Solutions B B B B B. ( B/.B œ œ B/ ( /.B œ B/ / G
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1 Solutions. (a) ( cosa& b. œ sina&b œ sina b sina& b œ Þ)$ & & & (b) Let? œ. Then.? œ., so.? % $ (. œ ( œ (?.? œ Œ? G œ G a % b?% $ ' a $ b (c) Let? œ and.@ œ /.. Then.? œ. œ /, so ( /. œ (?.@ œ?@ œ / ( /. œ / / G (d) ( (. œ. œ ( a $ b. œ G ' * a $ b $ 2. The equation for the line can be rewritten œ &, while the equation for the curve is œ %. You can set these equal to find the intersection points, which are at aß% b and a%ß b. Using vertical rectangles, the area is % ( Œa& b. œ & % ln œ (Þ& % ln% œ Þ*&%) % % 3. If? œ sin, then.? œ sin cos. (using the chain rule to take the derivative), so Î$ sin cos (. œ (.? lnasin b ln? Î' We must also change the bounds on the integral. If œ Î', then? œ sin aî' b œ aî b œ Î%. If œ Î$, then? œ sin aî$ b œ ˆ È$Î œ $Î%. Therefore, Î$ $Î% sin cos (. œ (.? lnasin b ln? Î'?? Î%
2 4. We must use vertical rectangles. (Using horizontal rectangles doesn't work here when you rotate a horizontal rectangle around the -axis, the result is neither a disc nor a washer.) ecause of the shape of the region, we must break it up into two parts: Ðß Ñ For each part, the vertical rectangles rotate to form discs. Therefore, the volume is ( ˆ ( Š È %.. œ (. ( a b. & œ œ Þ( & sin 5. The given equation says that Si ab is an antiderivative of. That is, the derivative of Siab sin is. sin (a) If? œ Si a b, then.? œ., so sin.? Î Î (. œ È ( œ È (?.? œ? G œ ÈSiab G Sia b? (b) Let? œ Si a b and.@ œ.. (This is the same strategy we used to evaluate ( ln. and sin ( tan..) Then.? œ. œ, so ( Sia b. œ (?.@ œ?@ sin œ Si ab (. œ Siab ( sin. œ Siab cos G
3 6. There are several different ways to find the equation of the line, including:. Use the point-slope formula, which is œ 7a b. 2. Assume the equation has the form œ 7,. Plug in both points to solve for 7 and,. 3. Find the slope (which is Î ), and then plug one point into the equation œ, to solve for,. & In any case, the equation of the line is œ & ( Š È ˆ & &. $. Using vertical rectangles, the area is 7. The formula for a normal distribution is 0ab œ / 5È a. b Î5 where. is the mean and 5 is the standard deviation. In this case,. œ &( and 5 œ ), so the probability distribution for the lengths of the fish is Therefore: 0ab œ / ) È a &( b Î) a &( b T a&& Ÿ Ÿ ' b œ ( / Î). && ) È According to Wolfram alpha, this translates to a probability of Þ%&, or %Þ&%. '
4 8. (a) Using the graphs, we can estimate the values of and for > œ ßßß$ß%: > : $ % : Þ Þ( $Þ Þ( Þ : %Þ $Þ' Þ Þ% Þ We plot the corresponding points and then connect them with a curve: 4 Ðß %Ñ ÐÞ(ß $Þ'Ñ 3 2 Ð$ß Ñ 0 Ðß Ñ ÐÞ(ß Þ%Ñ (b) There are many different ways to estimate the area. One simple possibility is to approximate the area using six squares and two triangles, as shown below: Each square has an area of, and each triangle has an area of Î, making the total area ' œ (. The actual area is about (Þ$$, and any answer between 'Þ& and )Þ would receive full credit on this problem.
5 9. We cut the circle into horizontal rectangles, which rotate around the line to make washers: œ The equation for the unit circle is œ. Solving for gives œ È. In particular, the right half of the circle has equation œ È, and the left half has equation œ È : V <. œ Thus < œ È and V œ ˆ È œ È. Therefore, the volume is ( ˆ V <. œ ( ˆ È ˆ È.
6 0. Let's try some different values for > : > : Î% Î $ Î% : Þ% Þ% : ased on these values, the parametric equation travels around the right loop between > œ and > œ : > œ > œ (a) This is simply asking for the arc length of the parametric curve during the right loop. The integral is:.. ( Ë Œ Œ.> œ ( Éa cos> b a cos> b.>.>.> Since the top half and bottom half have equal length, this integral can also be written ( Éa cos> b a cos > b.>. Î (b) This is twice the area under the curve between > œ and > œ Î : This is the same as ( sina> b cosa> b.>. (. œ %( sina> b cosa> b.> Î
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