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3 years ago

Trig questions:

Mathematics

1 answer:

navik [9.2K]3 years ago

8 0

Question 1
Because the period is 2π, and the amplitude is 1obtain
f(x) = sin(x)
Because the horizontal shift is π, obtain
f(x) = sin(x - π)
Because the vertical shift is -4, obtain
f(x) = sin(x - π) - 4

Answer: 1. f(x) = sin(x - π) - 4

Question 2
The radius is 36/2 = 18 in.
1 revolution (360°) is the circumference, which is
2π(18) = 36π in
When the revolution is 62°, the distance traveled is
(62/360)*(36π) = (31/5)π in

Answer: 3. (31π)/5

Question 3.
Consider f(x) = 3cos(2x-π) - 1
f(0) = 3cos(-π) - 1 = -4
f(π/2) = 3cos(0) - 1 = 2
Rate of change = (2+4)/(π/2) = 12/π

From the graph, the rate of change of g(x) is
3/(π/2) = 6/π

Consider h(x) = sin(x) - 4
h(0) = 0 - 4 = -4
h(π/2) = 1 - 4 = -3
Rate of change = (-3+4)/(π/2) = 2/π
Therefore h(x) has the smallest rate of change

Answer: h(x)

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2 years ago

Find the derivative of f(x)= (e^ax)*(cos(bx)) using chain rule

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$f(x) = e^{ax}\cos(bx)$

then by the product rule,

$f'(x) = \left(e^{ax}\right)' \cos(bx) + e^{ax}\left(\cos(bx)\right)'$

and by the chain rule,

$f'(x) = e^{ax}(ax)'\cos(bx) - e^{ax}\sin(bx)(bx)'$

which leaves us with

$f'(x) = \boxed{ae^{ax}\cos(bx) - be^{ax}\sin(bx)}$

Alternatively, if you exclusively want to use the chain rule, you can carry out logarithmic differentiation:

$\ln(f(x)) = \ln(e^{ax}\cos(bx)} = \ln(e^{ax})+\ln(\cos(bx)) = ax + \ln(\cos(bx))$

By the chain rule, differentiating both sides with respect to x gives

$\dfrac{f'(x)}{f(x)} = a + \dfrac{(\cos(bx))'}{\cos(bx)} \\\\ \dfrac{f'(x)}{f(x)} = a - \dfrac{\sin(bx)(bx)'}{\cos(bx)} \\\\ \dfrac{f'(x)}{f(x)} = a-b\tan(bx)$

Solve for f'(x) yields

$f'(x) = e^{ax}\cos(bx) \left(a-b\tan(bx)\right) \\\\ f'(x) = e^{ax}\left(a\cos(bx)-b\sin(bx))$

just as before.

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2 years ago