# Find the equation of the tangent line at the given t if x = 5cost, y =3sint at t=pi/4

justaguide | College Teacher | (Level 2) Distinguished Educator

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We have x = 5cos t, y = 3sin t.

x = 5cos t

dx/dt = -5sin t

y = 3sin t

dy/dt = 3cos t

Therefore dy/dx = (-3cos t / 5sin t)

At t = pi/4

dy/dx = -3/5

Also, x = 5cos t = 5/sqrt 2

y = 3sin t = 3/sqrt 2

Therefore the equation of the line is

y - 3/sqrt 2 = (-3/5)(x - 5/sqrt 2)

=> 5y - 15/sqrt 2 = -3x + 15/sqrt 2

=> 3x + 5y - 30/sqrt 2 = 0

Therefore the equation of the required line is 3x + 5y - 30/sqrt 2 = 0

neela | High School Teacher | (Level 3) Valedictorian

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To find the equation of the tangent line at the given t if x = 5cost, y =3sint at t=pi/4.

x = 5cost and y = 3sint.

To find the tangent at t = pi/4.

Therefore at t = pi/4, the coordintes :

x1 = 5cospi/4 = 5/qsrt2 and y1 = 5sinpi/4 = 5/sqrt2.

dy/dx = (sint)' = cost. dx/dt = (cost)' = -sint.

So the slope of the tangent at at x= pi, is dy/dx = (dy/dt)/(dx/dt) = {(cost/sint) at t = pi/4} = (1/sqrt2)/(-1/sqrt2) = -1.

Therefore the tangent at t = pi is given by:

y-y1 = {slope at t = pi/4}(x-x1).

y-5/sqrt2 = (-1)(x-5/sqrt2).

x+y =10/sqrt2 = 10sqrt2/2 = 5sqrt2.

x+y = 5sqrt2 is the equation of the tangent at to t =pi/4.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

To find the equation of the tangent line, first we'll ahve to find the derivative:

dy/dx = (dy/dt)/(dx/dt)

dx/dt = (d/dt)(5cost)

dx/dt = -5 sin t

dy/dt = (d/dt)(3sint)

dy/dt = 3 cos t

dy/dx = 3 cos t/-5 sin t

dy/dx = 3 cos(pi/4)/-5 sin (pi/4)

dy/dx = 3/-5

At t = pi/4, we'll get the point (x(t),y(t))= (5cos(pi/4),3sin(pi/4))

(x(t),y(t))= (5sqrt2/2,3sqrt2/2)

The equation of the tangent line is:

y - 3sqrt2/2 = m(x - 5sqrt2/2)

y - 3sqrt2/2 = (-3/5)(x - 5sqrt2/2)

y = (-3/5)(x - 5sqrt2/2) + 3sqrt2/2