Applications of antidifferentiation - AP Calculus AB

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Question

A ball is thrown into the air. It's height, after t seconds is modeled by the formula:

$h(t)=-15t^2$+30t feet.

At what time will the velocity equal zero?

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Answer

In order to find where the velocity is equal to zero, take the derivative of the function and set it equal to zero.

h(t) = –15t2 + 30t

h'(t) = –30t + 30

0 = –30t + 30

Then solve for "t".

–30 = –30t

t = 1

The velocity will be 0 at 1 second.

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Question

A ball is thrown into the air. It's height, after t seconds is modeled by the formula:

$h(t)=-15t^2$+30t feet.

At what time will the velocity equal zero?

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Answer

In order to find where the velocity is equal to zero, take the derivative of the function and set it equal to zero.

h(t) = –15t2 + 30t

h'(t) = –30t + 30

0 = –30t + 30

Then solve for "t".

–30 = –30t

t = 1

The velocity will be 0 at 1 second.

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Question

Find (dy/dx).

sin(xy) = x + cos(y)

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Answer

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))

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Question

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 .

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Answer

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 .

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Question

Find if

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

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Answer

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}$$}

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Question

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

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Answer

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}$x+\frac{1}{3}$

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Question

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

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Answer

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

$$\frac{dy}{dx}$=y$

$$\frac{dy}{y}$=dx$

$\int $\frac{dy}{y}$=\int dx$

, where is a constant

, where is a constant

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

Therefore, $y=2e^x$.

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

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Question

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

What is the derivative of ?

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Answer

Use the quotient rule.

${y}' = $$\frac{(1-x^2$$)(3x^2$$)-x^3$$(-2x)}{(1-x^2$$)^2$}$ = $$\frac{x^2$$(3-3x^2$+ $2x^2$$)}{(1-x^2$$)^2$}$ = $$\frac{x^2$$(3-x^2$$)}{(1-x^2$$)^2$}$$

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Question

The rate of a chemical reaction is given by the following differential equation:

$\frac{-d[A]}{dt}$ = $k[A]^{2}$ ,

where $\left [ A \right ]$ is the concentration of compound at a given time, . Which one of the following equations describes $\frac{1}{\left[ A \right]}$ as a function of time? Let $\left [ A \right $]_{0}$$ be the concentration of compound when .

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Answer

To begin with, the differential equation needs to be rearranged so that each variable is one side of the equation:

$$\frac{-d[A]}{dt}$ = $k[A]^{2}$ \Rightarrow $$\frac{d[A]}{[A]^{2}$$} = -kdt$ .

Then, integrate each side of the rate law, bearing in mind that $\left [ A \right ]$ will range from $\left [ A \right $]_{0}$$ to $\left [ A \right ]$ , and time will range from to :

After integrating each side, the equation becomes:

$$\frac{-1}{[A]}$ = -kt$ .

The left side has to be evaluated from $\left [ A \right $]_{0}$$ to $\left [ A \right ]$ , and the right side is evaluated from to :

$$\frac{-1}{[A]}$ - ( $$\frac{-1}{[A_{0}$$]}) = -kt - (-k(0))$ . This becomes:

.

Finally, rearranging gives:

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Question

Differentiate .

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Answer

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

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Question

If what is ?

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Answer

Taking the derivative of gives you .

Taking the derivative of gives you .

Finally taking the derivative of gives you .

Therefore .

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Question

Solve the differential equation: $$\frac{dy}{dx}$=-2xy$

Note that is on the curve.

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Answer

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

$$\frac{dy}{y}$=-2xdx$

$\int $\frac{dy}{y}$= \int -2xdx$

Since point is on the curve, .

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

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Question

Find the solution to the differential equation

$$\frac{dy}{dx}$ = $$\frac{x^{2}$$+ $4x}{y^{2}$}$ when .

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Answer

First, separate the variables of the original differential equation:

$$\frac{dy}{dx}$ = $$\frac{x^{2}$$+ $4x}{y^{2}$} \Rightarrow $y^{2}$dy = $(x^{2}$ + 4x)dx$ .

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

.

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Question

Find the derivative:

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Answer

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

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Question

Solve the following separable differential equation with initial condition .

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Answer

We proceed as follows

. Start

$\frac{dy}{dx}$ = $ye^x$ . Rewrite as $\frac{dy}{dx}$ .

. Multiply both sides by , and divide both sides by .

$\ln|y| $=e^x$ +C$ . Integrate both sides. Do not forget the on one of the sides.

Substitute the initial condition .

.

. Solve for .

$\ln|y| = $e^x$+1$

$y = $e^{e^x+1}$$ . Exponentiate both sides .

$y = $ee^{e^x}$$ . Rule of exponents.

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Question

Solve the separable, first-order differential equation for :

$(y+1)$\frac{dy}{dx}$=2x-8+\frac{dy}{dx}$$

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Answer

Solve the separable, first-order differential equation for :

$(y+1)$\frac{dy}{dx}$=2x-8+\frac{dy}{dx}$$

First collect all the terms with the derivative to one side of the equation.

$(y+1)$\frac{dy}{dx}$-$\frac{dy}{dx}$=2x-8$

$$\frac{dy}{dx}$[(y+1)-1]=2x-8$

$y$\frac{dy}{dx}$=2x-8$

Important Conceptual Note: often in texts on differential equations differentials often appear to have been rearranged algebraically as if $\frac{dy}{dx}$ is a "fraction," making it appear as if we "multiplied both sides" by to get: . This is not the case. The derivative is a limit by definition and, when the limit exists, can take on any real number which includes irrational numbers i.e. numbers which cannot be written as a ratio of two integers.

For instance, we cannot represent $\pi$ as a ratio, but some functions may have a derivative at a point such that the derivative is equal to $\pi$ , or a funciton may simply have an irrational number like $\pi$ as a derivative. For instance, if $y=\pi x$ we write the derivative $$\frac{dy}{dx}$=\pi$ . Claiming that and are representative of a "numerator" and "denominator" respectively, we would essentially be claiming to have found a way to write an irrational number, such as $\pi$ as a ratio, which is preposterous. The expression $\frac{dy}{dx}$ is simply notation.

Here is what we are really doing.

$\int y$\frac{dy}{dx}$dx=\int (2x-8)dx$

$\int y dy=\int (2x-8)dx$

Note that the constants of integration can just be combined into one constant by defining .

Solve for :

$y =\pm $$\sqrt{2x^2$ -16x+C }$$

Applying the initial condition:

$y(0)=\pm $\sqrt{C}$=8$

$y =\pm $$\sqrt{2x^2$ -16x+16 }$$

Here we have two possible solutions. However, because of the initial condition, we can easily rule out the negative solution. must be equal to positive .

$y(0)=8: : : : : : \Rightarrow: : : : : :$\sqrt{16}$=8$

$y $=$\sqrt{2x^2$ -16x+16 }$$

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Question

Solve the separable differential equation

$\frac{\mathrm{d}$ y}{\mathrm{d} $x}=x^2y$

given the condition

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Answer

To solve this equation, we must separate the variables such that terms containing x and y are on the same side as dx and dy, respectively:

Integrating both sides of the equation, we get

$\ln \left | y\right |+C_$1=\frac{x^3$}{3}$+C_2$

The integrals were found using the following rules:

$\int $\frac{dx}{x}$=\ln\left| x\right|+C$ , $\int $x^n$ $dx=\frac{x^{n+1}$$}{n+1}+C$

After combining the constants of integration into a single C, exponentiating both sides, and using the properties of exponents to simplify, we get

To solve for C, we use the condition given:

Our final answer is

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Question

is a function of . Solve for in this differential equation:

$$\frac{\mathrm{d}$ y}{\mathrm{d} x} = $\sqrt{\frac{ $e^{y}$$}{x}}$

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Answer

First, rewrite the expression on the right as the $\frac{1}{2}$ power of the radicand:

$$\frac{\mathrm{d}$ y}{\mathrm{d} x} = $\sqrt{\frac{ $e^{y}$$}{x}}$

$$\frac{\mathrm{d}$ y}{\mathrm{d} x} =\left( $\frac{ $e^{y}$$}{x} \right $)^{$\frac{1}${2}$}$

$$\frac{\mathrm{d}$ y}{\mathrm{d} x} = $\frac{ $(e^{y}$$$)^{$\frac{1}$${2}$}}{x^{$\frac{1}${2}$}}$

$$\frac{\mathrm{d}$ y}{\mathrm{d} x} = $\frac{ e^ ${\frac{y}{2}$}}{x^{$\frac{1}${2}$}}$

The expressions with can be separated from those with by multiplying both sides by $e^ {-$\frac{y}{2}$ }\mathrm{; d} x$ :

$$\frac{\mathrm{d}$ y}{\mathrm{d} x} \cdot e^ {-$\frac{y}{2}$ }\mathrm{; d} x = $\frac{ e^ ${\frac{y}{2}$}}{x^{$\frac{1}${2}$}} \cdot e^ {-$\frac{y}{2}$ }\mathrm{; d} x$

$e^ {-$\frac{y}{2}$ } \mathrm{d} y= $\frac{ $1}{x^{\frac{1}$${2}}} \mathrm{; d} x$

$e^ {-$\frac{1}{2}$ y} \mathrm{d} y= $x^{-$\frac{1}${2}$} \mathrm{; d} x$

Find the indefinite integral of both sides:

$\int e^ {-$\frac{1}{2}$ y} \mathrm{d} y= \int $x^{-$\frac{1}${2}$} \mathrm{; d} x$

The expression on the right can be integrated using the Power Rule. On the right, use some -substitution, setting $u = -$\frac{1}{2}$ y$ ; this makes $\mathrm{d} u = -$\frac{1}{2}$ \mathrm{d}y$ and $\mathrm{d}y = -2 \mathrm{; d} u$ :

$-2 \int e^ {u} \mathrm{d} u= $\frac{x\frac{1}{2}$}{$\frac{1}{2}$}+ C$

$-2 e^ {u} =2 $\sqrt{x}$+ C$

Apply some algebra to solve for :

$e^ {u} =- $\sqrt{x}$+ C$

$u =\ln (- $\sqrt{x}$+ C)$

Substitute $-$\frac{1}{2}$ y$ back for , and apply some algebra:

$-$\frac{1}{2}$ y =\ln (- $\sqrt{x}$+ C)$

$y =-2 \ln (- $\sqrt{x}$+ C)$

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Question

Solve the separable differential equation

$$\frac{\mathrm{d}$ y}{\mathrm{d} $x}=y(x^2$+5x+2)$

where

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Answer

To solve the separable differential equation, we must separate x and y to the same sides as their respective derivatives:

Next, we integrate both sides:

$\ln \left | y\right $|=\frac{x^3$$}{3}$+\frac{5x^2$}{2}$+2x+C$

The integrals were solved using the following rules:

$\int $\frac{dx}{x}$=\ln\left| x\right|+C$ , $\int $x^n$ $dx=\frac{x^{n+1}$$}{n+1}+C$

The two constants of integration were combined to make a single one.

Now, we exponentiate both sides to solve for y, keeping in mind rules for exponents which allow us to move the integration constant to the front:

To solve for the constant of integration, we use the condition given:

Our final answer is

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Question

Solve the following separable differential equation:

$$\frac{\mathrm{d}$ y}{\mathrm{d} x}= $y\sec^2$(x)$

given the condition that at $x=\pi, y=-2$

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Answer

To solve the separable differential equation, we must separate x and y and their respective derivatives to either side of the equal sign:

Now, we integrate both sides of the equation:

$\ln \left | y\right |=\tan(x)+C$

The integrals were found using their identical rules.

Exponentiating both sides of the equation to solve for y - and keeping in mind the rules of exponents - we get

Now, we solve for the integration constant by using the condition given:

Our final answer is

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