find the unit tangent to the curve r(t) = t cos(t) i - tsin(t) j at time t = pi/2

thanks for any help.

Printable View

- Feb 27th 2007, 09:28 PMrcmangofind the unit tangent to curve
find the unit tangent to the curve r(t) = t cos(t) i - tsin(t) j at time t = pi/2

thanks for any help. - Feb 27th 2007, 10:12 PMearboth
Hi,

first I've transformed your equation into:

x(t) = t*cos(t)

y(t) = -t*sin(t)

Then the point T where the tangent touches the curve has the coordinates:

T(0, -(π/2))

To calculate the tangent you need the gradient of the curve in T. I use the formula:

dy/dx = (dy/dt) / (dx/dt). Use product rule:

dy/dx = (sin(t)*(-1) + (-t)*cos(t)) / (cos(t) + t*(-sin(t)))

Now plug in t = π/2

therefore the slope of the tangent is 2/π.

Use the point-slope-formula to get the equation of the straight line:

t: y = (2/π)*x-(π/2)

I've attached an image of your curve and the tangent.

EB