1
151 WebCalc Fall 2002-copyright Joe Kahlig
In Class Questions
1.
lim
x→ π2 −
October 22
2
=0
1 + etan x
since as x →
2. Find
MATH 151-Fall 02
π−
we have that tan(x) → ∞. Thimeans that 1 + etan x → ∞.
2
dy
for these functions.
dx
2
2
(a) y = ex +5x e7−x
2
2
simplify first. y = ex +5x+7−x = e7+5x
y 0 = 5e7+5x
(b) y = tan(e3x
y0
=
sec2
2 +4
)
2
e3x +4
∗ 6xe3x
2 +4
(c) x = et + sec(t2 )
y = te + t4 + 1
dy
dy
ete−1 + 4t3
dt
= t
= dx
dx
e + 2t sec(t2 ) tan(t2 )
dt
3. Find the tangent line at the point (0, 2) for 2exy = x2 + y
2exy ∗ (1 ∗ y + x ∗ y 0 ) = 2x + y 0
now you have two choices: 1) solve for y 0 and then plug in the values for x and y, or 2) plug in
the values for x and y and then solve for y 0 . I’m going to do the second choice since it is a bit
easier.
2e0 ∗ (2 + 0) = 0 + y 0
4 = y0
Answer: y − 2 = 4(x − 0)
4. Find all the points where the tangent line is vertical or horizontal.
x = t3 − 3t2 + 5
y = 2t2 + t
If a tangent line is horizontal then the slope of the tangent line is zero. Since
means we are looking for
dy
=0
dt
dy
=
dx
dy
dt
dx
dt
, this
dy
= 4t + 1
dt
0 = 4t + 1
t = −1
4
Since the problem asks for points, plug the value of t back into the parametric formulas and
solve for x and y.
307 −1
Answer: at the point (4.796875, −0.125) or in fraction form
,
64 8
If the tangent line is vertical then the slope of the tangent line is undefined. That happens when
dx
=0
dt
2
151 WebCalc Fall 2002-copyright Joe Kahlig
dx
= 3t2 − 6t
dt
0 = 3t(t − 2)
so t = 0 or t = 2
Answers: at the point (5, 0) and at the point (1, 10)
5. Find the equation of the tangent line at the point (3, 1) for the curve
x = t2 + 2t
y = t3 − t + 1
If this is the point, that means that x = 3 and y = 1. We to know the value of t that makes this
happen.
3 = t2 + 2t
0 = t2 + 2t − 3
0 = (t + 3)(t − 1)
t = −3 or t = 1
1 = t3 − t + 1
0 = t3 − t
0 = t(t2 − 1)
0 = t(t + 1)(t − 1)
t = 0, t = −1, or t = 1
The only value of t that works for both x and y is t = 1.
dy 3−1
2
=
=
dx t=1 2 + 2
4
Answer: y − 1 = 24 (x − 3)