Recall the differential formula that we use for approximating the value of a function: f(x + ∆x)
≈ f(x) + f′(x)∆x. Here we want to approximate the value of , so we’ll use f(x) =
with x = 64 and ∆x = −0.03. First, we need to find f′(x): f′(x) = . Now, we
plug into the formula: f(x + ∆) ≈ ∆x. If we plug in x = 64 and ∆x = −0.03, we get
+ . If we evaluate this, we get ≈ 4 +
(−0.03) = 3.999375.
3. 1.802
Recall the differential formula that we use for approximating the value of a function: f(x + ∆x)
≈ f(x) + f′(x)∆x. Here, we want to approximate the value of tan 61°. Be careful! Whenever we
work with trigonometric functions, it is very important to work with radians, not degrees!
Remember that 60° = radians and 1° = radians, so we’ll use f(x) = tan x with x =
and ∆x = . First, we need to find f′(x): f′(x) = sec^2 x. Now, we plug into the formula: f(x +
∆x) ≈ tan x + sec^2 x∆x. If we plug in x = and ∆x = , we get tan ≈ tan +
sec^2 . If we evaluate this, we get ≈ 1.802.
- ±2.16 in.^3
Recall the formula that we use when we want to approximate the error in a measurement: dy =
f′(x) dx. Here we want to approximate the error in the volume of a cube when we know that it
has a side of length 6 in. with an error of ±0.02 in., where V(x) = x^3 (the volume of a cube of
side x) with dx = ±0.02. We find the derivative of the volume: V′(x) = 3x^2 . Now we can plug
into the formula: dV = 3x^2 dx. If we plug in x = 6 and dx = ±0.02, we get dV = 3(6)^2 (±0.02) =
±2.16.
- π ≈ 3.142 mm^3
Recall the formula that we use when we want to approximate the error in a measurement: dy =
f′(x) dx. Here we want to approximate the increase in the volume of a sphere when we know
