Answer
A toy boat is programmed to move according to this rule
\(\Large \frac{dy}{dx} = 3x + 2 – y\).
Each unit of \(x\) and \(y\) is a meter. The toy boat passes through the point (-2, 0.8).
(a) Copy and complete the table above with the values of \(\Large \frac{dy}{dx}\). [2]
(b) Select from the choices below the slope field that best represents the information from the table above. Provide a reason for your choice. [2]
(c) (i) Based on your choice, deduce whether or not the path of this toy boat has a turning point. Explain your reasoning. [2]
(ii) Hence, sketch a curve for y that passes through (-2,0.8). It is not necessary to reproduce the slope field in (b). [2]
(d) (i) Verify that \(y = 3x -1 + \large \frac{7.8}{e^2}e^{-x}\) satisfy the differential equation above. [3]
(ii) Verify that the point (0,2) is on \(y = 3x -1 + \large \frac{7.8}{e^2}e^{-x}\). [1]
As the toy boat moves further to the right of the origin, its path is very close to a line \(f(x) = ax + b\).
(e) (i) Use \(y = 3x -1 + \large \frac{7.8}{e^2}e^{-x}\) to find the values of \(a\) and \(b\). [2]
(ii) Show that the path of this toy boat never intersect with the line \(f(x) = ax + b\). [2]
Worked solution
(a) 
(b) Diagram C. If we concentrate on the vertical axis, diagrams A, B and D have gradients ranging from negative to positive as y increases. These gradients are not consistent with values in our table which are all non-negative in the given range.
(c) (i) 
If we imagine the tangent line segments above as current in a river then we can imagine point (-2,0.8 ) flows downward towards quadrant 3 until it hits a current to bring it towards quadrant 1.
Thus, the path of this toy boat that passes through (-2,0.8) has a turning point
somewhere in quadrant 2.
(ii) 
Note: The purple curve \(y\) should be smooth. The drawing above was limited by technology.
(d) (i), (ii) See the worked solution.
(e) (i) Thus, \( f(x) = 3x – 1 \) with\(a=3\) and \(b = -1\).
(ii) See the worked solution
(a) Notice that each row is a arithmetic sequence with a common difference of 3.
Each column is also an arithmetic sequence with a commond difference of 1.
These patterns can be used to complete the rest of the table.
A1A0 for any 2 correct entries. A1A1 for all correct entries.
(b) Diagram C. A1
R1 for an any reason to accept diagram C or reasons to reject the other diagrams.
e.g. If we concentrate on the vertical axis, diagrams A, B and D have gradients ranging from negative to positive as y increases. These gradients are not consistent with values in our table which are all non-negative in the given range.
(c) (i) If we imagine the tangent line segments above as current in a river then we can imagine point (-2,0.8 ) flows downward towards quadrant 3 until it hits a current to bring it towards quadrant 1. R1
Thus, the path of this toy boat that passes through (-2,0.8) has a turning point A1
somewhere in quadrant 2.
(ii) A1 for starting passing through (-2,0.8).
A1 for the correct shape.
Note: The curve y should be smooth. In Npire, press [Menu] then choose 3:Graph Entry/Edit and 7: Diff Eq. Enter the given differential equation and initial condition into the appropriate slots. Unfortunately, a student should not rely on this function in Npsire because it is disabled in the Press-to-test mode for examination purpose.
(d) (i)
Method 1: Consider taking the derivative of \(y\) with respect to \(x\).
\(y '= 3 - \large \frac{7.8}{e^2}e^{-x} \) A1
Notice that \(y'\) is very similar to \(y\). So the idea is to incorporate \(y\) into \(y'\).
\(y' = 3 – (3x – 1 +\large \frac{7.8}{e^2}e^{-x}) + 3x - 1 \) M1
\(y ‘ = 3 – (3x – 1 + \large{ \frac{7.8}{e^2}e^{-x}} ) + 3x - 1 \) A1
\(y ' = 3x +2 – y\) which is the given differential equation. AG
Method 2:
\(y ' = \large 3 - \frac{7.8}{e^2}e^{-x}\) A1
Thus, \(LHS = \large 3 - \frac{7.8}{e^2}e^{-x} \)
\(RHS = \large 3x + 2 – (3x -1 + \frac{7.8}{e^2}e^{-x}) \) M1
\(= 2 + 1 - \large \frac{7.8}{e^2}e^{-x} \)
\( = 3 - \large \frac{7.8}{e^2}e^{-x} \) A1
\( = LHS \)
(ii) We simply replace \(x = -2\) into the equation.
\(y(-2) = 3(-2) -1 +\large \frac{7.8}{e^2}e^{-2}\).
= -7 + 7.8 A1
= 0.8
which is the given y-coordinate in point (-2,0.8).
(e) (i) As \(x \) approaches positive infinity, \(\large \frac{7.8}{e^2}e^{-x} \)approaches zero. M1
So as \(x \) approaches positive infinity, \(y\) approaches \(y = 3x - 1\).
Thus, \(f(x) = 3x – 1\) with \(a = 3\) and \(b =-1\). A1
(ii) Let us assume that the path of the toy boat does intersect the line \(f(x) = 3x – 1\). M1
Then, we have \(3x – 1 = 3x -1 + \large \frac{7.8}{e^2}e^{-x}\)
which implies \(0 = \large \frac{7.8}{e^2}e^{-x}\)
but there is no real \(x \) that can satisfy the equation above. R1
Thus, it must be the case that the path of the toy boat does not intersect \(f(x) = 3x – 1\).
Answer
According to Stephen Hawking, the emission of a black hole follows this rule
\( \frac{dM}{dt} = \frac{- k }{M^{2}}\),
where \(t\) is time measured in years and mass \(M\).
Each unit of mass \(M\) is \(1 \times 10^{31}\) kg and \(k\) is a constant.
(a) Complete the table with appropriate values of \(\frac{dM}{dt}\).
Leave your answers in terms of \(k\). [2]
Let \(k = 1.26 \times 10^{23}\).
(b) Without sketching the slope field, deduce how all the tangent line segments in a slope field for the data in the table in (a) will look like. [2]
(c) (i) Find \(\frac{d^{2}M}{dt^2}\). [2]
(ii) Interpret the result for c(i) in this context. [2]
(d) Solve \(\frac{dM}{dt} = \frac{- k }{M^{2}}\) for \(M = 3 \times 10^{31}\). [5]
(e) Find the time for a black hole with an initial mass of \(3 \times 10^{31}kg\) to lose all its mass. [2]
Worked solution
(a) 
(b) The ratio of \(k\) and a square of \(M\) is a very small value that approaches zero. Thus all these tangent line segments are close to being horizontal with very slight negative gradients.
As the mass \(M\) gets smaller the rate of emission is faster and thus the tangent line segments with smaller mass are steeper than those with higher mass.
(c) (i) \(\frac{d^{2}M}{dt^2} = \frac{ 2k }{M^{3}}\)
(ii) The value of the second derivative is positive which means the growth in the emission rate increases as time progresses.
(d) \(M = ( 8.1 \times 10^{94} – 3kt )^ {\frac{1}{3} } \)
(e) \(t \approx 2.14 \times 10^{71} \) (3s.f.) years.
(a) Substitute the appropriate value of \(M\) into the given differential equation for each entry. Note that this particular calculation is independent of time \(t\). Thus, all we need to do is to ensure that all cells in a row have the same number.
A1 for any 2 correct entries, A1A1 for all correct entries.
(b) The ratio of \(k\) and a square of \(M\) is a very small value that approaches zero. Thus all these tangent line segments are close to being horizontal with a very slight negative gradient. A1.
As the mass \(M\) gets smaller the rate of emission is faster and thus the tangent line segments with smaller masses are steeper than those with higher masses. A1
(c) (i) \( \frac{dM}{dt} = \frac{- k }{M^{2}}\)
\(\frac{d^{2}M}{dt^2} = (-2) \frac{- k }{M^{3}} \) M1
\(\frac{d^{2}M}{dt^2} = \frac{ 2k }{M^{3}} \) A1
(ii) The value of the second derivative is positive which means the growth in the emission rate increases as time progresses.
A1 for the indicating the growth or the rate of change in the emission rate
A1 for increases.
(d) \(M^2 dM = -k dt\)
\(\int M^2 dM = \int (-k) dt\) M1
\(\frac{M^3}{3} = -kt + c \) A1
Use \((0, 3 \times 10^{31} )\) to solve for c. M1
\((3 \times 10^{31} )^3 = -k(0) + c\)
So \(c = (27 \times 10^{93} )\) -----Note that this value is not in scientific form and it is best to change it into the form below.
\(c = 2.7 \times 10^{94} \) A1
Thus, \(M = ( 3( 2.7 \times 10^{94} – k t ) )^{\frac{1}{3} }\)
\(M =( 8.1 \times 10^{94} – 3kt )^ {\frac{1}{3} }\) A1
(e) Set \(M = 0\) for the black hole to loss all its mass.
\(0 = ( 3( 2.7 \times 10^{94} – k t ) )^ {\frac{1}{3} } \) . Thus, the expression in the inner bracket must be zero.
Thus, \(2.7 \times 10^{94} – k t = 0 \) M1
and \(t = \frac{2.7 \times 10^{94} }{k}\)
Since \(k = 1.26 \times 10^{23}\) then
\(t = \frac{2.7 \times 10^{94} }{1.26 \times 10^{23} }\)
\(t \approx 2.14 \times 10^{71}\) years A1
which is a very long time indeed.
Answer
Stephanus simulated a satellite launch with this coupled differential equations
\(\frac{dx}{dt} = 3x + y \)
\(\frac{dy}{dt} = -x + 4y\)
(a) Show that \(\Large \frac{dy}{dx} = \frac{-x+4y}{3x + y }\). [1]
(b) Copy and complete the table above with the values of \(\frac{dy}{dx}\).
Entries can be left as fractions. [2]
(c) Select from the choices below the slope field that best represents the information from the table above. Provided two reasons for your choice. [3]
The simulated satellite is launched from (0,0) and passes through the point (0, 2).
(d) (i) Based on the slope filed, deduce into which quadrant was the satellite was being launched. Explain your reasoning. [2]
(ii) Hence, sketch a curve for \(y\) starting from (0, 0) and passes through (0,2). It is not necessary to reproduce the slope field in (c). [2]
(e) Use an analytical method to confirm that the satellite does not fall back to (0,0). [4]
Worked solution
(a) We wish to show that \(\Large \frac{dy}{dx} = \frac{-x+4y}{3x + y }\).
LHS \(= \Large \frac{dy}{dx}\)
\(= \Large \frac{dy}{dt} \times \frac{dt}{dx}\)
\(= (-x + 4y) \times \Large \frac{1}{3x+ y } \) A1
\(= \Large \frac{-x+4y}{3x + y }\)
= RHS and thus shown that \(\Large \frac{dy}{dx} = \frac{-x+4y}{3x + y }\).
(b) 
(c) Slope field D because it is the only one which has the same positive gradient when \(x =0\)
and very modest negative gradients when \(y =0\).
(d)(i) Since at the point (0,2), the tangent line segment is positive then the satellite must come from the second quadrant.
Thus, the satellite must be launched into the second quadrant.
(ii) 
The solution is the purple curve above. The slope field is not required for full marks.
(e) See the worked solution.
(a) We need to see this an application of chain rule where \(\Large \frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx} \)
and \(\Large \frac{dy}{dx} = \frac{dy}{dt} \times {1\over{ \frac{dt}{dx}}}\)
(b) Substitute the appropriate values of \(x\) and \(y\) into the given differentail equations.
A1A0 for any 2 correct entries. A1A1 for all correct entries.
(c) Slope field D A1
because it is the only one which has the same positive gradient when \(x =0\) R1
and very modest negative gradients when \(y =0\). R1
Accept any reasonable comments.
(d) (i) Since at the point (0,2), the tangent line segment is positive then the satellite must come from the second quadrant. R1
Thus, the satellite must be launched into the second quadrant. A1
(ii) A1 for starting from (0,0) and passing through (0,2).
A1 for the correct shape.
Note: The purple curve y should be smooth. The drawing above was limited by technology.
The slope field is not required for A1A1.
(e) \(\begin {Vmatrix} 3- \lambda & 1 \\ -1 & 4 - \lambda \end{Vmatrix} \) M1
\(= ( 3 - \lambda)(4 - \lambda) + 1 \) A1
\(= 12 – 7 \lambda + \lambda^2 + 1 \)
\(= \lambda^2 – 7 \lambda + 13 \)
Thus, by setting \( \lambda^2 – 7 \lambda + 13 = 0\), we obtain
\(\Large \lambda = \frac{7 \pm \sqrt{3}i }{2 } \) A1
Since the \(Re(\lambda) > 0\) then the system spiral out R1
and the satellite does not fall back to (0,0) or earth. AG
Answer
The rate of change in the charge held in a capacitor \(Q\) in a RC circuit is given by
\( \frac{dQ}{dt} = \frac{-1}{40}Q\), where t is time measured in seconds.
The capacitor has an initial charge of \( Q =330 \).
(a) Complete the table with appropriate values of \( \frac{dQ}{dt}\). [2]
(b) Select from the choices below the diagram that best represents the information from the table above. Explain your reason. [2]
(c) Find the initial rate of change in the capacitor. [2]
(d) (i) Describe the growth rate as time increases. [2]
(ii) Hence, sketch the curve \(Q\) in your chosen diagram above. [2]
Worked solution
(a)
(b) Diagram C.
(c) (i) -8.75
(ii) 
(d) (i) The capacitor loses its charge (at a decreasing rate).
(ii) See the worked solution.
(a) A1 for any 2 correct entries, A1A1 for all correct entries.
The gradient at each \(Q\) is independent of the time \(t \). Thus, you should notice that all columns have identical results. Thus, all we need to do is to ensure each row has the same entry. Thus, this is an exercise of copying the numbers into the appropriate cell.
(b) Diagram C. A1
R1 for an any reason to accept diagram C or reasons to reject the other diagrams.
e.g. Studying the table, we see each column only has non-positive gradients. If we concentrate on the vertical axis, diagrams A, B and D all have positive or zero gradients which are not consistent with values in our table.
(c) Initially, the time is \(t =0\) but this does not affect the calculation because the differential equation has not variable \(t\) in it.
\(\frac{dQ}{dt} \vert_{Q=350}= \frac{-1}{40}(350)\) M1
\(= - 8.75 \) A1
(d)
(i) The capacitor loses its charge (at a decreasing rate) A1
and eventually approaches zero (or in practice has not charge) at time progress. A1
(ii) The solution is given as the red curve. A1 for starting from (0,330).
A1 for the correct shape. The curve for solution Q should be as smooth as possible. Note: the shape drawn here is limited by technology. Note: The differential equation function is disabled under the Press-to-test mode.
Answer
Each tangent line segment in the slope field above represent the change in velocity \(v\) of a cart with respect to time \(t\). The time \(t\) is measured in seconds and the velocity \(v\) is measured in meters per second.
(a) Use the slope field above to complete the table below with appropriate values of \(\frac{dv}{dt}\). [2]
(b) Hence, suggest an appropriate differential equation that generate the slope field above. [1]
(c) (i) Sketch the curve for \(v\) that has an initial velocity of 2.5 on the slope field above. [2]
(ii) Hence, find the time when the cart is momentarily at a stop. [1]
Worked solution
(a) The completed table for \(\frac{dv}{dt}\).
(b) By comparing the ratios in each column with the \(v\) values, we deduce that \(\frac{dv}{dt} = \frac{-1}{v}\). A1
(c) 
(a) Start from (0,3) count the steps across and the steps down to the end of the tangent line segment. In this case, we would count 3 units to the right and 1 unit down. Thus, the ratio is \(\frac{1}{3}\). Similarly, we do the same for the tangent line segments along the \(v\)-axis.
Notice that gradients in each row are independent of time \(t\). Thus, we just need to copy the same value in each cell in a row.
A1A0 for 2 correct entries. A1A1 for all correct entries.
(b) By comparing the ratios in each column with the \(v\) values, we see that the denominator in each entry is actually the velocity \(v\) in the first column. Since all these segments are downward sloping, the gradient values are all negatives.
Thus, we deduce that \(\frac{dv}{dt} = \frac{-1}{v}\). A1
(c) (i) A1 for starting from (0,2.5) and A1 for the correct shape. Accept as A1 if the curve does not continue beyond \(t = 2\). Note: The red curve \(v\) should be smooth. The drawing above was limited by technology.
(ii) The time \(t\) is about 3.2 or 3.3 seconds. A1
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HL Slope fields