Unformatted text preview: Math 1A — Worksheet — February 13, 2008
2. Differentiate the following functions. a) y = sint + ncost
y’= cost — Mint b) v=t2~ m... Write v = t2 —(t3)7 Then v t = 2, _ (_ 3(9)“ W)
e) y = 47:2 y ‘ = 0 3‘ The position function ofa particle is given by s = t" — 4.5!2 — 7!, t2 0 .
a) When does the particle reach a velocity of 5 m/s?
We ﬁrst must ﬁnd the derivative of s:
v=s/=3tz—9t—7
So v=5 —> Biz—9t—7=5 —r 3tz—9t— 12:0
—>t2—3t—4=0 —> (t—4)(t+1)=0 —> t=4sec We do not consider I = ~ 1 a solution because time must
be apositive quantity b) When is the acceleration zero? What is the signiﬁcance of this value of t? The equation of acceleration of the particle can be found by
diﬁhrentiating the velocity ﬁtnction  a = v ’ = 9! — 9 So a=0 —> 9t—9=0 —> t—1=0 —>t=lsec 4. Find equations ofthc tangent line and normal line to y = 4x2 +4): + 1 at (1,9) Following the methods used in the notecard bonus question ﬁmm
Quiz 3 we get:
Equation oft/1e tangent line: y7 9 = 12 (x 7 l) 1 Equation ofthe normal linezy — 9 =— E (x 4 1) 5. Sketch the palabulas y3 = x2 and y1 = x2 — 2x + 2 . Do you think there is a
line tangent to both curves? If so, ﬁnd its equation. If not why not?
Let us assume thata line tangent to both curves exists and let us try to ﬁnd it ~ So we must diﬂerentiate each ﬁmction to build the tangent line for
each function: Wegetyj=2x and y2’=2x—2 N aw let us call thepoint where this assumed tangent line touches yl , (a, yl (11)) And let us call the pointwhere this assum2d tangent line touches yz , (b, y2 (b))
Then the respective equations for the tangent lines would be:
Fory,: y—y,(a)=2a(x—a)—>y=2wc—2a‘+y,(a)=cht—Zaz+a2
so y = 2413: — aZ
Foryz: y—y2(b)=(2b — 2) (x—b)> y=2bx—2bz—Zx+2b +y2(b)
so y=2bx+2x— b2+2
Now we must ﬁnd out of if there are specific values for (a, y1 ((1)) and (b, y2 (b)) that allow these two lines to be the same 
Since we would like these to be the same line ,. . then their slopes should equal: So 2a=2b—2—>a=b—l Plugging this into the tangent line for yI gives: y = 2 (b — 1) x (b —— 1)2
Equating the two tangentlines gives: 2bx—2x—b‘+2b—1=2bx+2x—b“+2
Simpltﬁling yields: 4:: = 2b — 3 Since we cannot ﬁnd an explicit answer for b (asx changes, it will change)
then we know it is not possible for the simultaneous tangent line to exist » ...
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 Spring '08
 WILKENING
 Calculus

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