Circle A, with radius a, and circle B , with radius b , are tangent to each other and to P Q ¯ (see figure). P R ¯ passes through the center of each circle. Let x be the distance from point P to a point S where P R ¯ intersects circle A on the left. Let θ denote ∠ R P Q . a. Show that sin θ = a x + a and sin θ = b x + 2 a + b . b. Use the results from part (a) to show that sin θ = b − a b + a .
Circle A, with radius a, and circle B , with radius b , are tangent to each other and to P Q ¯ (see figure). P R ¯ passes through the center of each circle. Let x be the distance from point P to a point S where P R ¯ intersects circle A on the left. Let θ denote ∠ R P Q . a. Show that sin θ = a x + a and sin θ = b x + 2 a + b . b. Use the results from part (a) to show that sin θ = b − a b + a .
Solution Summary: The author proves the following values of the trigonometric function, using the given figure.
Circle A, with radius a, and circle
B
, with radius
b
, are tangent to each other and to
P
Q
¯
(see figure).
P
R
¯
passes through the center of each circle. Let
x
be the distance from point
P
to a point
S
where
P
R
¯
intersects circle
A
on the left. Let
θ
denote
∠
R
P
Q
.
a. Show that
sin
θ
=
a
x
+
a
and
sin
θ
=
b
x
+
2
a
+
b
.
b. Use the results from part (a) to show that
sin
θ
=
b
−
a
b
+
a
.
Given lim x-4 f (x) = 1,limx-49 (x) = 10, and lim→-4 h (x) = -7 use the limit properties
to find lim→-4
1
[2h (x) — h(x) + 7 f(x)] :
-
h(x)+7f(x)
3
O DNE
17. Suppose we know that the graph below is the graph of a solution to dy/dt = f(t).
(a) How much of the slope field can
you sketch from this information?
[Hint: Note that the differential
equation depends only on t.]
(b) What can you say about the solu-
tion with y(0) = 2? (For example,
can you sketch the graph of this so-
lution?)
y(0) = 1
y
AN
(b) Find the (instantaneous) rate of change of y at x = 5.
In the previous part, we found the average rate of change for several intervals of decreasing size starting at x = 5. The instantaneous rate of
change of fat x = 5 is the limit of the average rate of change over the interval [x, x + h] as h approaches 0. This is given by the derivative in the
following limit.
lim
h→0
-
f(x + h) − f(x)
h
The first step to find this limit is to compute f(x + h). Recall that this means replacing the input variable x with the expression x + h in the rule
defining f.
f(x + h) = (x + h)² - 5(x+ h)
=
2xh+h2_
x² + 2xh + h² 5✔
-
5
)x - 5h
Step 4
-
The second step for finding the derivative of fat x is to find the difference f(x + h) − f(x).
-
f(x + h) f(x) =
= (x²
x² + 2xh + h² -
])-
=
2x
+ h² - 5h
])x-5h) - (x² - 5x)
=
]) (2x + h - 5)
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