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AP Calculus AB : Applications of Derivatives

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Example Questions

Example Question #1 : Applications Of Derivatives

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

a) Find the rate at which the angle $\theta$ opposite y(t) is changing with respect to time.

Possible Answers:

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

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

$\frac{d\theta}{dt}=\sin(t)+\frac{1}{1+t}$

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

$\frac{d\theta}{dt}=\cos(t)-\frac{1}{1+t}$

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 $\theta$ opposite y(t) is changing with respect to time.

Quesstion 9 related rates triangle

a) First we need to write an expression for the angle $\theta$ as a function of . Because the angle is opposite the side we know that the tangent is simply 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 we need to differentiate with respect to .

Recall the the general derivative for the inverse tangent function 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, we obtain:

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

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

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Example Question #1 : Modeling Rates Of Change, Including Related Rates Problems

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

Possible Answers:

$\frac{1}{80\pi} mm^2/sec$

1 mm^2/sec

80 mm^2/sec

$80\pi mm^2/sec$

Correct answer:

1 mm^2/sec

Explanation:

To determine the rate of change of the surface area of the spherical bubble, we must relate it to something we do know the rate of change of - the volume.

The volume of a sphere is given by the following:

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

The rate of change of the volume is given by the derivative with respect to time:

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

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 can later solve for the rate of change of surface area:

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

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

Next, we must find the surface area and rate of change of the surface area of the sphere the same way as above:

$A=4\pi r^2$

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

Plugging in the known rate of change of the surface area at the specified radius, and this radius into the rate of surface area change function, we get

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

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Example Question #2 : Modeling Rates Of Change, Including Related Rates Problems

A pizzeria chef is flattening a circular piece of dough. The surface area of the dough (we are only considering the top of the dough) is increasing at a rate of 0.5 inches/sec. How quickly is the diameter of the pizza changing when the radius of the pizza measures 4 inches?

Possible Answers:

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

inches/sec

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

inches/sec

Correct answer:

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

Explanation:

To find the rate of change of the diameter, we must relate the diameter to something we do know the rate 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, then, is found 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 rate 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 relate the diameter to the radius of the pizza dough:

2r=d

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

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

Plugging in the known rate of change of the radius at the given radius, we get

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

We could have found this directly by writing our surface area formula in terms of diameter, however the process we used is more applicable to problems in which the related rate of change is of something not as easy to manipulate.

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Example Question #4 : Modeling Rates Of Change, Including Related Rates Problems

A spherical balloon is increasing in volume at a constant 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?

Possible Answers:

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

$\frac{1}{90\pi} cm/min$

$\frac{1}{360} cm/min$

$\frac{1}{360\pi} cm/min$

$\frac{1}{90} cm/min$

Correct answer:

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

Explanation:

To determine the rate of change of the circumference at a given radius, we must relate 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 balloon,

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

we take the derivative of the function with respect to time, giving us the rate 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 found using the following 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 function 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, we get

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

Now, we use this rate of change and apply it to the rate of change of the circumference, which we get by taking the derivative of the circumference with respect to time:

$c=2\pi r$

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

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

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

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Example Question #1 : Modeling Rates Of Change, Including Related Rates Problems

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

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

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

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

Correct answer:

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

Explanation:

To determine the rate of the change of the angle opposite to the base of the given right triangle, we must relate it to the rate of change of the base of the triangle when the triangle is a certain area.

First 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

Now, we must find something that relates the angle opposite of the base to the length of the base and height - the tangent of the angle:

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

To find the rate of change of the angle, we take the derivative of both sides with respect to time, keeping in mind that the base 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 change 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 information in and solving for the rate of change of the angle adjacent to the base, 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

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Example Question #2 : Applications Of Derivatives

The position of a car is given by the equation

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

Find the car's acceleration when t = 4 .

Possible Answers:

$\pi$ + 14

$-\pi^{3}$

Correct answer:

Explanation:

To find the car's acceleration, take 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:

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

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Example Question #1 : Applications Of Derivatives

A point on a circle of radius 1 unit is orbiting counter-clockwise around the circle's center. It makes one full orbit every 8 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 x axis?

Possible Answers:

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

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

$\frac{1}{2}$

$-\frac{\sqrt{3}\pi}{8}$

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

Correct answer:

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

Explanation:

This is a related rates problem.

Since the problem gives the time for one orbit, we 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 by dividing the amount of radians in one revolution, $2\pi$ , by the time it takes to travel one revolution, 8 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 need to relate the position to the angle, $\theta$ . Recall that

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

Since the radius is given as 1 unit, we can write this equation as

$x = \cos \theta$ .

Now we take the derivative of both sides with respect to time, using implicit differentiation. Remember 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 have a formula that relates the horizontal speed of the particle at a instant in time, $\frac{\mathrm{d} x}{\mathrm{d} t}$ , to the angle above the positive x axis and angular speed at that same instant. We are told to find how fast the x coordinate is changing when the angle, $\theta$ is $\frac{\pi}{4}$ radians above the positive x axis. So we will plug in $\frac{\pi}{4}$ for $\theta$ . However we also need to know $\frac{\mathrm{d} \theta}{\mathrm{d} t}$ . Fortunately, we already found it. It is the angular speed, $\frac{\pi}{4}$ radians/second.

Plugging this information in, we 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, because the point is traveling counter-clockwise. Hence, it is moving left when the angle is $\frac{\pi}{4}$ radians.

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Example Question #2 : Applications Of Derivatives

A man is standing on the top of a 10 ft long ladder that is leaning against the side of a building, when the bottom of the ladder begins to slide out from under it. How fast is the man standing on the top of the ladder falling when the bottom of the ladder is 6 ft from the building and is sliding at 2ft/sec?

Possible Answers:

$-1/4 \text{ feet/sec}$

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

$32 \text{ feet/sec}$

$3/4 \text{ feet/sec}$

Correct answer:

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

Explanation:

This is a related rates problem. The ladder leaning against the side of a building forms a right triangle, with the 10ft ladder as 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 be 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 right 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 man standing on the top of the ladder is falling whenthe ladder's base is 6ft from the building and is sliding away at 2 ft/sec. These two values, x=6 and $\frac{\mathrm{d} x}{\mathrm{d} t}=2$ , only happen at a single instant in time. So we will find the derivative of the equation at this point in time.

Using implicit differentiation to find the derivative with respect to time, we get

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

We only care about the instant that x=6 and $\frac{\mathrm{d} x}{\mathrm{d} t}=2$ . We need to find the rate that the top of the ladder, and thus the man, is falling. So we want to solve for $\frac{\mathrm{d} y}{\mathrm{d} t}$ . However, we will need to know what is at this instant in order to find an answer. Fortunately, the Pythagorean Theorem applies at all points in time, so we can use 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 only use 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 down, so the height is getting smaller. Thus our answer is $-3/2 \text{ feet/sec}$

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Example Question #4 : Modeling Rates Of Change, Including Related Rates Problems

The velocity of a car is given by the equation:

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

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

Possible Answers:

$98 + \frac{25}{e^{4}}$

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

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

$145 - \frac{25}{e^{4}}$

$95 + \frac{25}{e^{4}}$

Correct answer:

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

Explanation:

To find the total distance 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 be from home. It was 3 miles from home when t=0 , so at t=4 , it will be:

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

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Example Question #1 : Applications Of Derivatives

Given the function $\inline y=sin(2x)+x^2z+z^3$ ,

find $\frac{\mathrm{d} y}{\mathrm{d} x}$ at $(x,z)=(\frac{\pi}{2},3)$ .

Possible Answers:

$-2-3\pi$

$2-3\pi$

$2+3\pi$

$-2+3\pi$

$-3\pi$

Correct answer:

$-2+3\pi$

Explanation:

We are differentiating with respect to . So we can treat as a constant.

Therefore, the derivative becomes $y'=2cos(2x)+2xz$ .

By plugging in $\frac{\pi }{2}$ and $z=3$ , we get $-2+3\pi$ .

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AP Calculus AB : Applications of Derivatives

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #1 : Applications Of Derivatives

Example Question #1 : Modeling Rates Of Change, Including Related Rates Problems

Example Question #2 : Modeling Rates Of Change, Including Related Rates Problems

Example Question #4 : Modeling Rates Of Change, Including Related Rates Problems

Example Question #1 : Modeling Rates Of Change, Including Related Rates Problems

Example Question #2 : Applications Of Derivatives

Example Question #1 : Applications Of Derivatives

Example Question #2 : Applications Of Derivatives

Example Question #4 : Modeling Rates Of Change, Including Related Rates Problems

Example Question #1 : Applications Of Derivatives

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AP Calculus AB : Applications of Derivatives

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #1 : Applications Of Derivatives

Example Question #1 : Modeling Rates Of Change, Including Related Rates Problems­

Example Question #2 : Modeling Rates Of Change, Including Related Rates Problems­

Example Question #4 : Modeling Rates Of Change, Including Related Rates Problems­

Example Question #1 : Modeling Rates Of Change, Including Related Rates Problems­

Example Question #2 : Applications Of Derivatives

Example Question #1 : Applications Of Derivatives

Example Question #2 : Applications Of Derivatives

Example Question #4 : Modeling Rates Of Change, Including Related Rates Problems­

Example Question #1 : Applications Of Derivatives

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Example Question #1 : Modeling Rates Of Change, Including Related Rates Problems

Example Question #2 : Modeling Rates Of Change, Including Related Rates Problems

Example Question #4 : Modeling Rates Of Change, Including Related Rates Problems

Example Question #1 : Modeling Rates Of Change, Including Related Rates Problems

Example Question #4 : Modeling Rates Of Change, Including Related Rates Problems