MAT 21A, Spring 2017
Solutions to HW 5
Section 3.6: 60. (10 points) Compute the derivative of the function f (t) =
−5
3t−4
.
5t+2
Solution: By Quotient Rule
3t − 4
5t + 2
0
=
3(5t + 2) − (3t − 4)5
15t + 6 − 15t + 20
26
=
=
.
2
2
(5t + 2)
(5t + 2)
(5t + 2)2
Now by Chain Rule
3t − 4
f (t) = −5
5t + 2
0
−6
·
26
.
(5t + 2)2
94.
√ (10 points) Find the equation of the tangent line to the graph of y =
x2 − x + 7 at x = 2.
√
√
Solution: At x = 2 we have y(2) = 4 − 2 + 7 = 9 = 3. The slope of
the tangent is equal to the derivative at x = 2. By Chain Rule the derivative
equals
1
y 0 (x) = √ 2
· (2x − 1),
2 x −x+7
so the derivative at x = 2 equals
1
1
1
· (4 − 1) = √ · 3 = .
y 0 (2) = √
2
2 4−2+7
2 9
Therefore the equation of the tangent has the form y = 12 x + b. To find b, we
plug in x = 2 and y = y(2) = 3:
3=
1
· 2 + b = 1 + b ⇒ b = 2.
2
Finally, the equation of the tangent line is y = 21 x + 2.
Section 3.7: 46. (10 points) a) Find the slope of the tangent line to the
folium of Descartes
x3 + y 3 − 9xy = 0
at points (4, 2) and (2, 4).
1
b) At what point other than the origin does the folium have a horizontal
tangent?
c) Find the coordinates of the points where it has vertical tangents.
Solution: We use implicit differentiation to find the derivative. Write
the equation as:
x3 + y(x)3 − 9xy(x) = 0,
take the derivative of both sides and solve for y 0 :
3x2 + 3y(x)2 y 0 (x) − 9 · 1 · y(x) − 9xy 0 (x) = 0,
3x2 − 9y(x) = 9xy 0 (x) − 3y 2 (x)y 0 (x) = y 0 (x)(9x − 3y(x)2 ),
y0 =
x2 − 3y
3x2 − 9y
=
.
9x − 3y 2
3x − y 2
(a) For x = 2 and y = 4 we get
y0 =
4 − 12
−8
4
=
= .
6 − 16
−10
5
For x = 4 and y = 2 we get
y0 =
16 − 6
10
5
=
= .
12 − 4
8
4
(b) The tangent is horizontal if y 0 = 0, so x2 − 3y = 0, y =
this into the original equation:
0 = x3 + y 3 − 9xy = x3 + (
x2
.
3
Let us plug
x2
x6
x6
x2 3
) − 9x = x3 +
− 3x3 =
− 2x3 ,
3
3
27
27
√
x3
x6
3
, 2 = , x3 = 2 · 27, x = 3 2.
27
27
√
√
3
2
3
Then y = x3 = 9 3 2 = 3 2.
(c)√The tangent
is vertical if y 0 = ∞, so 3x = y 2 . Similarly to (b), we get
√
x = 3 3 4, y = 3 3 2.
2x3 =
Section 4.1: (10 points) Find the absolute maximum and the absolute
2
minimum of the function f (x) = e−x on the interval [−2, 1].
2
2
Solution: By Chain Rule we have f 0 (x) = e−x · (−2x), so the only
critical point is x = 0. Now
f (−2) = e−4 =
Since e4 > e > 1, we have
minimal value is e14 .
1
1
0
−1
,
f
(0)
=
e
=
1,
f
(1)
=
e
=
e4
e
1
e4
<
1
e
< 1, so the maximal value is 1 and the
3