Unformatted text preview: which the
.
dx dy
Gathering the solution
which gives terms involving dx to one side, the equation becomes
Solution: Diﬀerentiating the equation4with 4
y 2 − respect to x, we have
dy
=
.
dy
dy
2y dx − 82y − 8xy 4y 2 − 4dy
=2
dy xy
dx + 4 dx4y + 8xy .
2y
=
dx
dx
which gives the solution
dy
Gathering terms involving dx to one side,ythe equation becomes
4 2−4
dy
=
.
dx
2y − 8xy
dy
dy this curve at the point ( 1 , 2).
b. [4 points] Find the equation for the 2y
tangent 8xy to = 4y 2 − 4
− line
3
dx
dx
Solution: The slope is
which gives the solution dy
4 · 22 − 4
2
= dy
= −9,
1
dx ( 1 ,2) 2 · 2 − 8 4y · − 4 .
= ·3 2
3
dx
2y to this
1
b. [4 points] Find the equation for the tangent line− 8xy curve at the point ( 3 , 2).
!
so by the pointslope formula, the equation is
Solution: The slope is
y = −9x + 5. 4
dy
4 · 22 −
=
= −9,
dx ( 1 ,2) 2 · 2 − 8 · 1 · 2
3
The actual value is page 4 26dy 5.099.
= 3 1
b. so by the pointslope equation for the tangent line to this curve at the point ( 3 , 2).
[4 points] Find the formula, the equation is Solution: The slope is
y = −9x + 2 .
5
dy
c. [4 points] Find the x and y coordinates = all points 4 which 9, tangent line to this
of 4 · 2 − at = − the
dx ( 1 ,2)
2·2−8· 1 ·2
curve is vertical.
!
3
3 Solution:the The slope isformula, the equation is
so by
pointslope undeﬁned as these points, so we must have 2y − 8xy = 0.
Factoring out a 2y we get
2y (1 − = −=x0+ 5.
y 4x) 9
c. [4 points] Find the x and y coordinates of all points at which the tangent line to this
which gives the solutions y = 0 or x = 1 . Plugging into the equation for the implicit
4
curve is vertical.
function, y = 0 gives the point (0, 0). However, when we plug in x = 1 , we get the
4
Solution:
equation y 2 +The y 2 , which undeﬁned as these points, so , 0) is the only point− 8which 0.
1 = slope is has no solutions. Therefore, (0 we must have 2y at xy =
Factoring line 2 vertical.
the tangentout a is y we get
!
2y (1 − 4x) = 0
c. [4 points] Find the x and y coordinates of all points at which the tangent line to this
1
which is vertical.
curve gives the solutions y = 0 or x = 4 . Plugging into the equation for the implicit
function, y = 0 gives the point (0, 0). However, when we plug in x = 1 , we get the
4
Solution:
The slope is undeﬁned as these points, so we must have 2y − 8xy = 0.
equation y 2 + 1 = y 2 , which has no solutions. Therefore, (0, 0) is the only point at which
Factoring out a 2y we get
the tangent line is vertical.
2y (1 − 4x) = 0 which gives the solutions y = 0 or x = 1 . Plugging into the equation for the implicit
4
function, y = 0 gives the point (0, 0). However, when we plug in x = 1 , we get the
4
equation y 2 + 1 = y 2 , which has no solutions. Therefore, (0, 0) is the only point at which actual
5. [12 points] value of f (x)? (Show your work.)
a. [3 points] Find locallocal linearization Lunderestimates of the actual value when f (0) >
Solution: The the linearization...
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 Fall '11
 fisherselin
 Derivative, Differential Calculus, Slope

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