Name:
MATH 151 - Quiz 7
Instructions. Time: 20 minutes. Please, no notes, no calculator, no cellphone, no headphones, no
electronic devices. Use the back of this sheet for Exercises 2 and 3 and copy the nal answer on this side.
Exercise 1
(3pts). Compute the derivatives of the following functions1 :
· f (x) = (x2 + 2x) log(x3 );
Note: f (x) = 3(x2 + 2x) log(x).
d
3(x2 + 2x)
f (x) = 3(2x + 2) log(x) +
= 6(x + 1) log(x) + 3(x + 2).
dx
x
· g(x) = ex cos(x)+2x ;
d
g(x) = (−x sin(x) + cos(x) + 2)ex cos(x)+2x .
dx
· h(x) = 3x arcsin(x + 4).
d
1
h(x) = 3 arcsin(x + 4) + 3x p
.
dx
1 − (x + 4)2
Exercise 2
(4pts). Use the linear approximation of the function f (x) =
approximation of
√1
4.1
√1
5−x
at a = 1 to nd an
.
The tangent line to the graph of f at a = 1 is L(x) =
1
approximation is L(0.9) = 12 + 16
(0.9 − 1) =
79
160
1
2
+
1
16 (x
− 1). Since
√1
4.1
= f (0.9), its linear
.
(3pts). Consider a pyramid with square basis of area A = 144cm2 and height H = 6cm,
vertex at the bottom. Water is poured into the pyramid at a rate of 40cm3 /sec. Denote by h(t) the level
of the water at time t. Find h and its rate of change when the pyramid contains exactly 36cm3 .
Exercise 3
Denote by `(t) the edge of the piramid dened by the water level. From relations of similar triangles we
obtain 2h(t) = `(t). Therefore the volume of poured water at time t is V (t) = 43 h3 (t). When V = 36cm3
we get h = 3cm. By dierentiating V (t) we get V 0 = 4h2 h0 ; so 40cm3 /sec = 4 · (3cm)2 h0 and solving for
h0 we conclude h0 = 10
9 cm/sec.
1 log
is the natural logarithm;
arcsin
is the inverse function of the sine function.