d y t x t t y t t dx 1 Find the first derivative dydx dy dy dt y dx dx dt 2 1 3

D y t x t t y t t dx 1 find the first derivative dydx

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d ytxttyttdx=-=-1.Find the first derivative (dy/dx).dydydtydxdxdt==21312tt-=-
2. Find the derivative of dy/dxwith respect to t.21312dydtdtdtt-=-(292226612ttt-+=-Quotient Rule
3. Divide by dx/dt.22d ydxdxdtdydt=(29222661212tttt-+-=-(292326612ttt-+=-
Ldxdtdydtdtab=FHGIKJ+FHGIKJz22The equation for the length of a parametrized curve is similar to our previous “length of curve” equation:(Notice the use of the Pythagorean Theorem.)
Likewise, the equations for the surface area of a parametrized curve are similar to our previous “surface area” equations:(29Revolution about the -axis 0xyf222badxdySydtdtdtπ=+(29Revolution about the -axis 0yxf222badxdySxdtdtdtπ=+(29Revolution about the -axis 0xyf222badxdySydtdtdtπ=+
This curve is:πx(t) = t + 2 sin (2t)y(t) = t + 2 cos (5t)

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