Find T, N, and k for the plane curve r(t) = ln(sect)i + tj, (-π)/(2) < t < (π)/(2) T(t) = ( )i +

Step 1: Find the unit tangent vector T(t). T(t) = r'(t)/||r'(t)|| r'(t) = (tan t)i + j ||r'(t)||

Find T, N, and k for the plane curve r(t) = ln(sect)i + tj,...

Find T, N, and k for the plane curve r(t) = \ln(\sec t)i + tj, $\frac{-\pi}{2} < t < \frac{\pi}{2}$ T(t) = ( )i + ( )j (Type exact answers, using radicals as needed.) N(t) = ( )i - ( )j (Type exact answers, using radicals as needed.) \kappa(t) = (Type an exact answer, using radicals as needed.)

Transcript

00:01 We have this curve that is given by r of t is equal to natural algorithm of secant of t along the i duration plus t along the j direction and this is our position so and we have the t is between is between pi halves and minus pi halves and smaller than by halves.

00:48 So why do we do that condition? because secant is one of a cosign and well we want that to be we're putting a lot natural logarithm so secant has to be positive and all cosine of t as you can see on the graph it is it is positive between minus by halves and by halves so that this is positive also one or cosine t will be positive see can to t is positive within that range so we're good so now we have that what is so the velocity of the ove r prime t that is positive that is so it would be 1 over secant of t times the derivative of secant of t and that along the i direction plus the derivative of t that is 1 with respect to t so it times 1 of the derivative of secant is second times tangent so that this velocity will turn into second times second so the second tens out to be tangent of t i plus j so this is a velocity now the norm of that is going to be equal to the component to squared and square rid of that so you have tangent squared plus one but there is a very nice trigger identity that also that that is secant of t squared and the second is positive within that range or t line there second is positive so that this will be equal to second of t so uh so that if you will be equal to second of t so that if we want to compute what is what is t the tangent unit vector will be the velocity divided by its norm so it's going to be equal to tangent of t along the i plus j divided by secant of t that is the norm but s divin by second by second of t is the same as multiplying by cosine of t so that this would be equal to tangent of t times cosine of t i plus cosine of t j that this tangent is sine over cosine so that we multiply with cosine we cancel the cosines and this is written very nicely so it will be sign of t i plus cosine of t j so that is t now if we want to if you want to compute what is kappa kappa will be equal to 1 over b d t d t so uh this one over b that is 1 over b that is 1 over this sequence of 1 over 2x5 1 over 2nd of t which is equal to cosine of t and then this that one has norm will differentiate that we would get what we'll first before doing the norm so the derivative of t with respect to t with respect to this derivative of sign is cosine of t i minus sign of t i minus sign of t j so that is d t and this one has known the components squared added together and a square root of that which happens to be one so that this would be equal to cosine of t times one cosine of t that is kappa and so we have the tangent normal kappa and the tangent vector and they are to complete the normal so n this one is can be computed as d t b t b t divided by it's norm...

Question

Find T, N, and k for the plane curve r(t) = \ln(\sec t)i + tj, $\frac{-\pi}{2} < t < \frac{\pi}{2}$ T(t) = ( )i + ( )j (Type exact answers, using radicals as needed.) N(t) = ( )i - ( )j (Type exact answers, using radicals as needed.) \kappa(t) = (Type an exact answer, using radicals as needed.)

Please give Ace some feedback