MATH 4 (R) Winter 8 Intermediate Calculus I Solutions to Problem Set #5 Completion Date: Frida Februar 5, 8 Department of Mathematical and Statistical Sciences Universit of Alberta Question. [Sec.., # ] Given the parametric equations = t, = t 3 (a) Sketch the curve b using the parametric equations to plot points. Indicate with an arrow the direction in which the curve is traced as t increases. (b) Eliminate the parameter to find a Cartesian equation of the curve. (a) Plotting points for integer values of t with 3 t 3, we have and the curve is plotted below. t 3 3 9 4 4 9 7 8 8 7 (4,8) t= t= t= (4, 8) t= (b) Solving for the parameter t in the first equation, we have t = ±, so that = ± 3/,, R. The Cartesian equation of the curve is given b = /3,, R.
Question. [Sec.., # ] Given the parametric equations = 4 cos θ, = 5 sin θ, π/ θ π/ (a) Eliminate the parameter to find a Cartesian equation of the curve. (b) Sketch the curve and indicate with an arrow the direction in which the curve is traced as the parameter increases. (a) Eliminating the parameter, since cos θ + sin θ =, we have that is, (/4) + (/5) =, 6 + 5 = which is the equation of an ellipse with the -intercepts = ±4, the -intercepts = ±5. However, since π/ θ π/, we have cos θ so the graph consists of onl the portion on the right side of the -ais. (b) The curve is plotted below. 5 θ= π 4 θ= 5 θ= π Question 3. [Sec.., # 6] Given the parametric equations = ln t, = t, t (a) Eliminate the parameter to find a Cartesian equation of the curve. (b) Sketch the curve and indicate with an arrow the direction in which the curve is traced as the parameter increases. (a) Eliminating the parameter, we have = ln t, which implies that t = e, so that = e = e /,. (b) The curve is plotted below.
Question 4. [Sec.., # ] Describe the position of a particle with position (, ) where as t varies in the given interval. = cos t, = cos t, t 4π The curve = is a parabola opening to the right with the verte (, ). The particle starts at the point (, ) (t = ). It then moves along the parabola to the point (, ) (t = π/) down to (, ) (t = π), then back to (, ) (t = 3π/) and to (, ) (t = π). The same motion is repeated from π to 4π. (,) (, ) Question 5. [Sec.., # 8] Find an equation of the tangent to the curve = tan θ, = sec θ at the point (, ) b methods: (a) without eliminating the parameter and (b) b first eliminating the parameter. (a) Differentiating, we have d d = d/dθ sec θ tan θ = d/dθ sec = tan θ = sin θ, θ sec θ and at (, ), = tan θ, so that = sec θ, so that θ = π or θ = π/4 + nπ, (since tan θ > in 4 quadrant I and III while sec θ > in quad. I and IV). Hence the slope of the tangent at (, ) is ( π ) π = sin 4 4 = = and the equation of the tangent line is = ( ). (b) Since tan θ + = sec θ implies + =, differentiating implicitl we obtain = which implies that which in turn implies that d d =, (, ) = and this gives the same equation as in part (a).
Question 6. [Sec.., # 6] Find d d and d if = cos t, = cos t, < t < π. For which values of d t is the curve concave upward? d d = d/ d/ = d d = sin t sin t = ( ) d d d sec t tan t = 4 = d/ sin t 6 = 6 cos 3 t sin t 4 sin t cos t = 4 cos t = sec t, 4 sin t cos t sin t cos t or d d = 6 sec3 t. The curve is concave upward if () > on < t < π, that is, if 6 sec3 t >, or equivalentl or so that sec t <, cos t <, t (π/, π). Question 7. [Sec.., # 8] Find the points on the curve where the tangent is horizontal or vertical. Differentiating, we have = t 3 + 3t t, = t 3 + 3t + d d = 6t + 6t 6t + 6t = 6t(t + ) 6(t + )(t ) = t(t + ) (t + )(t ). Horizontal tangents occur when =, that is, when t =,. Therefore the points where the tangent line if horizontal are t = : =, =, that is, (, ), t = : = + 3 + = 3, = + 3 + =, i.e., (3, ). Vertical tangents occur at t =, ( does not eist there), and the points where the tangent line is vertical are t = : = 6 + + 4 =, = 6 + + = 3, that is, (, 3), t = : = + 3 = 7, = + 3 + = 6, that is, ( 7, 6).
Question 8. [Sec.., # 34] Find the area of the region enclosed b the astroid = a cos 3 θ, = a sin 3 θ. The graph of the astroid is a a a a Since /3 + /3 = a /3, using smmetr we have A = 4 a d = 4 a ( a /3 /3) 3/ d π/ π/ = 4a 3 cos t sin 4 t = a 4 ( sin t cos t) sin t = 3 π/ a sin t( cos t) [ = 3 π/ a sin t = 3 a π/ = 3a π/ = 3πa 8. π/ sin t 3 a sin3 t 6 ( ) cos 4t sin t cos t π/ ] Question 9. [Sec.., # 44] Find the length of the curve = e t + e t, = 5 t, t 3. Differentiating, d/ = e t e t and d/ =, so that (d ) (d ) + = (e t e t ) + 4 = e t + e t + 4 = e t + + e t = (e t + e t ) L = 3 ( d ) + (d ) 3 3 = (e t + e t ) = e t e t = e 3 e 3 + = e 3 e 3.
Question. [Sec.., # 6] Find the area of the surface obtained b rotating the curve about the -ais. = 3t t 3, = 3t, t We have S = π b a ds = π 3t ( d ) (d ) + where (d (d ) = (3 3t ) = 9 8t + 9t 4 (d ) = (6t) = 36t ) (d ) + = 9 8t + 9t 4 + 36t = 9 + 8t + 9t 4 = (3 + 3t ) S = π = 8π 3t (3 + 3t ) = π ( t (t + t 4 3 ) = 8π 3 + t5 5 9t ( + t ) = 8π ( 3 + ) 48π = 5 5. Question. [Sec.., # 66] Find the surface area generated b rotating the curve about the -ais. We have S = π = e t t, = 4e t/, t ds where ds = (d/) + (d/), and (d ) = (e t ) = e t e t +, (d ) = (e t/ ) = 4e t (d ) (d ) + = e t e t + + 4e t = e t + e t + = (e t + ) S = π = π (e t t)(e t + ) (e t + e t te t t) ( = π et + e t (te t e t ) t (Here to do te t, integration b parts was used.) = π ( e + e e + e ) = π(e + e 6).