How To Find Rate Of Change Of An Angle

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A right triangle has sides of length and which are both increasing in length over time such that:

x(t)=2t

y(t)=4t^2

a) Discover the charge per unit at which the bending $\theta$ opposite y(t) is changing with respect to time.

Correct answer:

$\frac{d\theta}{dt}=\frac{2}{1+4t^2}$

Explanation:

A right triangle has sides of lenght and which are both increasing in length over time such that:

x(t)=2t

y(t)=4t^2

Find the rate at which the angle $\theta$ reverse y(t) is irresolute with respect to time.

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a) Offset, nosotros need to write an expression for the angle $\theta$ as a office of. Because the angle is opposite the side we know that the tangent is just y/x . Take the inverse of the tangent:

$\theta (t)\, \, =\, \, \tan^{-1}\left(\frac{y(t)}{x(t)} \right )\: \: =\: \: \tan^{-1}\left(\frac{4t^2}{2t}\right)\, \, =\, \, \tan^{-1}\left(2t \right )$

$\theta(t)=\tan^{-1}(2t)$

Now nosotros demand to differentiate with respect to.

Recall the general derivative for the inverse tangent office is:

$\frac{d}{dx}\tan^{-1}(x)=\frac{1}{1+x^2}$

Applying this to our function for $\theta(t)$ , and remembering to use the chain rule, nosotros obtain:

$\frac{d\theta}{dt}=\frac{d}{dt}\tan^{-1}(2t)=\frac{2}{1+(2t)^2}=\frac{2}{1+4t^2}$

$\frac{d\theta}{dt}=\frac{2}{1+4t^2}$

Soap is sometimes used to determine the location of leaks in industrial pipes. A perfectly spherical lather chimera is growing at a charge per unit of 5 mm^3/sec . What is the charge per unit of change of the surface area of the bubble when the radius of the bubble is 10 mm ?

Correct answer:

1 mm^2/sec

Explanation:

To determine the charge per unit of change of the surface area of the spherical bubble, nosotros must chronicle it to something nosotros do know the rate of change of - the volume.

The book of a sphere is given by the following:

$V=\frac{4\pi}{3}r^3$

The rate of alter of the volume is given by the derivative with respect to fourth dimension:

$\frac{\mathrm{d} V}{\mathrm{d} t}=4\pi r^2 \frac{\mathrm{d} r}{\mathrm{d} t}$

The derivative was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x)) \cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$

We must now solve for the rate of change of the radius at the specified radius, so that we tin later solve for the rate of alter of expanse:

$5=4\pi (100) \frac{\mathrm{d} r}{\mathrm{d} t}$

$\frac{\mathrm{d} r}{\mathrm{d} t}=\frac{1}{80\pi}$

Next, we must find the expanse and rate of alter of the surface expanse of the sphere the same way equally higher up:

$A=4\pi r^2$

$\frac{\mathrm{d} A}{\mathrm{d} t}=8 \pi r \frac{\mathrm{d} r}{\mathrm{d} t}$

Plugging in the known charge per unit of modify of the expanse at the specified radius, and this radius into the rate of surface area change function, we become

$\frac{\mathrm{d} A}{\mathrm{d} t}=8\pi (10)\cdot \frac{1}{80\pi}= 1mm^2/sec$

A pizzeria chef is flattening a circular slice of dough. The surface surface area of the dough (we are only considering the superlative of the dough) is increasing at a rate of 0.five inches/sec. How chop-chop is the diameter of the pizza irresolute when the radius of the pizza measures iv inches?

Possible Answers:

$\frac{1}{8 \pi}$ inches/sec

$\frac{1}{16 \pi}$ inches/sec

inches/sec

Correct answer:

$\frac{1}{8 \pi}$ inches/sec

Caption:

To find the rate of change of the diameter, nosotros must relate the bore to something we practise know the charge per unit of change of: the surface area.

The surface area of the top side of the pizza dough is given by

$A=\pi r ^2$

The rate of change, and then, is plant by taking the derivative of the function with respect to time:

$\frac{\mathrm{d} A}{\mathrm{d} t}=2\pi r \frac{\mathrm{d} r}{\mathrm{d} t}$

Solving for the charge per unit of change of the radius at the given radius, we get

$\frac{1}{2}=2\pi (4)\frac{\mathrm{d} r}{\mathrm{d} t}$

$\frac{\mathrm{d} r}{\mathrm{d} t}=\frac{1}{16\pi}$ inches/sec

Now, we chronicle the diameter to the radius of the pizza dough:

2r=d

Taking the derivative of both sides with respect to time, we become

$2 \frac{\mathrm{d} r}{\mathrm{d} t}=\frac{\mathrm{d} d}{\mathrm{d} t}$

Plugging in the known rate of change of the radius at the given radius, nosotros become

$\frac{\mathrm{d} d}{\mathrm{d} t}=\frac{1}{8\pi}$ inches/sec

We could have found this direct by writing our surface area formula in terms of diameter, however the procedure we used is more than applicative to problems in which the related rate of change is of something not as like shooting fish in a barrel to manipulate.

A spherical airship is increasing in volume at a abiding rate of 0.1 cm^3/min . At a radius of 3 cm, what is the rate of change of the circumference of the balloon?

Right reply:

$\frac{1}{180} cm/min$

Explanation:

To decide the charge per unit of change of the circumference at a given radius, we must chronicle the circumference rate of change to the rate of change we know - that of the volume.

Starting with the equation for the volume of the spherical airship,

$V=\frac{4}{3}\pi r^3$

we take the derivative of the function with respect to fourth dimension, giving us the charge per unit of change of the volume:

$\frac{\mathrm{d} V}{\mathrm{d} t}=4\pi r^2 \frac{\mathrm{d} r}{\mathrm{d} t}$

The derivative was establish using the post-obit rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$

The chain rule was used when taking the derivative of the radius with respect to time, because we know that it is a office of time.

Solving for $\frac{\mathrm{d} r}{\mathrm{d} t}$ using our known $\frac{\mathrm{d} V}{\mathrm{d} t}$ at the given radius, nosotros get

$\frac{\mathrm{d} r}{\mathrm{d} t}=\frac{1}{360\pi} cm/min$

At present, we use this rate of change and utilise it to the charge per unit of change of the circumference, which we get by taking the derivative of the circumference with respect to fourth dimension:

$c=2\pi r$

$\frac{\mathrm{d} c}{\mathrm{d} t}=2\pi \frac{\mathrm{d} r}{\mathrm{d} t}$

Solving for the rate of change of the circumference by plugging in the known rate of change of the radius, we get

$\frac{\mathrm{d} c}{\mathrm{d} t}=2\pi (\frac{1}{360 \pi})=\frac{1}{180} cm/min$

Determine the rate of change of the angle opposite the base of a right triangle -whose length is increasing at a rate of 1 inch per minute, and whose height is a constant 2 inches - when the area of the triangle is 2 square inches.

Possible Answers:

$\frac{1}{4\sqrt{2}}$ radians per minute

$\frac{1}{\sqrt{2}}$ radians per minute

$\frac{1}{4}$ radians per minute

$\frac{1}{2\sqrt{2}}$ radians per minute

Right answer:

$\frac{1}{\sqrt{2}}$ radians per minute

Caption:

To decide the charge per unit of the change of the bending opposite to the base of the given correct triangle, nosotros must relate it to the rate of alter of the base of operations of the triangle when the triangle is a sure area.

Beginning, we must determine the length of the base of the right triangle at the given area:

$A=\frac{bh}{2}$

4=b(2)

b=2

At present, nosotros must detect something that relates the angle opposite of the base to the length of the base and summit - the tangent of the angle:

$\tan \theta=\frac{b}{h}$

To observe the rate of modify of the angle, we take the derivative of both sides with respect to time, keeping in heed that the base of operations of the triangle is dependent on time, while the height is constant:

$\sec^2(\theta)\frac{\mathrm{d} \theta}{\mathrm{d} t}=\frac{2}{2}\frac{\mathrm{d} b}{\mathrm{d} t}$

We know the rate of modify of the base, and we can find the angle from the sides of the triangle:

$\tan(\theta)=1$

$\theta=\frac{\pi}{4}$

Plugging this and the other known data in and solving for the rate of modify of the angle adjacent to the base of operations, we get

$\sec^2(\frac{\pi}{4})\frac{\mathrm{d} \theta}{\mathrm{d} t}=1$

$\frac{\mathrm{d} \theta}{\mathrm{d} t}=\frac{1}{\sqrt{2}}$ radians per minute

The position of a motorcar is given by the equation

$f(t) = sin(\pi t) + 3t^{2} - 10t + 7$ .

Find the motorcar'due south dispatch when t = 4 .

Correct answer:

Explanation:

To find the car's acceleration, accept the SECOND derivative of f(t) .

$f'(t) = \pi cos(\pi t) + 6t - 10$ , and

$f''(t) = -\pi ^{2}sin(\pi t) + 6$ .

The car's position at t = 4 is then given by:

$f''(4) = -\pi ^{2}sin(\pi *4) + 6 = 6$

A point on a circle of radius 1 unit is orbiting counter-clockwise around the circle's heart. It makes one total orbit every eight seconds. How fast is the coordinate changing when the line segment from the origin to the point, (x,y) , forms an angle of $\frac{\pi}{4}$ radians above the positive ten-axis?

Correct answer:

$\frac{-\sqrt{2}\pi}{8}$

Explanation:

This is a related rates problem.

Since the problem gives the time for one orbit, we tin can find the angular speed of the point. The angular speed is simply how many radians the particle travels in one second. We find this past dividing the number of radians in ane revolution, $2\pi$ , by the time information technology takes to travel 1 revolution, eight seconds.

$\frac{2\pi}{8} = \frac{\pi}{4}$

This gives us the change in the angle with respect to time, $\frac{\mathrm{d} \theta}{\mathrm{d} t} = \frac{\pi}{4}$ .

Now we demand to relate the position to the bending, $\theta$ . Call back that

$\cos \theta = \frac{x}{r}$ , where r is the radius.

Since the radius is given every bit ane unit of measurement, we tin write this equation as

$x = \cos \theta$ .

Now nosotros take the derivative of both sides with respect to time , using implicit differentiation. Retrieve that we use the chain rule for any variable that is not. This gives,

$\frac{\mathrm{d} x}{\mathrm{d} t} = -\sin\theta \frac{\mathrm{d} \theta}{\mathrm{d} t}$ .

Now we accept a formula that relates the horizontal speed of the particle at an instant in fourth dimension, $\frac{\mathrm{d} x}{\mathrm{d} t}$ , to the angle above the positive x-axis and angular speed at that aforementioned instant. We are told to find how fast the 10 coordinate is changing when the angle, $\theta$ is $\frac{\pi}{4}$ radians above the positive x-axis. And so we will plug in $\frac{\pi}{4}$ for $\theta$ . Even so, we also need to know $\frac{\mathrm{d} \theta}{\mathrm{d} t}$ . Fortunately, we already plant it. It is the athwart speed, $\frac{\pi}{4}$ radians/second.

Plugging this information in, nosotros get

$\frac{\mathrm{d} x}{\mathrm{d} t}= - \sin{(\frac{\pi}{4})} (\frac{\pi}{4})$

$\frac{\mathrm{d} x}{\mathrm{d} t} = -(\frac{\sqrt{2}}{2})(\frac{\pi}{4})$

$\frac{\mathrm{d} x}{\mathrm{d} t}= \frac{-\sqrt{2}\pi}{8}$

This is the answer. The negative makes sense considering the bespeak is traveling counter-clockwise. Hence, it is moving left when the angle is $\frac{\pi}{4}$ radians.

A homo is continuing on the top of a 10 ft long ladder that is leaning against the side of a building when the lesser of the ladder begins to slide out from under information technology. How fast is the human continuing on the top of the ladder falling when the lesser of the ladder is half dozen ft from the building and is sliding at 2ft/sec?

Correct reply:

$-3/2 \text{ feet/sec}$

Explanation:

This is a related rates problem. The ladder leaning against the side of a edifice forms a correct triangle, with the 10ft ladder equally its hypotenuse. The Pythagorean Theorem, a^2 + b^2 = c^2 relates all three sides of this triangle to each other. Let be the height from the top of the ladder to the ground. Let exist the distance from the bottom of the ladder to the building. Since 10 is the hypotenuse, we have the following equation.

x^2 + y^2 = 10^2

Simplifying the correct side gives us

x^2 + y^2 = 100

Since and are variables, we will wait to plug values into them until after we take the derivative.

The question asks how fast the human standing on the summit of the ladder is falling when the ladder'southward base is 6ft from the building and is sliding away at 2 ft/sec. These 2 values, x=6 and $\frac{\mathrm{d} x}{\mathrm{d} t}=2$ , only happen at a unmarried instant in time. So we will find the derivative of the equation at this point in time.

Using implicit differentiation to discover the derivative with respect to time, we go

$2x \frac{\mathrm{d} x}{\mathrm{d} t} + 2y\frac{\mathrm{d} y}{\mathrm{d} t} = 0$

We only care about the instant that x=6 and $\frac{\mathrm{d} x}{\mathrm{d} t}=2$ . Nosotros need to discover the rate that the top of the ladder, and thus the man, is falling. And then nosotros want to solve for $\frac{\mathrm{d} y}{\mathrm{d} t}$ . Notwithstanding, nosotros will need to know what is at this instant in order to find an reply. Fortunately, the Pythagorean Theorem applies at all points in fourth dimension, so we tin can apply it for this particular instant to find.

6^2 + y^2 = 100

36 + y^2 = 100

y^2 = 64

$y = \pm 8$

Since we are dealing with physical distances, we will but apply the positive 8.

Plugging all the information into our derivative equation gives us

$2x \frac{\mathrm{d} x}{\mathrm{d} t} + 2y\frac{\mathrm{d} y}{\mathrm{d} t} = 0$

$2(6)(2) + 2(8)\frac{\mathrm{d} y}{\mathrm{d} t}=0$

$24+ 16\frac{\mathrm{d} y}{\mathrm{d} t}=0$

$16\frac{\mathrm{d} y}{\mathrm{d} t}= -24$

$\frac{\mathrm{d} y}{\mathrm{d} t}= -3/2$

The negative makes sense because the man is falling downward, so the top is getting smaller. Thus our answer is $-3/2 \text{ feet/sec}$

The velocity of a machine is given by the equation:

$v(t) = 25e^{-t} + 15t$ , where is the time in hours.

If the auto starts out at a distance of 3 miles from its home, how far will it be afterward 4 hours?

Correct answer:

$148 -\frac{25}{e^{4}}$

Explanation:

To observe the full altitude traveled, the velocity function has to be integrated from t=0 to t=4 hours:

$distance = \int_{t=0}^{t=4} v(t) dt = \int_{t=0}^{t=4} (25e^{-t} + 15t) dt$

$distance = -25e^{-4} + \frac{15(4)^{2}}{2} - (-25e^{0} + \frac{15(0)^{2}}{2})$

$distance = \frac{-25}{e^{4}} + 120 - (-25 + 0) = \frac{-25}{e^{4}} + 145$

Finally, the question is asking how far the car will exist from domicile. Information technology was 3 miles from habitation when t=0 , so at t=4 , it will be:

$3 + \frac{-25}{e^{4}} + 145 = 148 -\frac{25}{e^{4}}$ miles from habitation.

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