Functions - AP Calculus AB

Card 0 of 43260

Question

The initial position of a particle is 0. If its velocity is described by v(t) = 4t + 5, what is its position at the time when the velocity is equal to 45?

Answer

The position s(t) is equal to ∫v(t) dt = ∫ 4t + 5 dt = 2t2 + 5t + C

Now, since we know that the initial position of the particle (at t = 0) is 0, we know C is 0. Therefore, s(t) = 2t2 + 5t + C

To find the time t when at which the velocity is 45, set v(t) equal to 45. 45 = 4t + 5 → 40 = 4t → t = 10

The position of the particle is s(10) = 2 * 102 + 50 = 200 + 50 = 250

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Question

The position of a particle is defined by s(t) = 4t3 – 3t2 + 2t. At what time (to the nearest hundreth) is its velocity equal to 384?

Answer

The position of a particle is defined by s(t) = 4t3 – 3t2 + 2t. At what time (to the nearest hundreth) is its velocity equal to 384?

Start by finding the velocity function:

v(t) = s'(t) = 12t2 – 6t + 2

To find the time, set v(t) equal to 384: 384 = 12t2 – 6t + 2

To solve, set equal the equation equal to 0: 12t2 – 6t – 382 = 0

Use the quadratic formula:

(–(–6) ± √(36 – 412(–382)))/(2 * 12)

= (6 ± √(36 + 18336))/24 = (6 ± √(18372))/24 = (6 ± 2√(4593))/24

= (3 ± √(4593))/12

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Question

Find the derivative at x=3.

Answer

First, find the derivative using the power rule:

$\frac{d}{dx}6x^3=18x^2$

Now, substitute 3 for x.

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Question

The initial position of a particle is 44.5. If its velocity is described by v(t) = 3t + 12, what is its position at the time when the velocity is equal to 8391?

Answer

The initial position of a particle is 44.5. If its velocity is described by v(t) = 3t + 12, what is its position (to the nearest hundreth) at the time when the velocity is equal to 8391?

To solve for t, set v(t) equal to 8391: 3t + 12 = 8391; 3t = 8379; t = 2793

Now, the position function is equal to ∫v(t)dt = ∫ 3t + 12 dt = (3/2)t2 + 12t + C

Now, since the initial position is 44.5, C is 44.5; therefore, s(t) = (3/2)t2 + 12t + 44.5

To solve for the position, solve s(2793) = (3/2)27932 + 122793 + 44.5 = 1.5 7800849 + 33516 + 44.5 = 11701273.5 + 33516 + 44.5

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Question

Find the derivative.

$\frac{2}{3x^4}$

Answer

Use the quotient rule to find the derivative.

$\frac{3x^4(0)-2(12x^3)}{(3x^4)^2}=\frac{-24x^3}{9x^8}$

Simplify.

The derivative is $\frac{-8}{3x^5}$ .

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Question

Find the derivative.

$\frac{1}{x^5}$

Answer

Use the quotient rule to find the derivative.

$\frac{x^5(0)-1(5x^4)}{(x^5)^2}=\frac{-5x^4}{x^{10}}=\frac{-5}{x^6}$

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Question

Find the derivative.

$12x^{12}$

Answer

Use the power rule to find the derivative.

$\frac{d}{dx}12x^{12}=144x^{11}$

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Question

Find $\frac{dy}{dx}$ for the equation:

Answer

Note that:

Product Rule:

Take the derivative of each term in the equation twice: with respect to and then with respect to . When taking the derivative with respect to one variable, treat the other variable as a constant.

For the function

The derivative is then found using the product rule to be:

Notice how the chain rule needs to be utilized an additional time when taking the derivative of the term with respect to .

Now bring and terms to opposite sides of the equation:

Now rearraging variables gives $\frac{dy}{dx}$ :

$\frac{x(2sin(xy^2)+xy^2cos(xy^2))}{1-2x^3ycos(xy^2)}=\frac{dy}{dx}$

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Question

Find $f_{xyz}$ if $f(x,y,z)=e^{yz}cos(x^2z)$

Answer

For this problem, note that:

Product Rule

To solve this problem, differentiate the expression one variable at a time, treating other variables as constants:

If we're looking for $f_{xyz}$ for the function $f(x,y,z)=e^{yz}cos(x^2z)$ then we'll begin by differentiating with respect to first:

$f_x=-2xze^{yz}sin(x^2z)$

Next, differentiate with respect to :

$f_{xy}=-2xz^2e^{yz}sin(x^2z)$

Now finally we'll differentiate with respect to ; remember to use the product rule:

$f_{xyz}=-4xze^{yz}sin(x^2z)-2xz^2(ye^{yz}sin(x^2z)+x^2e^{yz}cos(x^2z))$

$f_{xyz}=-4xze^{yz}sin(x^2z)-2xyz^2e^{yz}sin(x^2z)-2x^3z^2e^{yz}cos(x^2z))$

$f_{xyz}=-2xz(2e^{yz}sin(x^2z)+yze^{yz}sin(x^2z)+x^2ze^{yz}cos(x^2z)))$

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Question

Find given:

Answer

To solve, simply find the first derivative and let . Thus,

$f'(x)=2x-16\Rightarrow f'(-1)=2(-1)-16=-2-16=-18$

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Question

Find the area between and the -axis.

Answer

First we find that the equation crosses the y-axis at y=-1, 1.

Partitioning the y-axis, we get the area is equal to

$\int_{-1}^{1}(1-y^2) dy=\frac{4}{3}$ .

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Question

Differentiate:

$f(x) = \left(\frac{1}{x}\right)\cos^2(x)$

Answer

To find the derivative of this function we must use the Product Rule and the Chain Rule. First we set

$h(x) = \frac{1}{x}$

and

$g(x) = \cos^2(x)$

Now differentiating both of these functions gives

$h'(x) = \frac{-1}{x^2}$

$g'(x) = 2\cos(x)(-\sin(x)) = -2\cos(x)\sin(x)$

Applying this to the Product Rule gives us,

$f'(x) = \frac{-1}{x^2}\cos^2(x)-\frac{2}{x}\cos(x)\sin(x)$

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Question

Differentiate the function

$f(x) = \sin(\cos(2x))$

Answer

To differentiate the function properly, we must use the Chain Rule which is,

$f(x) = h(g(x)) \rightarrow f'(x) = h'(g(x))g'(x)$

Therefore the derivative of the function is,

$f'(x) = \cos(\cos(2x))(-\sin(2x))(2) = -2\cos(\cos(2x))\sin(2x)$

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Question

Find the derivative.

$-\frac{5}{x^3}$

Answer

Use the quotient rule to find this derivative.

Recall the quotient rule:

$\ \ h(x)=\frac{f(x)}{g(x)} \ \ h'(x)=\frac{g(x)f'(x)-g'(x)f(x)}{(g(x))^2}$

$h'(x)=\frac{x^3(0)-(-5)3x^2}{x^6}=\frac{15x^2}{x^6}=\frac{15}{x^4}$

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Question

Differentiate

$f(x) = \sin^2(x^2)$

Answer

To differentiate this equation we use the Chain Rule.

$f(x) = h(g(x)) \rightarrow f'(x) = h'(g(x))\cdot g'(x)$

Using this throughout the equation gives us,

$f'(x) = 2\sin(x^2)\cos(x^2)(2x)$

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Question

Find the derivative of

$f(x) = 4\sin(x+3x^2)$

Answer

To find the derivative of the function we must use the Chain Rule, which is

$f(x) = h(g(x)) \rightarrow f'(x) = h'(g(x))\cdot g'(x)$

Applying this to the function we get,

$f'(x) = 4\cos(x+3x^2)(1+6x)$

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Question

Find the slope of the function at .

Answer

To consider finding the slope, let's discuss the topic of the gradient.

For a function , the gradient is the sum of the derivatives with respect to each variable, multiplied by a directional vector:

$\frac{\delta f}{\delta x_1}\widehat{e_1}+\frac{\delta f}{\delta x_2}\widehat{e_2}+...+\frac{\delta f}{\delta x_n}\widehat{e_n}$

It is essentially the slope of a multi-dimensional function at any given point

Knowledge of the following derivative rules will be necessary:

The approach to take with this problem is to simply take the derivatives one at a time. When deriving for one particular variable, treat the other variables as constant.

at

x:

$\frac{\delta f}{\delta x}=12x^2+6xy+2y^2;12(1)^2+6(1)(1)+2(1)^2=20$

y:

$\frac{\delta f}{\delta y}=3x^2+4xy+3y^3;3(1)^2+4(1)(1)+3(1)^3=10$

The slope is

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Question

Find the derivative of

$f(x) =8\sin(4x+3)$

Answer

To find the derivative of the function we must use the Chain Rule

$f(x) = h(g(x)) \rightarrow f'(x) = h'(g(x))\cdot g'(x)$

Applying this to the function we are given gives,

$f'(x) = 8\cos(4x+3)(4) = 32\cos(4x+3)$

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Question

Using the method of midpoint Reimann sums, approximate the area of the region between the functions $f(x)=10(1-e^{-x})$ and over the interval using three midpoints.

Answer

The Reimann sum approximation of an integral of a function with subintervals over an interval takes the form:

$\int_a^b f(x)dx \approx \frac{b-a}{n}[f(x_1)+f(x_2)+...+f(x_n)]$

Where $\frac{b-a}{n}$ is the length of the subintervals.

For this problem, since there are three midpoints, the subintervals have length $\frac{3-0}{3}=1$ , and the midpoints are .

Furthermore, since the question is asking for the area between the two functions, it's asking for the difference between the larger function over the interval, f(x), and the smaller function, g(x).

The integral is thus:

${\int_0^{3} (f(x)-g(x))dx \approx [(10(1-e^{-0.5})-2(0.5))+(10(1-e^{-1.5})-2(1.5))+(10(1-e^{-2.5})-2(2.5))]}$

${\int_0^{3} (f(x)-g(x))dx \approx 11.8825$

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Question

A Riemann Sum approximation of an integral $\int_a^b (f(x))dx$ follows the form

$\frac{b-a}{n}\sum_{i=1}^{n}f(x_i)$ .

Where n is number of points/subintervals used to approximate the integral.

Knowing this, imagine a modified style of Riemann Sum, such that the subintervals are not of uniform width.

Denoting a particular subinterval's width as ,

the integral approximation becomes

$\sum_{i=1}^{n}I_if(x_i)$ .

Which of the following parameters would give the closest integral approximation of the function:

$\int_{10}^{100} \frac{e^x}{x^2}dx$ ?

Answer

The more points/intervals that are selected, the closer the Reimann sum approximation becomes to the actual integral value, so n=20 will give a closer approximation than n=10:

Now, assuming the amount of points/intervals are fixed, but the actual widths can differ, it would be prudent to have narrower intervals at those regions where the graph is steepest. Relatively flatter regions of the function do not lead to the approximation overshooting or undershooting as much as the steeper areas, as can be seen in the above figure.

Consider the function $\frac{e^x}{x^2}$ over the interval . The steepness of this graph at a point can be found by taking the function's derivative.

The quotient rule of derivatives states:

$d\frac{u}{v}=\frac{vdu-udv}{v^2}$

So the derivative is

$\frac{x^2e^x-2xe^x}{x^4}$

$\frac{e^x(x-2)}{x^3}$

It might be intuitive to see that the steepness is positive and that it gets progressively greater over the specified interval, although to be precise, we may take the derivative once more to find the rate of change of this steepness and see if it's postive or negative:

$\frac{x^4(2xe^x+x^2e^x-2e^x-2xe^x)-4x^3(x^2e^x-2xe^x)}{x^8}$

$\frac{x^6e^x+6x^4e^x-4x^5e^x}{x^8}$

$\frac{e^x(x^2-4x+6)}{x^4}$

Since this derivative is positive over the specified interval, , and since the slope was shown to be positive as well, the slope is ever-increasing and growing steeper.

The intervals should in turn grow increasingly thinner.

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