Determine the first derivative of each of the following functions. Print your answer in the box provided. You do not have to simplify your answers.
- \(\displaystyle g(x) = 3x^5 - \frac{20}{x^2} - 900\).
- \(\displaystyle h(r) = 5r \cos(r - 3) - 4r + 10\)
- \(\displaystyle f(t) = \arctan(t)\)
- \(\displaystyle R(x) = 5x - \frac{e^x}{x^2 - 1}\)
- \(\displaystyle P(x) = 5x^2 \tan\left(\sqrt{x} - 1\right)\)
Solution:
- Rewrite \(-\dfrac{20}{x^2}\) as \(-20x^{-2}\), then differentiate term by term: \[ \frac{d}{dx}(3x^5)=15x^4,\qquad \frac{d}{dx}(-20x^{-2})=40x^{-3}. \] Therefore \[ \boxed{g'(x)=15x^4+\frac{40}{x^3}}. \]
- Use the product rule on \(5r\cos(r-3)\), and differentiate \(-4r+10\) term by term. Since \[ \frac{d}{dr}\cos(r-3)=-\sin(r-3), \] we get \[ h'(r)=5\cos(r-3)+5r(-\sin(r-3))-4. \] Hence \[ \boxed{h'(r)=5\cos(r-3)-5r\sin(r-3)-4}. \]
- This is the standard inverse-trig derivative: \[ \boxed{\frac{d}{dt}\arctan t=\frac{1}{1+t^2}}. \]
- Differentiate \(\dfrac{e^x}{x^2-1}\) by the quotient rule. With \(N(x)=e^x\) and \(D(x)=x^2-1\), we have \(N'(x)=e^x\) and \(D'(x)=2x\). Thus \[ \frac{d}{dx}\left(\frac{e^x}{x^2-1}\right)=\frac{e^x(x^2-1)-2xe^x}{(x^2-1)^2}. \] Since \(R(x)=5x-\dfrac{e^x}{x^2-1}\), we subtract this result: \[ \boxed{R'(x)=5-\frac{e^x(x^2-2x-1)}{(x^2-1)^2}}. \]
- This is a product of \(5x^2\) and \(\tan(\sqrt{x}-1)\). Apply the product rule: \[ P'(x)=10x\tan(\sqrt{x}-1)+5x^2\frac{d}{dx}\tan(\sqrt{x}-1). \] Now use the chain rule on \(\tan(\sqrt{x}-1)\): \[ \frac{d}{dx}\tan(\sqrt{x}-1)=\sec^2(\sqrt{x}-1)\cdot\frac{1}{2\sqrt{x}}. \] Substituting gives \[ \boxed{P'(x)=10x\tan(\sqrt{x}-1)+\frac{5}{2}x^{3/2}\sec^2(\sqrt{x}-1)}. \]
Determine the the most general anti-derivative of each of the following expressions. Print your answer in the box provided.
- \(\displaystyle 10x^2 + 4x - 100\)
Most general anti-derivative:
- \(\displaystyle 4\sin(t) - 8\)
Most general anti-derivative:
- \(\displaystyle x \cos \left(x^2 + 1\right)\)
Most general anti-derivative:
out of a possible 5 points
Solution:
- \(\displaystyle \int(10x^2+4x-100)dx=\boxed{\frac{10}{3}x^3+2x^2-100x+C}\).
- \(\displaystyle \int(4\sin t-8)dt=\boxed{-4\cos t-8t+C}\).
- Let \(u=x^2+1\), so \(du=2x\,dx\). Then \[ \int x\cos(x^2+1)dx=\frac12\int\cos u\,du=\boxed{\frac12\sin(x^2+1)+C}. \]
Evaluate the following definite integrals. Print your answer in the box provided.
- \(\displaystyle \int_{1}^{2} (e^{t} - t^{-1}) dt\)
Value:
- \(\displaystyle \int_{0}^{\pi / 8} \sec^{2}(2x) dx\)
Value:
- \(\displaystyle \int_{0}^{1/2} \frac{x}{\sqrt{1 - x^{4}}} dx\)
Value:
Solution:
- \[ \int_1^2(e^t-t^{-1})dt=\bigl[e^t-\ln t\bigr]_1^2=e^2-e-\ln2. \] So the value is \(\boxed{e^2-e-\ln2}\).
- Let \(u=2x\). Then \(du=2dx\), so \[ \int_0^{\pi/8}\sec^2(2x)dx=\frac12\int_0^{\pi/4}\sec^2u\,du=\frac12\tan\left(\frac\pi4\right)=\boxed{\frac12}. \]
- Let \(u=x^2\), so \(du=2x\,dx\). Then \[ \int_0^{1/2}\frac{x}{\sqrt{1-x^4}}dx=\frac12\int_0^{1/4}\frac{du}{\sqrt{1-u^2}}=\boxed{\frac12\arcsin\left(\frac14\right)}. \]
Make a sketch of \(\displaystyle f(x)\) given that \(\displaystyle f(0) = 1\), where the graph of \(\displaystyle \frac{df}{dx}\) is shown below.
(The grid lines are one unit apart.)
(Graph of \(\displaystyle \frac{df}{dx}\) vs \(\displaystyle x\) with a wavy curve and labeled axes)
(Empty grid for sketching \(\displaystyle f\) vs \(\displaystyle x\) with labeled axes)
IMAGE
Solution: Start with the information \(f(0)=1\), so the graph of \(f\) must pass through \((0,1)\).
Next, use the sign of \(f'\). Where the graph of \(f'\) is above the x-axis, \(f\) is increasing; where the graph of \(f'\) is below the x-axis, \(f\) is decreasing. From the picture, \(f'\) is positive on about \((0,1)\) and again on about \((5,6)\), and negative on about \((1,5)\).
Therefore \(f\) increases from \((0,1)\) until about \(x=1\), so it has a local maximum there when \(f'\) changes from positive to negative. The derivative is zero again near \(x=3\), but it does not change sign there, so \(f\) has a horizontal tangent at \(x=3\) without a local extremum. Near \(x=5\), \(f'\) changes from negative to positive, so \(f\) has a local minimum there. After that, \(f\) rises again.
A correct sketch must reflect exactly that increasing/decreasing behavior and those turning points.
Make a sketch of \(\displaystyle g'(x)\) where the graph of \(\displaystyle g\) is shown below. (The grid lines are one unit apart.)
Graph grid labeled \(\displaystyle g'\) vs \(\displaystyle x\) (empty for student to sketch).
Graph grid labeled \(\displaystyle g\) vs \(\displaystyle x\) with a plotted curve (specific to the problem).
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Solution: To sketch \(g'\), read slopes from the graph of \(g\). Where \(g\) is decreasing, \(g'\) is negative; where \(g\) is increasing, \(g'\) is positive; and where \(g\) is flat, \(g'\) is zero.
From the graph, \(g\) decreases on about \((0,1)\), increases on about \((1,2)\), decreases again on about \((2,4)\), and is constant for \(x\ge4\). Therefore a correct sketch of \(g'\) must be below the x-axis on \((0,1)\), equal to \(0\) at \(x=1\), above the x-axis on \((1,2)\), equal to \(0\) at \(x=2\), below the x-axis on \((2,4)\), and exactly \(0\) on the flat interval \((4,6)\).
The exact height is not determined, but the sign and zeros must match the slope behavior of \(g\).
The derivatives of two different functions are described in parts (a) and (b).
- The graph of the derivative of a function, \(\displaystyle g\), is given below. Determine all values of \(\displaystyle t\) where a local maximum or a local minimum occurs for the original function. **Provide a brief justification for each value of \(\displaystyle t\) you give.**
This is the derivative of \(\displaystyle g\).
\(\displaystyle t\) (sec.)
- The derivative of a function, \(\displaystyle h\), is given by
where \(\displaystyle x\) is any real number. Determine all values of \(\displaystyle x\) where a local maximum or a local minimum occurs for the original function. **Give a brief justification for each value of \(\displaystyle x\).**
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Solution:
- A local extremum of \(g\) occurs at a point where \(g'\) changes sign. Looking at the graph of \(g'\), there is no sign change at \(t=1\), because the curve only touches the axis there and stays on the same side. At \(t=3\), however, \(g'\) changes from negative to positive, so \(g\) changes from decreasing to increasing; therefore \(g\) has a local minimum at \(t=3\). At \(t=5\), \(g'\) changes from positive to negative, so \(g\) changes from increasing to decreasing; therefore \(g\) has a local maximum at \(t=5\). Thus \[ \boxed{\text{local minimum at }t=3,\qquad \text{local maximum at }t=5.} \]
- For \[ h'(x)=\frac{x^2-1}{x^2+4}, \] the denominator \(x^2+4\) is always positive, so the sign of \(h'(x)\) is determined entirely by the numerator \(x^2-1\). We have \(x^2-1>0\) when \(|x|>1\), and \(x^2-1<0\) when \(|x|<1\). Therefore \(h\) is increasing on \((-\infty,-1)\) and \((1,\infty)\), and decreasing on \((-1,1)\). So \(h\) changes from increasing to decreasing at \(x=-1\), giving a local maximum there, and from decreasing to increasing at \(x=1\), giving a local minimum there. Hence \[ \boxed{\text{local maximum at }x=-1,\qquad \text{local minimum at }x=1.} \]
A car moves along a straight track. The car starts at the origin and has an initial velocity of 2 m/sec. The car’s acceleration is
where \(\displaystyle t\) is the time in seconds and \(\displaystyle t \geq 0\).
- Determine the position of the car at time \(\displaystyle t\) where \(\displaystyle t \geq 0\).
- What is the car’s position after a very long time?
Solution:
- Since acceleration is the derivative of velocity, we start from \[ v'(t)=-6e^{-3t}. \] Integrating gives \[ v(t)=\int -6e^{-3t}\,dt=2e^{-3t}+C. \] Use the initial velocity \(v(0)=2\): \[ 2e^{0}+C=2 \quad\Longrightarrow\quad 2+C=2 \quad\Longrightarrow\quad C=0. \] So \[ v(t)=2e^{-3t}. \] Now integrate velocity to get position: \[ s(t)=\int 2e^{-3t}\,dt=-\frac23e^{-3t}+C. \] Use the initial position \(s(0)=0\): \[ -\frac23+C=0 \quad\Longrightarrow\quad C=\frac23. \] Therefore \[ \boxed{s(t)=\frac23\bigl(1-e^{-3t}\bigr)}. \]
- As \(t\to\infty\), the factor \(e^{-3t}\to0\). Therefore \[ \lim_{t\to\infty}s(t)=\frac23(1-0)=\boxed{\frac23\text{ m}}. \] So after a very long time, the car approaches the position \(2/3\) meter from the origin.
Information about two functions and their derivatives are given in the tables below. Use the information to determine the values of each of the quantities below.
| t | f(t) |
|---|---|
| 0 | 4 |
| 1 | 8 |
| 2 | 3 |
| 3 | 1 |
| t | f'(t) |
| 0 | 7 |
| 1 | 6 |
| 2 | 4 |
| 3 | 0 |
| t | g(t) |
| 0 | 2 |
| 1 | 3 |
| 2 | 5 |
| 3 | 7 |
| t | g'(t) |
| 0 | 3 |
| 1 | 8 |
| 2 | 2 |
| 3 | 1 |
- \(\displaystyle \frac{d}{dt} \left( 2f(t) - 4g(t) \right)\) at \(\displaystyle t = 1\).
- \(\displaystyle \frac{d}{dt} \left( 2f(t) \cdot g(t) \right)\) at \(\displaystyle t = 3\).
- \(\displaystyle \frac{d}{dt} \left( -g(f(t)) \right)\) at \(\displaystyle t = 2\).
Solution:
- \(\displaystyle \frac{d}{dt}(2f-4g)\big|_{t=1}=2f'(1)-4g'(1)=2(6)-4(8)=\boxed{-20}.\)
- For \(2f(t)g(t)\), \[ \frac{d}{dt}(2fg)=2(f'g+fg'). \] At \(t=3\), this gives \(2(0\cdot7+1\cdot1)=\boxed{2}\).
- For \(-g(f(t))\), the chain rule gives \(-g'(f(t))f'(t)\). At \(t=2\), we have \(f(2)=3\), \(g'(3)=1\), and \(f'(2)=4\). Hence the value is \(\boxed{-4}\).
Answer the questions below for the graph of the equation \(\displaystyle y = 2 \cos (y - \pi x)\).
- Determine \(\displaystyle \frac{dy}{dx}\).
- Determine an equation of the tangent line to the curve at the point \(\displaystyle \left( \frac{1}{2}, 0 \right)\).
- Find a point, \(\displaystyle (x, 2)\), on the curve where the tangent line is horizontal.
[Graph appears here]
An experiment is performed by dropping two identical steel balls simultaneously from the same height onto different surfaces, as shown below. One bounces off the steel plate; the other remains in contact with the carpet.
(Image: Table with two balls labeled "Steel Plate" and "Carpet")
Will the impulse delivered to the steel ball by the steel plate be greater than, less than, or equal to the impulse delivered to the steel ball by the carpet? Explain your answer.
IMAGE
IMAGE
Solution:
- Differentiate implicitly: \[ y'=-2\sin(y-\pi x)(y'-\pi). \] Collecting the \(y'\)-terms gives \[ y'\bigl(1+2\sin(y-\pi x)\bigr)=2\pi\sin(y-\pi x), \] so \[ \boxed{\frac{dy}{dx}=\frac{2\pi\sin(y-\pi x)}{1+2\sin(y-\pi x)}}. \]
- At \(\left(\tfrac12,0\right)\), we have \(y-\pi x=-\tfrac\pi2\), so \(\sin(y-\pi x)=-1\). Hence the slope is \(2\pi\). The tangent line is \[ \boxed{y=2\pi\left(x-\frac12\right)=2\pi x-\pi}. \]
- A horizontal tangent requires \(y'=0\), hence \(\sin(y-\pi x)=0\). If we also want \(y=2\), the curve equation gives \(2=2\cos(2-\pi x)\), so \(\cos(2-\pi x)=1\). Taking the simplest case \(2-\pi x=0\) gives \(x=\dfrac{2}{\pi}\). Thus one such point is \(\boxed{\left(\frac{2}{\pi},2\right)}\).
For the second prompt that was merged into this block by OCR: impulse equals change in momentum. A ball bouncing off the steel plate reverses its velocity, so its momentum changes by a larger amount than the ball that comes to rest on the carpet. Therefore the impulse from the steel plate is \(\boxed{\text{greater}}\).
A rectangle is inscribed in a circle of radius two centered at the origin. Determine the dimensions of the rectangle with the largest possible area, and determine the largest possible area. (Use calculus to show that the dimensions really give the largest possible area.)
Solution: Let the upper-right corner of the rectangle be \((x,y)\). Since the rectangle is inscribed in the circle of radius \(2\), we have \(x^2+y^2=4\), so \(y=\sqrt{4-x^2}\). The area is \[ A(x)=4xy=4x\sqrt{4-x^2}. \] Differentiate: \[ A'(x)=4\sqrt{4-x^2}-\frac{4x^2}{\sqrt{4-x^2}}=\frac{16-8x^2}{\sqrt{4-x^2}}. \] Set \(A'(x)=0\): \(16-8x^2=0\), so \(x^2=2\), hence \(x=\sqrt2\) and \(y=\sqrt2\). Thus the maximizing rectangle is a square of dimensions \(2\sqrt2\times2\sqrt2\), and the largest area is \[ \boxed{8}. \] Because the endpoint areas are \(0\) and the interior critical point gives a larger value, this is indeed the maximum.
Determine the values of each of the limits given below.
- \(\displaystyle \lim_{x \to 5} (x^3 - 7x + 10)\)
- \(\displaystyle \lim_{x \to \infty} \frac{7|x| - 3}{4x - 12}\)
- \(\displaystyle \lim_{x \to 0^+} (\sin(x))^{8x}\)
Solution:
- Direct substitution gives \(5^3-7\cdot5+10=125-35+10=\boxed{100}\).
- As \(x\to\infty\), \(|x|=x\), so \[ \frac{7|x|-3}{4x-12}=\frac{7x-3}{4x-12}\to\boxed{\frac74}. \]
- Let \(L=\lim_{x\to0^+}(\sin x)^{8x}\). Then \[ \ln L=\lim_{x\to0^+}8x\ln(\sin x). \] Since \(\sin x\sim x\) and \(x\ln x\to0\), we get \(\ln L=0\). Hence \(L=e^0=\boxed{1}\).
The power in an electrical circuit is given by
where \(\displaystyle P\) is the power (Watts), \(\displaystyle I\) is the current (Amps) and \(\displaystyle R\) is the resistance (Ohms) in the circuit. The circuit will be designed to have a constant resistance of \(\displaystyle R = 5,000\) Ohms.
- Determine the linearization of the power in terms of the current for a level of 0.1 Amps of current.
- For the final design the power should be about 50 Watts, but it can vary by \(\displaystyle \pm 0.1\) Watts. Use the linearization or the differential to estimate the possible change in the current.
Solution:
- With \(R=5000\), the power is \(P(I)=5000I^2\). At \(I_0=0.1\), we have \(P(0.1)=50\) and \(P'(I)=10000I\), so \(P'(0.1)=1000\). Therefore the linearization at \(I=0.1\) is \[ \boxed{L(I)=50+1000(I-0.1)}. \]
- A change \(dP\) in power produces approximately \[ dP\approx P'(0.1)\,dI=1000\,dI. \] If \(|dP|\le0.1\), then \(|dI|\approx\dfrac{0.1}{1000}=0.0001\). So the possible change in current is approximately \(\boxed{\pm0.0001\text{ A}}\).