dy/dx = (-sin t / -2sin 2t)

dy/dx = (sin t / 2sin 2t)

Slope of the tangent at t = π/4 .

slope = (sin π/4 / 2sin 2(π/4))

slope = (sin π/4 / 2sin (π/2))

slope = [(1/√2) / 2(1)]

slope = [(1/√2) / 2]

slope = 1 / 2√2

Now find the point at t = π/4 .

x = cos 2(π/4 ) = cos π/2

x = 0

y = cos (π/4)

y = 1/√2

So the point is (0,1/√2)

The equation of tangent with slope 1 / 2√2  and point (0,1/√2) using slope point form .

y – y1 = m(x – x1)

y – 1/√2 = 1 / 2√2(x – 0)

y – 1/√2 = 1 / 2√2(x)

2√2[y – 1/√2 ]= (x)

2√2y – = x

x -2√2y – = 0

So the equation of tangent is x -2√2y – = 0 .

 

answered Nov 12, 2014 by yamin_math Mentor
0 votes

Equation for normal :

The functions are x= cos2t and y= cost .

x= cos 2t

Derivative with respect to t .

dx/dt = -sin 2t (2)                                 [ derivative of cos at = -a sin at ]

dx/dt = -2sin 2t 

y= cost

Derivative with respect to t .

dy/dt = -sin t 

Slope of the tangent dy/dx = (dy/dt)*(dt/dx)

dy/dx = (-sin t )*(1/-2sin 2t)

dy/dx = (-sin t / -2sin 2t)

dy/dx = (sin t / 2sin 2t)

Slope of the tangent at t = π/4 .

slope = (sin π/4 / 2sin 2(π/4))

slope = (sin π/4 / 2sin (π/2))

slope = [(1/√2) / 2(1)]

slope = [(1/√2) / 2]

slope = 1 / 2√2

Slope of tangent is 1 / 2√2 .

Slope of normal is -2√2                        [Sinec m1*m2 = -1]

Now find the point at t = π/4 .

x = cos 2(π/4 ) = cos π/2

x = 0

y = cos (π/4)

y = 1/√2

So the point is (0,1/√2)

The equation of normal with slope -2√2 and point (0,1/√2) using slope point form .

y – y1 = m(x – x1)

y – 1/√2 = -2√2(x – 0)

y – 1/√2 = -2√2(x)

(√2y – 1)/√2 = -2√2(x)

(√2y – 1)= √2*[-2√2(x)]

(√2y – 1)= -2*2(x)

(√2y – 1)= -4x

√2y – 1 = -4x

4x + √2y – 1 = 0

So the equation of normal is 4x + √2y – 1 = 0 .
answered Nov 12, 2014 by yamin_math Mentor

Related questions

asked May 1, 2014 in PRECALCULUS by anonymous
...