sum10.2.14

sum10.2.14 - on the homework. Problem 10.2.14: Find the...

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Week 5 Homework: Pre-view 10.2 Calculus on parametric curves 10.3 Polar coordinates: a. translations ( x, y ) to ( r, θ ) b. translations ( r, θ ) to ( x, y ) c. standard curves ( r = a cos θ ; r = a + b cos θ ; r = cos nθ. ) 10.4 Area and arc length in polar coordinates 11.1, 11.2 Sequences and Series
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2 If x = x ( t ) , y = y ( t ) is a parametric curve, we have dy dx = dy dt dx dt . This formula is used in Problem 10.2.5 to find the equation of the tangent line to the curve at a point. To clarify, we still have x = x ( t ) , but a new parametric value h ( t ) = dy dx ( t ) . We read the above formula for the first derivative as x = x ( t ) , y = g ( t ) having d dx ( g ( t )) = d dt ( g ( t )) dx dt , and apply this to the function h, and get d dx ( dy dx ) = d dt ( dy dx ) dx dt as the formula for the 2nd derivative. We use this
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Unformatted text preview: on the homework. Problem 10.2.14: Find the points on the parametric curve x ( t ) = t + ln t, y ( t ) = t-ln t where the curve is concave up ( y > 0). Solution: dy dx = 1-1 t 1 + 1 t = t-1 t + 1 . So d dx dy dx = d dt t-1 t + 1 d dt ( t + ln t ) , which simplies to 2 t ( t + 1) 3 . Now we check signs of the numerator (changes from minus to plus at t = 0) and the denominator (changes from minus to plus at t =-1) to see that the curve is concave up at the points where t <-1 and where t > ....
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sum10.2.14 - on the homework. Problem 10.2.14: Find the...

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