AP Calculus AB : Applications of antidifferentiation

Study concepts, example questions & explanations for AP Calculus AB

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Example Questions

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Example Question #1 : Solving Separable Differential Equations And Using Them In Modeling

Find (dy/dx). 

sin(xy) = x + cos(y)

Possible Answers:

dy/dx = (cos(xy) + sin(y))/(1 – cos(xy))

dy/dx = (1 – ycos(xy))/(xcos(xy) + sin(y))

dy/dx = (xcos(xy) + sin(y))/(1 – ycos(xy)) 

dy/dx = (1 – cos(xy))/(cos(xy) + sin(y))

None of the above

Correct answer:

dy/dx = (1 – ycos(xy))/(xcos(xy) + sin(y))

Explanation:

The first step of the problem is to differentiate with respect to (dy/dx):

cos(xy)[(x)(dy/dx) + y(1)] = 1 – sin(y)(dy/dx)

*Note: When differentiating cos(xy) remember to use the product rule. (xy' + x'y)

Step 2: Clean the differentiated problem up 

cos(xy)(x)(dy/dx) + cos(xy)y = 1 – sin(y)(dy/dx)

cos(xy)(x)(dy/dx) + sin(y)(dy/dx) = 1 – cos(xy)y

Step 3: Solve for (dy/dx)

dy/dx = (1 – ycos(xy))/(xcos(xy) + sin(y))

 

Example Question #2 : Solving Separable Differential Equations And Using Them In Modeling

Find the equation of the normal line at  on the graph y=x^{3}-6x+4.

Possible Answers:

y=\frac{-1}{6}x+\frac{1}{3}

y'=3x^{2}-6

y=\frac{-1}{6}x+2

Correct answer:

y=\frac{-1}{6}x+\frac{1}{3}

Explanation:

The answer is y=\frac{-1}{6}x+\frac{1}{3}.

 

y=x^{3}-6x+4

y'=3x^{2}-6   

Now plug in .

y'=3(2)^{2}-6 = 6  now we know 6 is the slope for the tangent line. However, we aren't looking for the slope of the tangent line. The slope of the normal line is the negative reciprocal of the tangent's slope; meaning the slope of the normal is \frac{-1}{6}. Now find the equation of the normal line.

y-0=\frac{-1}{6}(x-2)

y=\frac{-1}{6}x+\frac{1}{3}

Example Question #1 : Solving Separable Differential Equations And Using Them In Modeling

f(x) = \frac{x^3}{1-x^2}

What is the derivative of

Possible Answers:

-x

\frac{x^2(3-5x^2)}{1-x^2}

\frac{-2x^4-3 x^2(1-x^2)}{(1-x^2)^2}

\frac{x^2(3-x^2)}{(1-x^2)^2}

Correct answer:

\frac{x^2(3-x^2)}{(1-x^2)^2}

Explanation:

Use the quotient rule. 

Example Question #1 : Applications Of Antidifferentiation

Find  if

 y=\frac{ln(x)}{x^{3}}

Possible Answers:

y'=\frac{1-3ln(x)}{x^{4}}

\frac{ln(x)-1}{x^{4}}

\frac{3}{x^{3}}

\frac{1}{x^{4}}

\frac{1+3ln(x)}{x^{4}}

Correct answer:

y'=\frac{1-3ln(x)}{x^{4}}

Explanation:

The answer is

 y'=\frac{1-3ln(x)}{x^{4}}

 

y=\frac{ln(x)}{x^{3}}

y'=\frac{(\frac{1}{x})x^{3}-ln(x)(3x^{3})}{x^{6}}

y'=\frac{x^{2}(1-3ln(x))}{x^{6}}

y'=\frac{1-3ln(x)}{x^{4}}

Example Question #4 : Solving Separable Differential Equations And Using Them In Modeling

Find the derivative: 

Possible Answers:

Correct answer:

Explanation:

To find the derivative, multiply the exponent by the coefficent in front of the x term and then decrease the exponent by 1:

 

Example Question #2 : Solving Separable Differential Equations And Using Them In Modeling

Find the solution to the equation y'=y at x=2 with initial condition y(0)=2.

Possible Answers:

1

2e^2

e

2e

e^2

Correct answer:

2e^2

Explanation:

First, we need to solve the differential equation of y'=y.

, where is a constant

, where is a constant

To find , use the initial condition, , and solve:

Therefore, y=2e^x.

Finally, at , y(2)=2e^2.

Example Question #3 : Solving Separable Differential Equations And Using Them In Modeling

Solve the differential equation: 

Note that  is on the curve. 

Possible Answers:

Correct answer:

Explanation:

In order to solve differential equations, you must separate the variables first. 

Since point  is on the curve, .

To get rid of the log, raise every term to the power of e:

Example Question #21 : Integrals

Suppose $1000 is invested in an account that pays 4.3% interest compounded continuously. Find an expression for the amount in the account after time .

Possible Answers:

y(t)=43e^{1000t}

y(t)=1000e^{t}

y(t)=4.3e^{1000t}

y(t)=1000e^{0.043t}

y(t)=4.3e^{t}

Correct answer:

y(t)=1000e^{0.043t}

Explanation:

The differential equation is \frac{dy}{dt}=0.043y, with boundary condition y(0)=1000.

 

This is a separable first order differential equation.

\frac{1}{y}dy=0.043dt

Integrate both sides.

ln(y)=0.043t+c

y=Ce^{0.043t}

Plug in the initial condition above to see that .

Example Question #4 : Solving Separable Differential Equations And Using Them In Modeling

Find the solution to the differential equation

 when .

Possible Answers:

Correct answer:

Explanation:

First, separate the variables of the original differential equation:

.  

Then, take the antiderivative of both sides, which gives

Use the given condition , plugging in 

 and , to solve for .  This gives , so the correct answer is

 

.

Example Question #31 : Integrals

Differentiate .

Possible Answers:

Correct answer:

Explanation:

While differentiating, multiply the exponent with the coefficient then subtract the exponent by one. 

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