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Expert: - 9/9/2008


Question
Consider the curve r(t)=(sin(t)cos(t))i + (sin(t)^2)j + cos(t)k

a)Show that the curve lies on a sphere centered at the origin.
b)Where does the tangent line at t=pi/6 intersect with the xy-plane?

Thanks in advance for the help. I'm totally lost.

Answer
a) To lie on a plane, any point must follow the equationx²+y²+z²=K.
In this case, K=1.

Square the coefficient of i, j, and k, then add them up to get x²+y²+z².  The variable x is the factor for i, the variable y is the factor for j, and the variable z is the factor for k.

When we square the terms, we get sin²(t)cos²(t) + sin^4(t) + cos²(t).
Factoring sin²(t) out of the first two terms gives
sin²(t)(cos²(t)+sin²(t)) + cos²(t).

It is known that sin²(t)+cos²(t) is 1, and that reduces the first term to sin²(t) and then solves the equation.

b) Since we are only interested in the xy plane, the curve in this plane is r(t)=sin(t)cos(t)i + sin²(t)j.

To find a tangent line, we need the slope.  To get the slope, we need the derivative.  
The derivative is [cos²(t)-sin²(t)]i + [2sin(t)cos(t)]j.
Putting t=π/6 in, we get cos(π/6)=1/2 and sin(π/6) = √3/2.
This means our derivative is (1/4-3/4)i + (2√3/4)j =(-1/2)i +(√3/2)j.

The slope is given by dy/dt, or, in this case, the j coefficient on the derivative over the i coefficient on the derivative.

You get (√3/2)/(-1/2) = -√3.

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