Assignment 5 Math 101 Spring 2009

Transcription

1 Assignment 5 Math 11 Spring 9 1. Find an equation of the tangent line(s) to the given curve at the given point. (a) x 6 sin t, y t + t, (, ). (b) x cos t + cos t, y sin t + sin t, ( 1, 1). Solution. (a) The point (,) corresponds to t, so the slope of the tangent line at that point is /dt t /dt t + 1 t 6 cos t 1 t 6. Therefore, an equation of the required tangent line is y 1 6 x. (b) We need to find the value(s) of t corresponding to the point ( 1, 1). x 1 cos t + cos t 1 cos t + cos t cos t(1 + cos t) cos t or cos t 1. Since the interval [, π] gives the complete curve, we only need to find values in this interval. This way we get t π or t 3π or t π or t 4π. Substituting those values 3 3 into the formula y sin t + sin t, we see that only t π gives y 1. thus the point ( 1, 1) corresponds to t π. Now let s find the slope of the tangent line at that point: /dt cos t + cos t /dt sin t sin t 1. t π t π t π Therefore, an equation of the required tangent line is y 1 (x + 1), or y x Find the points of the curve where the tangent is horizontal or vertical. (a) x t 3 + 3t 1t, y t 3 + 3t + 1. (b) x cos 3θ, y sin θ. Solution. (a) By direct computations, dt 6t + 6t 6t(t + 1), To find the points where the tangent line is horizontal, dt 6t + 6t 1 6(t + )(t 1). 6t(t + 1) t or t 1.

2 The corresponding points are (x, y) (, 1) or (13, ). Similarly, to find the points where the tangent line is vertical, 6(t + )(t 1) t or t 1. The corresponding points are (x, y) (, 3) or ( 7, 6). To summarize, the curve has horizontal tangents at (, 1) and (13, ), and vertical tangents at (, 3) and ( 7, 6). (b) In this case, direct computation of the derivatives gives dθ cos θ, dθ 3 sin 3θ. Thus dθ θ π + nπ (n an integer) (x, y) (, ±). Similarly, dθ 3θ nπ (n an integer) θ nπ 3 (x, y) (±1, ) or (±1, ± 3). The curve has horizontal tangents at (, ±) and vertical tangents at (±1, ), (±1, 3) and (±1, 3). 3. Find equations of the tangent lines to the curve x 3t +1, y t 3 +1 that pass through the point (4, 3). Solution. By direct computations, dt 6t, dt 6t, so 6t 6t t. This means that at a point of the curve corresponding to the value t, an equation of the tangent line is y (t 3 + 1) t ( x (3t + 1) ). That the line passes thought the point (4, 3) means that the coordinates of this point must satisfy the equation of the tangent line. This way, 3 (t 3 + 1) t ( 4 (3t + 1) t 3 3t + (t 1) (t + ) t 1 or t. Notice that t 1 gives the point of the curve (4, 3), and t gives the point of the curve (13, 15). Hence, the desired equations are y 3 1(x 4) and y + 15 (x 13).

3 4. Find the area of the region that lies inside the curve r 3 cos θ and outside the curve r cos θ. Solution. First thing you should do is sketch a graph of these curves! Now let s start by fiding the points of intersection of the two curves: 3 cos θ cos θ cos θ 1 θ ±π 3 (Recall that r 3 cos θ is a circle, so we only need to look at values of θ in the interval [, π]). Now let s find the required area: A π/3 1[ (3 cos θ) ( cos θ) ] dθ 1[ (3 cos θ) ( cos θ) ] dθ (symmetry about polar axis) [ 8 cos θ + 4 cos θ 4 ] dθ [ 4 cos θ + 4 cos θ ] dθ [ sin θ + 4 sin θ] π/ Find an equation of the plane passing through the point (, 8, 1) and perpendicular to the line x 1 + t, y t, z 4 3t. Solution. Since the line is perpendicular to the plane, its direction vector 1,, 3 is a normal vector to the plane. Then an equation of the plane is ( ) ( ) ( ) x ( ) + y 8 3 z 1, or x + y 3z Find an equation of the plane passing through the point ( 1, 6, 5) and parallel to the plane x + y + z +. Solution. What does it mean for two planes to be parallel? It means that they have the same normal vectors! A normal vector the the given plane is n 1, 1, 1, so an equation of the plane having this normal vector and passing through the point ( 1, 6, 5) is 1 ( x ( 1) ) + 1 ( y 6 ) + 1 ( z ( 5) ), or x + y + z. 7. Find parametric equations for the tangent line to the curve x e t, y te t, z te t at the point (1,, ). Solution. The vector equation (position vector) for the curve is r(t) e t, te t, te t, so the tangent vector (velocity vector) is r (t) e t, e t + te t, e t + t e t. The point (1,, )

4 thus corresponds to the value t, so the tangent vector there is r () 1, 1, 1. Thus the tangent line passes through the point (1,, ) and has direction vector 1, 1, 1, so its parametric equations are x t, x 1 + t, y + 1 t, or y t, z + 1 t. z t. 8. Find the point of intersection of the tangents to the curve r(t) sin πt, sin πt, cos πt at the points where t and t.5. Solution. This curve has tangent vector r (t) π cos πt, π cos πt, π sin πt. For t, we get the point with position vector r(), 1, 1 and having direction vector r () π, π,. So an equation of this line is x, y, z, 1, 1 + u π, π,, u R. For t.5 1, we get the point with position vector r(1/) 1,, and having direction vector r (1/),, π. So an equation of this line is These lines intersect at the point where x, y, z 1,, + v,, π, v R., 1, 1 + u π, π, 1,, + v,, π u 1 π & v 1 π. Thus the point of intersection is (1,, 1). 9. Find the exact length of the curve. (a) x e t + e t, y 5 t, t 3. (b) x 3 cos t cos 3t, y 3 sin t sin 3t, t π. (c) r e θ, θ π. Solution. (a) In this case the position vector is r(t) e t +e t, 5 t, so the velocity vector is r (t) e t e t,, and therefore the length of the curve is L r (t) dt (et e t ) + ( ) dt e t + e t + 4 dt e t + + e t dt (et + e t ) dt ( e t + e t) dt ( e t e t) 3 e3 e 3.

5 (b) In this case the position vector is r(t) 3 cos t cos 3t, 3 sin t sin 3t, so the velocity vector is r (t) 3 sin t + 3 sin 3t, 3 cos t 3 cos 3t. To make the computations a bit organized let s find first the speed function: r (t) ( 3 sin t + 3 sin 3t) + (3 cos t 3 cos 3t) 9 sin t 18 sin t sin 3t + 9 sin (3t) + 9 cos t 18 cos t cos 3t + 9 cos (3t) 9 ( sin t + cos t ) + 9 ( sin (3t) + cos (3t) ) 18 ( cos t cos 3t + sin t sin 3t ) cos(t 3t) 18 ( 1 cos( t) ) 18 ( 1 cos(t) ) 18 ( sin t ) Therefore the length of the curve is L π r (t) dt 36 sin t 6 sin t. π 6 sin t dt 6 π sin t dt 6 cos t π 1. (c) In this case since the curve r e θ is a polar curve, π ( ) dr π π L r + dθ (e dθ θ ) + (e θ ) dθ e 4θ + 4e 4θ dθ π 5e 4θ dθ π 5 e θ dθ 5 eθ π 5 (e4π 1). Alternatively, we may have used the position vector r(θ) e θ cos θ, e θ sin θ, so the correspoiding velocity vector is r (θ) e θ cos θ e θ sin θ, e θ sin θ + e θ cos θ, and then apply the usual formula for arclength: L Of course the result would be the same. π r (θ) dθ. 1. Find the area of the surface generated by rotating the given curve about the y-axis. (a) x 3t, y t 3, t 5. (b) x e t t, y 4e t/, t 1. Solution. (a) For this curve the position vector is r(t) 3t, t 3, so the velocity vector is r (t) 6t, 6t. Since we are rotating about the y-axis, to compute the surface area we use the formula S πx(t) r (t) dt.

6 This way you have S π3t (6t) + (6t ) dt π3t 36t (1 + t ) dt Using your favorite integration technique (really do this!!!) you find that t t dt 5 (1 + t ) 5/ 3 (1 + t ) 3/ + c Therefore ] 5 [ S 18π 5 (1 + t ) 5/ 3 (1 + t ) 3/ [ 18π 5 (6)5/ 3 (6)3/ 5 + ] 3 4π 5 ( ). π18t t dt (b) For this curve the position vector is r(t) e t t, 4e t/, so the velocity vector is r (t) e t 1, e t/. Keeping in mind that we are rotating about the y-axis, S πx(t) r (t) dt π π(e t t) (e t 1) + (e t/ ) dt π(e t t) e t e t e t dt π(e t t) e t + e t + 1 dt π(e t t) (e t + 1) dt π(e t t)(e t + 1) dt [ 1 et + e t (t 1)e t 1 ] 1 t π(e + e 6).