1. By the Product Rule, in conjunction with the Chain Rule, we compute the derivative as follows: and. So the slopes of the tangent lines to the curve

Transcription

1 MAT 11 Solutions TH Eam 3 1. By the Product Rule, in conjunction with the Chain Rule, we compute the derivative as follows: Therefore, d 5 5 d d 5 5 d and ( ) 3 d ( ) So the slopes of the tangent lines to the curve y 5 at the points 1, and, and -3. This implies that the equations of these two tangent lines are respectively given by 3 1 y and y 3 8. Setting the equations of these two tangent lines equal to each other yields: y 3 8 y , 9 We therefore conclude that the two tangent lines intersect at the point 17 7, 1.89, (See attached graph.)

2 . a) First, note that the bullet-nose curve is defined as y 0. By the Quotient Rule, in conjunction with the Chain Rule, we compute the derivative for 0 as follows: d 1 d for So the slope of the tangent line to this bullet-nose curve at the point 1,1 is - and its equation is given by y 1. This implies that the slope of the normal line to this curve at the point 1,1 is y. and its equation is given by b) This bullet-nose curve does not have a tangent line at the origin. To see why, we show that the derivative at zero does not eist as a two-sided limit as follows: y(0 h) y(0) h lim lim d h0 h 0 0 h h h 1, 0 if h 1 h lim h0 h 1, if h 0 The non-differentiability at the origin results from the fact that these two onesided limits are not equal to one another. (See attached graph.)

3 3. a) If y f ( g( )), then f '( g( )) g '( ) by the Chain Rule. d Applying the Product Rule, in conjunction with the Chain Rule again, we compute the second derivative as follows: d y d f g g '( ( )) '( ) d d f ''( g( )) g '( ) f '( g( )) g ''( ) 1, we can then apply the formula above using e f ( ) sin and g( ) e. Since f '( ) cos, f ''( ) sin, g '( ) e, b. If ysin sin e and g ''( ) e, we then get sin cos cos sin d 1 1 e cos sin e e e e e e e e e e e. Let d 1 and y cos sin d. 180 We then have the following approimation based on the differentials d, : cos 9 cos cos 6 sin Comparing this estimate with the calculator value of cos 9, which is , yields a relative error of 0.015%. Thus, this estimate is remarkably accurate. Note that you could also compute this estimate by evaluating is the linearization of y cos at the point 3, 6. L 6 180, where L

4 5. a) Repeated use of the Chain Rule here yields 1 d d d 1 1 d 1 d d 8 b) y y ln ln 3 ln y ln ln ln 3 ln ln3 ln ln By implicit differentiation, we then have d d ln ln y ln 3 ln ln d d y ' ln ln 3 yln y (since ln ln is a constant) y 3 ' 3 ln ln3 ln 3 ln ln3 d 1 3

5 c) 1 ln 0 y log 15 y ln 15 [by the Change-of-Base formula] 0 t t d ln Quotient + Chain Rule ln 0 dt ln 0 dt ln t ln d) tan tan ysin ln yln sin tan ln sin By implicit differentiation, we then have y' d cos tan ln sin sec ln sin tan y d sin y' y sec ln sin 1 tan y ' sin sec ln sin 1 d e) Using implicit differentiation, we get 3 7 y sin y 100 y d d d d d d 6 1 yy ' sin y cos y y ' 100y 100 y ' 7 3 y yy ' 3 y sin y y ' 100y 100 y ' y ' y 3 y sin y y 100y 1 y ' y 100y sin 100 y y y y 3 y 1 y ' d 7 y 3 y sin y 100

6 f) By the Quotient Rule and Chain Rule, we have: d 1 0e 1 dt 015e 300e e dt e e 1 0e y g) By the Quotient Rule and Chain Rule, we have: 1 1 d 1 d 1 1 d d 1 h) Using implicit differentiation, we get d tan d y y d d 1 y ' 1 y sec y 1 y ' 1 1 sec y 1 y ' y ' 1 y y ' 1 sec y 1 y 1 sec y y 1 sec y y ' d 1 1 sec y 1

7 6. Using implicit differentiation yields y yy yy ' 5 ', Since horizontal tangents to the lemniscate must occur when y ' 0, we get y 5 y 5 (provided 0 y (1) 5 ) Plugging this result back into the equation that implicitly describes the lemniscate, we get 5 5 y Adding equations (1) and () yields 5 5 y 5 8 () and y y We therefore conclude that the lemniscate has four horizontal tangent lines at the following points, all symmetrically located at each quadrant: 5 3 5,, 5 3 5,, 5 3 5,, 5 3 5, [See attached graph.]

8 7. First we compute the derivative function of f using the Quotient Rule as follows: f '( ) To find the critical numbers of the continuous function f on 0, we solve the following equation for 0 : 1 3 f '( ) Since f (0) 0, f , and f () 0.88, 5 3 we conclude that the absolute maimum of f on the interval 0, is and the absolute minimum of f on the interval 0, is 0. (See attached graph.) 3 8. The volume of the hemisphere is given by V r. Therefore, we get the 3 following differential for the volume: dv r dr. Plugging into this differential the values dr 0.05cm m and r 5m (half the diameter of the hemisphere), we get an estimate for the amount of paint needed: dv 5m m m 1.96m 8 Therefore, about cubic meters of paint would be needed for this paint job.

9 B1. By implicit differentiation we get the following derivative:. d y Let ab, be a point on the ellipse whose tangent line passes through the point a 1,3. Then y3 1 is the equation of this tangent line. b Plugging in the coordinates of ab, to this equation yields a b 3 a 1 b b 3 1a a a b 1 a b b Since ab, belongs to the ellipse, this implies the following: 1 a b 36 a b 3 b 3 a Therefore, 0 a 3 a 36 5a a 0 or We conclude from this that tangent line at the points 0,0 and, 5 5 pass through the point 1,3. The equations of these tangent lines are, respectively, given by y 3 and (See attached graph.) y 5. 3

10 B. Let a and denote, respectively, the distance (in meters) and the angle (in radians) between the two tips. Then, by the Law of Cosines, we have a 8 8 cos 80 6cos. Implicit differentiation then yields da d a 6sin dt dt d 6sin da dt dt a Now, at 1:00 o clock the angle between the hands is given by and so we 6 1 have sin sin and a 80 6cos mms. 6 6 d 11 Moreover, rad / hr since is measured clockwise from dt 6 6 the minute hand to the hour hand. This implies that da dt 1:00 am/ pm mm / hr