Lecture for Week 6 Secs 3 6 9 Derivative Miscellany I Implicit differentiation We want to answer questions like this 1 What is the derivative of tan 1 x dy if 2 What is dx x3 y 3 xy 2 x2 y 25x 25y 0 x3 y 3 xy 2 x2 y 25x 25y 0 Here we don t know how to solve for y as a function of x but we expect that the formula defines a function implicitly if we consider a small enough window on the graph to pass the vertical line test y x Temporarily assuming this is so we differentiate the equation with respect to x remembering that y is a function of x d 3 0 x y 3 xy 2 x2 y 25x 25y dx 3x2 3y 2 y y 2 2xyy 2xy x2 y 25 25y 3x2 y 2 2xy 25 y 3y 2 2xy x2 25 25 3x2 y 2 2xy y 2 3y 2xy x2 25 To use this formula you need to know a point x y on the curve You can check that 3 4 does satisfy x3 y 3 xy 2 x2 y 25x 25y 0 Plug those numbers into 25 3x2 y 2 2xy y 2 3y 2xy x2 25 to get y 3 4 But x y 3 3 also satisfies the equation and it gives y 1 And 3 4 satisfies the equation and gives y 43 Three different functions are defined near x 3 by our equation and each has a different slope 4 3 3 4 The curve in this problem is the union of a circle and a line 0 x3 y 3 xy 2 x2 y 25x 25y x2 y 2 25 x y 4 3 3 4 We can clearly see the three points of intersection with the line x 3 Two other interesting points are 1 x 5 y 0 vertical tangent The denominator of the formula for y equals 0 but the numerator does not 2 x 5 2 y 5 2 intersection Both numerator and denominator vanish because the slope is finite but not unique An inportant application of implicit differentiation is to find formulas for derivatives of inverse functions such as u tan 1 v This equation just means v tan u together with the branch condition that 2 u 2 without which u would not be uniquely defined So d du d dv tan u tan u 1 dv dv du dv But d tan u sec2 u 1 tan2 u 1 v 2 du Putting those two equations together we get d du 1 tan v dv dv d tan u du 1 1 1 v2 Generally speaking the derivative of an inverse trig function is an algebraic function We will see more of this in Sec 4 2 Exercise 3 6 39 Show that the curve families y cx2 x2 2y 2 k are orthogonal trajectories of each other That means that every curve in one family each curve labeled by c is orthogonal to every curve in the other family labeled by k dy 2cx For the curves y cx we have dx No implicit differentiation was needed in this case For the curves x2 2y 2 k we have dy x dy 0 2x 4y dx dx 2y If the curves are orthogonal the product of the slopes must be 1 and vice versa Well the product is x cx2 2cx 1 2y y 2 Derivatives of vector functions No big surprise here Conceptually we are subtracting the arrows for two nearby values of the parameter then dividing by the parameter difference and taking the limit r r a r t r r t r a lim r a lim t a h 0 h t a And calculationally since our basis vectors do not depend on t we just differentiate each component i d h2 t i 3t j 5 k 2t i 3 j dt Second and higher derivatives This is fairly obvious too The second derivative is the derivative of the first derivative s t At2 Bt C s t 2At B s t 2A This was essentially Exercise 3 8 37 The most important application of second derivatives is acceleration the derivative of velocity which is the derivative of position Exercise 3 8 49 A satellite completes one orbit of Earth at an altitude 1000 km every 1 h 46 min Find the velocity speed and acceleration at each time Earth radius 6600 km 46 The period is 1 60 1 767 hr Therefore the angular speed is 2 1 767 3 557 radians per hour The radius of the circle is 7600 so the speed in the orbit is 7600 3 557 27030 km h at all times To represent the velocity we must choose a coordinate system say that the satellite crosses the x axis when t 0 and moves counterclockwise so it crosses the y axis after a quarter period Then v t 27030h sin 3 557t cos 3 557t i When t 0 v is in the positive y direction after a quarter period it is in the negative x direction The acceleration is the derivative of that a t 27030 3 557h cos 3 557t sin 3 557t i Finally let s find the position function Its derivative must be v so a good first guess is r t 27030 hcos 3 557t sin 3 557t i 3 557 To this we could add any constant vector but a quick check shows that r 0 is in the positive x direction as we wanted and this orbit is centered at the origin as it should be So this is the right answer Notice that a points in the direction opposite to r i e toward the center of the orbit as always for uniform circular motion Slopes and tangents of parametric curves What is the slope of a curve defined by parametric equations x x t y y t If we had y as a function of x we would just caldy But we can find the slope without culate dx eliminating t from the equations It may come as no surprise that the answer is obtained by dividing numerator and denominator by dt dy dx dy dt dx dt The valid proof of this formula is simply an application of the chain rule dy dx dy dt dx dt But you should be shouting what if the dedy 0 and nominator is 0 If dx dt dt 6 0 the curve is vertical at that point so the slope is properly undefined If both derivatives are 0 we need to consider another parametrization to get an answer the moving point has slowed to a standstill at the time of interest so the parametric derivatives give no information To apply the formula you may need to do some work to determine the correct value of t to plug in Exercise 3 9 19 At what point does the curve x t t2 3 y 3 t2 3 cross itself Find equations of both tangents at that point If the curve crosses …
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