Q.1 Invasive Species Problem: Step-by-Step Solution
Note: All calculations requiring a calculator are performed with the calculator in radian mode, as specified in the problem. Results are typically rounded to three decimal places.
Part A: Average Number of Acres Affected
To find the average number of acres affected, we need to find the average value of the function \(C(t) = 7.6 \arctan(0.2t)\) on the time interval \([0, 4]\). The formula for the average value of a function \(f(t)\) on an interval \([a, b]\) is:
Applying this formula to our function \(C(t)\) over the interval \([0, 4]\), we set up the following integral:
This integral is best evaluated using a calculator.
Part B: Instantaneous Rate of Change = Average Rate of Change
This question is an application of the Mean Value Theorem. We first need to find the average rate of change of \(C\) over the interval \([0, 4]\). The formula for the average rate of change is:
Let's calculate the values of \(C(4)\) and \(C(0)\):
\( C(4) = 7.6 \arctan(0.2 \cdot 4) = 7.6 \arctan(0.8) \approx 5.1281 \)
\( C(0) = 7.6 \arctan(0.2 \cdot 0) = 7.6 \arctan(0) = 0 \)
Now, we can find the average rate of change:
Next, we set the instantaneous rate of change, \(C'(t)\), equal to this average rate of change and solve for \(t\). We are given \(C'(t) = \frac{38}{25 + t^2}\).
We can solve this equation for \(t\) using a calculator's solver feature or by finding the intersection of the graphs of \(y = \frac{38}{25 + x^2}\) and \(y = 1.2820\).
Part C: End Behavior of the Rate of Change
The "end behavior" of the rate of change describes what happens to \(C'(t)\) as time \(t\) increases without bound (approaches infinity). We need to evaluate the following limit:
To evaluate this limit, we observe the behavior of the denominator. As \(t \to \infty\), the term \(t^2\) also approaches \(\infty\). Therefore, the entire denominator, \(25 + t^2\), grows infinitely large. Since the numerator is a constant (38), the value of the fraction approaches 0.
This means that as time goes on, the rate at which the invasive species spreads (the number of new acres affected per week) approaches zero.
Part D: Maximum Value of A(t)
We need to find the absolute maximum value of the function \(A(t) = C(t) - \int_4^t 0.1 \ln(x) \, dx\) on the closed interval \([4, 36]\). We will use the Extreme Value Theorem, which states that the absolute maximum must occur at either a critical point or an endpoint of the interval.
Step 1: Find the derivative, A'(t).
We use the Fundamental Theorem of Calculus, Part 2, which states that \(\frac{d}{dt} \int_a^t f(x) \, dx = f(t)\).
Step 2: Find critical points.
We set \(A'(t) = 0\) and solve for \(t\) within the interval \((4, 36)\).
This equation is best solved using a calculator's graphing or solver utility. Solving for \(t\) gives us one critical point in the interval:
Step 3: Test the candidates.
The candidates for the absolute maximum are the endpoints (\(t=4, t=36\)) and the critical point (\(t \approx 13.922\)). We evaluate \(A(t)\) at each of these values.
At \(t=4\): $$ A(4) = C(4) - \int_4^4 0.1 \ln(x) \, dx = C(4) - 0 $$ $$ A(4) = 7.6 \arctan(0.8) \approx 5.128 $$
At \(t \approx 13.922\): $$ A(13.922) = C(13.922) - \int_4^{13.922} 0.1 \ln(x) \, dx $$ $$ A(13.922) \approx 7.6 \arctan(0.2 \cdot 13.922) - 0.1 \int_4^{13.922} \ln(x) \, dx $$ $$ A(13.922) \approx 9.771 - 2.039 \approx 7.732 $$
At \(t=36\): $$ A(36) = C(36) - \int_4^{36} 0.1 \ln(x) \, dx $$ $$ A(36) \approx 7.6 \arctan(0.2 \cdot 36) - 0.1 \int_4^{36} \ln(x) \, dx $$ $$ A(36) \approx 11.660 - 5.700 \approx 5.960 $$
Step 4: Justify and Conclude.
By comparing the values of \(A(t)\) at the endpoints and the critical point, we have:
- \(A(4) \approx 5.128\)
- \(A(13.922) \approx 7.732\)
- \(A(36) \approx 5.960\)
Therefore, the function \(A\) attains its maximum value at \(t \approx 13.922\) weeks. This is justified by the Candidates Test, which shows that the value of \(A\) at this critical point is greater than its value at the endpoints of the interval.
Q.2 Region Bounded by Curves: Step-by-Step Solution
Note: Your calculator should be in radian mode. The functions are \(f(x) = x^2 - 2x\) and \(g(x) = x + \sin(\pi x)\).
Part A: Find the area of R
The area of a region bounded by two curves is found by integrating the difference between the upper curve and the lower curve over the interval defined by their intersection points.
Step 1: Find the bounds of integration.
From the graph, the region R starts at \(x=0\) and ends where the curves intersect again. We can verify the intersection points. At \(x=0\), \(f(0)=0\) and \(g(0)=0\). The graph shows another intersection at \(x=3\). Let's verify:
\(f(3) = 3^2 - 2(3) = 9 - 6 = 3\)
\(g(3) = 3 + \sin(3\pi) = 3 + 0 = 3\)
Since \(f(3) = g(3)\), the curves intersect at \(x=3\). The interval of integration is \([0, 3]\).
Step 2: Set up the integral.
From the graph, for \(x\) in \((0, 3)\), \(g(x)\) is the upper function and \(f(x)\) is the lower function. The area, \(A\), is given by:
Step 3: Evaluate the integral.
We use a calculator to evaluate the definite integral:
Part B: Volume of a Solid with Rectangular Cross-Sections
The volume of a solid with known cross-sectional area \(A(x)\) is found by integrating \(A(x)\) over the length of the base.
Step 1: Find the area of a single cross-section, A(x).
The cross-section is a rectangle perpendicular to the x-axis.
- The base of the rectangle lies in region R, so its length is the vertical distance between the curves: \(g(x) - f(x)\).
- The height of the rectangle is given as \(x\).
Step 2: Set up the volume integral.
The volume, \(V\), is the integral of the cross-sectional area from \(x=0\) to \(x=3\).
Step 3: Evaluate the integral.
Using a calculator to compute the value:
Part C: Volume of Revolution (Washer Method)
We need to write an integral for the volume of the solid generated when region \(R\) is rotated about the horizontal line \(y = -2\). Since there is a gap between the axis of rotation and the region, we must use the washer method. The formula is \(V = \pi \int_a^b (R(x)^2 - r(x)^2) \, dx\).
Step 1: Define the Outer Radius, R(x).
The outer radius is the distance from the axis of rotation (\(y=-2\)) to the farthest edge of the region, which is the upper curve \(y=g(x)\).
Step 2: Define the Inner Radius, r(x).
The inner radius is the distance from the axis of rotation (\(y=-2\)) to the closest edge of the region, which is the lower curve \(y=f(x)\).
Step 3: Write the integral expression.
The problem asks only for the setup, not the evaluation. The volume \(V\) is:
Part D: Parallel Tangent Lines
Two lines are parallel if they have the same slope. The slope of a tangent line to a function at a point is given by the derivative of the function at that point. We need to find the value of \(x\) where the tangent to \(f\) is parallel to the tangent to \(g\), which means we must find where their derivatives are equal: \(f'(x) = g'(x)\).
Step 1: Find the derivatives.
For \(f(x) = x^2 - 2x\), the derivative is:
$$ f'(x) = 2x - 2 $$
We are given the derivative of \(g(x)\):
$$ g'(x) = 1 + \pi \cos(\pi x) $$
Step 2: Set the derivatives equal and solve.
We set up the equation and solve for \(x\) in the specified interval \(0 < x < 1\).
This equation is difficult to solve algebraically. We use a calculator to find the intersection of the graphs of \(y_1 = 2x-2\) and \(y_2 = 1 + \pi \cos(\pi x)\) on the interval \((0, 1)\).
By graphing these two functions and finding their intersection point, we get the value:
Q.3 Reading Rate Problem: Step-by-Step Solution
Part A: Approximate \(R'(1)\)
The derivative \(R'(1)\) represents the instantaneous rate of change of the reading speed at \(t=1\) minute. Since we don't have the function \(R(t)\), we can approximate this value using the average rate of change over the smallest available interval that contains \(t=1\). From the table, the best interval is \([0, 2]\).
The formula for the average rate of change is:
Using the values from the table:
Units: The units of \(R(t)\) are "words per minute" and the units of \(t\) are "minutes". Therefore, the units for \(R'(t)\) are (words per minute) per minute.
Part B: Must there be a value \(c\) such that \(R(c) = 155\)?
This question involves the Intermediate Value Theorem (IVT). The IVT states that if a function is continuous on a closed interval \([a, b]\), then it must take on every value between \(f(a)\) and \(f(b)\) at some point within the interval.
Justification using the Intermediate Value Theorem:
- Continuity: The problem states that \(R\) is a differentiable function. Since every differentiable function is also continuous, we know that \(R(t)\) is continuous on the interval \([0, 10]\).
- Interval and Values: We need to find an interval where the value 155 lies between the function values at the endpoints. Looking at the table, consider the interval \([8, 10]\).
- At \(t=8\), \(R(8) = 150\).
- At \(t=10\), \(R(10) = 162\).
- Conclusion: Because \(R(t)\) is continuous on \([8, 10]\) and 155 is between \(R(8)\) and \(R(10)\), the IVT guarantees that yes, there must be at least one value \(c\) in the interval \((8, 10)\) such that \(R(c) = 155\).
Part C: Approximate \(\int_{0}^{10} R(t) dt\) using a trapezoidal sum
The definite integral \(\int_{0}^{10} R(t) dt\) represents the total number of words the student read in the 10 minutes. We can approximate this value using a trapezoidal sum with the three subintervals given by the table: \([0, 2]\), \([2, 8]\), and \([8, 10]\).
The area of a trapezoid is \(\frac{1}{2}(b_1 + b_2)h\). For our subintervals, the "height" is the width of the interval (\(\Delta t\)), and the "bases" are the function values.
Now, we substitute the values from the table:
Part D: Total words read by the teacher
The teacher's reading rate is given by the function \(W(t) = -\frac{3}{10}t^2 + 8t + 100\). To find the total number of words the teacher read from \(t=0\) to \(t=10\) minutes, we must calculate the definite integral of the rate function \(W(t)\) over this interval.
First, we find the antiderivative of \(W(t)\):
Next, we use the Fundamental Theorem of Calculus to evaluate the definite integral:
Q.4 Graph of f and a Function g: Step-by-Step Solution
We are given the function \(g(x) = \int_6^x f(t) \, dt\) and the graph of \(f\).
Part A: Find \(g'(8)\)
Step 1: Find the derivative of g(x).
According to the Second Fundamental Theorem of Calculus, if \(g(x) = \int_a^x f(t) \, dt\), then \(g'(x) = f(x)\).
Step 2: Evaluate the derivative at x = 8.
To find \(g'(8)\), we simply need to find the value of \(f(8)\) from the graph.
Looking at the graph, the line segment from \((6, 0)\) to \((12, 3)\) passes through the point \((8, 1)\). We can also find the equation of the line: slope \(m = \frac{3-0}{12-6} = \frac{3}{6} = \frac{1}{2}\). The equation is \(y - 0 = \frac{1}{2}(x - 6)\), so at \(x=8\), \(y = \frac{1}{2}(8-6) = 1\).
Part B: Find points of inflection for g
A point of inflection for \(g\) occurs where the concavity of its graph changes. This happens when \(g''(x)\) changes sign.
Since \(g'(x) = f(x)\), we have \(g''(x) = f'(x)\). So, we are looking for points where \(f'(x)\) (the slope of the graph of \(f\)) changes sign.
- At \(\boldsymbol{x=-3}\), the slope of \(f\) changes from positive (increasing) to negative (decreasing).
- At \(\boldsymbol{x=0}\), the slope of \(f\) changes from negative (decreasing) to positive (increasing).
- At \(\boldsymbol{x=3}\), the slope of \(f\) changes from positive (increasing) to negative (decreasing).
- At \(\boldsymbol{x=6}\), the slope of \(f\) changes from negative (decreasing) to positive (increasing).
Part C: Find \(g(12)\) and \(g(0)\)
The value of \(g(x)\) is the net signed area under the curve of \(f\) from 6 to \(x\).
To find g(12):
This is the area of the triangle from \(x=6\) to \(x=12\). The base is \(12-6=6\) and the height is \(3\).
To find g(0):
The integral \(\int_0^6 f(t) \, dt\) is the area of the semicircle above the x-axis. This semicircle has a radius of \(r=3\). The area is \(\frac{1}{2}\pi r^2\).
Part D: Find the absolute minimum of g
To find the absolute minimum of \(g\) on the closed interval \([-6, 12]\), we use the Candidates Test. The candidates are the endpoints of the interval (\(-6\) and \(12\)) and any critical points inside the interval.
Step 1: Find critical points.
Critical points occur where \(g'(x) = f(x) = 0\). From the graph, this happens at \(x=0\) and \(x=6\).
Step 2: List the candidates.
The candidates for the location of the absolute minimum are the endpoints and the critical points: \(x = -6, 0, 6, 12\).
Step 3: Evaluate g(x) at each candidate.
- \(\boldsymbol{g(-6)} = \int_6^{-6} f(t) \, dt = -\int_{-6}^6 f(t) \, dt\)
This integral is the sum of two areas: a semicircle below the axis (radius 3) and a semicircle above the axis (radius 3).
\(\int_{-6}^6 f(t) \, dt = \text{Area}_{(-6,0)} + \text{Area}_{(0,6)} = -\frac{1}{2}\pi(3)^2 + \frac{1}{2}\pi(3)^2 = 0\)
So, \(g(-6) = -0 = \boldsymbol{0}\). - \(\boldsymbol{g(0)} = \span class="answer">-\frac{9\pi}{2} \approx -14.137\) (from Part C)
- \(\boldsymbol{g(6)} = \int_6^6 f(t) \, dt = \boldsymbol{0}\)
- \(\boldsymbol{g(12)} = \boldsymbol{9}\) (from Part C)
By comparing the values of \(g(x)\) at all the candidates:
\(g(-6) = 0\)
\(g(0) = -9\pi/2 \approx -14.137\)
\(g(6) = 0\)
\(g(12) = 9\)
The smallest value is \(-\frac{9\pi}{2}\). Therefore, the absolute minimum of \(g\) on the interval \([-6, 12]\) occurs at \(x=0\).
Q.5 Particle Motion Problem: Step-by-Step Solution
Part A: Find the velocity of particle H at time \(t=1\)
Velocity is the derivative of the position function. We are given the position of particle H, \(x_H(t) = e^{t^2 - 4t}\). To find its velocity, \(v_H(t)\), we need to find the derivative \(x_H'(t)\). We will use the chain rule.
Now, we evaluate the velocity at \(t=1\):
Part B: When are the particles moving in opposite directions?
Particles move in opposite directions when their velocities have opposite signs. We need to find the intervals where \(v_H(t)\) and \(v_J(t)\) have different signs.
Step 1: Analyze the sign of \(v_H(t)\)
\(v_H(t) = (2t - 4)e^{t^2 - 4t}\). The term \(e^{t^2 - 4t}\) is always positive. So, the sign of \(v_H(t)\) is determined by the sign of \((2t-4)\).
- \(2t - 4 = 0\) when \(t=2\).
- For \(0 < t < 2\), \(v_H(t)\) is negative.
- For \(t > 2\), \(v_H(t)\) is positive.
Step 2: Analyze the sign of \(v_J(t)\)
\(v_J(t) = 2t(t^2 - 1)^3\). For \(0 < t \le 5\), the term \(2t\) is positive. So, the sign of \(v_J(t)\) is determined by the sign of \((t^2 - 1)^3\), which is the same as the sign of \(t^2-1\).
- \(t^2 - 1 = 0\) when \(t=1\) (since t > 0).
- For \(0 < t < 1\), \(t^2-1\) is negative, so \(v_J(t)\) is negative.
- For \(t > 1\), \(t^2-1\) is positive, so \(v_J(t)\) is positive.
Step 3: Compare the signs and give a reason.
We can summarize the signs in a table:
| Interval | Sign of \(v_H(t)\) | Sign of \(v_J(t)\) | Same/Opposite? |
|---|---|---|---|
| \((0, 1)\) | - | - | Same |
| \((1, 2)\) | - | + | Opposite |
| \((2, 5)\) | + | + | Same |
The velocities have opposite signs only on the interval \((1, 2)\).
Reason: The particles are moving in opposite directions on the interval \((1, 2)\) because on this interval, \(v_H(t) < 0\) and \(v_J(t) > 0\).
Part C: Is the speed of particle J increasing or decreasing at \(t=2\)?
The speed of a particle increases when its velocity and acceleration have the same sign. It decreases when they have opposite signs. The acceleration of particle J is \(a_J(t) = v_J'(t)\).
Step 1: Find the sign of the velocity \(v_J(2)\).
The velocity \(v_J(2)\) is positive.
Step 2: Find the sign of the acceleration \(a_J(2)\).
We are given that \(v_J'(2) > 0\). This means the acceleration \(a_J(2)\) is positive.
Part D: Find the position of particle J at \(t=2\)
To find the position at a later time, we use the "final position = initial position + displacement" formula. The displacement is the definite integral of the velocity function.
We are given \(x_J(0) = 7\). Now we must evaluate the integral \(\int_0^2 2t(t^2-1)^3 \, dt\). We use u-substitution.
Let \(u = t^2 - 1\). Then \(du = 2t \, dt\).
We can also change the limits of integration:
- When \(t=0\), \(u = 0^2 - 1 = -1\).
- When \(t=2\), \(u = 2^2 - 1 = 3\).
The integral becomes:
Now, we find the final position:
The position of particle J at time \(t=2\) is 27.
Q.6 Implicit Differentiation and Related Rates Problem
Part A: Show that \(\frac{dy}{dx} = \frac{-x}{2(3y^2 - 2y - 1)}\)
We start with the equation of the curve G: \(y^3 - y^2 - y + \frac{1}{4}x^2 = 0\). We use implicit differentiation, taking the derivative of each term with respect to \(x\).
Applying the chain rule to the terms with \(y\):
Now, we isolate the terms with \(\frac{dy}{dx}\) and factor it out.
Finally, we solve for \(\frac{dy}{dx}\).
This matches the expression we were asked to show.
Part B: Tangent Line Approximation
We will use the tangent line to the curve at the point \((2, -1)\) to approximate the y-coordinate of point P where \(x=1.6\).
Step 1: Find the slope of the tangent line at \((2, -1)\).
We evaluate \(\frac{dy}{dx}\) at \(x=2\) and \(y=-1\).
Step 2: Write the equation of the tangent line.
Using the point-slope form \(y - y_1 = m(x - x_1)\):
Step 3: Approximate the y-value at \(x=1.6\).
Part C: Find the y-coordinate of the vertical tangent
A vertical tangent line occurs where the slope \(\frac{dy}{dx}\) is undefined. This happens when the denominator of the derivative is zero (and the numerator is non-zero).
Step 1: Set the denominator of \(\frac{dy}{dx}\) to zero.
Step 2: Solve the quadratic equation for y.
The problem states that for point S, \(y > 0\). Therefore, we must have \(y=1\). To be sure this point exists, we find the corresponding x-value from the original equation.
Since the problem also states \(x>0\), the point S is \((2, 1)\). At this point, the numerator \(-x = -2 \neq 0\), so the tangent is indeed vertical.
Part D: Related Rates
A particle moves along the curve H defined by \(2xy + \ln y = 8\). We need to find \(\frac{dy}{dt}\) given \(x=4, y=1\), and \(\frac{dx}{dt} = 3\).
Step 1: Differentiate the equation with respect to time \(t\).
We use the product rule for \(2xy\) and the chain rule for \(\ln y\).
Step 2: Substitute the known values into the differentiated equation.
Step 3: Solve for \(\frac{dy}{dt}\).