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Viktor [21]
1 year ago
12

Suppose that the derivable functions x=x(t) and y=y(t) satisfy xcosy=2.

Mathematics
1 answer:
ololo11 [35]1 year ago
4 0

Applying implicit differentiation, it is found that dy/dt when y=π/4 is of:

a-) -√2 / 2.

<h3>What is implicit differentiation?</h3>

Implicit differentiation is when we find the derivative of a function relative to a variable that is not in the definition of the function.

In this problem, the function is:

xcos(y) = 2.

The derivative is relative to t, applying the product rule, as follows:

\cos{y}\frac{dx}{dt} - x\sin{y}\frac{dy}{dt} = 0

\frac{dy}{dt} = \frac{\cos{y}\frac{dx}{dt}}{x\sin{y}}

Since dx/dt=−2, we have that:

\frac{dy}{dt} = -2\frac{\cos{y}}{x\sin{y}}

When y = π/4, x is given by:

xcos(y) = 2.

x = \frac{2}{\cos{\frac{\pi}{4}}} = \frac{2}{\frac{\sqrt{2}}{2}} = \frac{4}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = 2\sqrt{2}

Hence:

\frac{dy}{dt} = -2\frac{\cos{y}}{x\sin{y}}

\frac{dy}{dt} = -\frac{1}{\sqrt{2}}\cot{y}

Since cot(pi/4) = 1, we have that:

\frac{dy}{dt} = -\frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = -\frac{\sqrt{2}}{2}

Which means that option a is correct.

More can be learned about implicit differentiation at brainly.com/question/25608353

#SPJ1

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A competitive knitter is knitting a circular place mat. The radius of the mat is given by the formula
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Answer: A. A'(t)=\frac{4356\pi}{(t+11)^{3}}-\frac{527076\pi}{(t+11)^{5}}

              B. A'(5) = 1.76 cm/s

Step-by-step explanation: <u>Rate</u> <u>of</u> <u>change</u> measures the slope of a curve at a certain instant, therefore, rate is the derivative.

A. Area of a circle is given by

A=\pi.r^{2}

So to find the rate of the area:

\frac{dA}{dt}=\frac{dA}{dr}.\frac{dr}{dt}

\frac{dA}{dr} =2.\pi.r

Using r(t)=3-\frac{363}{(t+11)^{2}}

\frac{dr}{dt}=\frac{726}{(t+11)^{3}}

Then

\frac{dA}{dt}=2.\pi.r.[\frac{726}{(t+11)^{3}}]

\frac{dA}{dt}=2.\pi.[3-\frac{363}{(t+11)^{2}}].\frac{726}{(t+11)^{3}}

Multipying and simplifying:

\frac{dA}{dt}=\frac{4356\pi}{(t+11)^{3}} -\frac{527076\pi}{(t+11)^{5}}

The rate at which the area is increasing is given by expression A'(t)=\frac{4356\pi}{(t+11)^{3}} -\frac{527076\pi}{(t+11)^{5}}.

B. At t = 5, rate is:

A'(5)=\frac{4356\pi}{(5+11)^{3}} -\frac{527076\pi}{(5+11)^{5}}

A'(5)=\frac{4356\pi}{4096} -\frac{527076\pi}{1048576}

A'(5)=\frac{2408693760\pi}{4294967296}

A'(5)=1.76268

At 5 seconds, the area is expanded at a rate of 1.76 cm/s.

5 0
3 years ago
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