Moderate -0.3 This is a standard linearization problem requiring students to take logarithms of an exponential equation to obtain ln y = ln A - kx², recognize the linear relationship, use two points to find gradient and intercept, then convert back to find A and k. While it involves multiple steps, it follows a well-practiced technique taught explicitly in P3 syllabuses with no novel problem-solving required.
\includegraphics{figure_3}
The variables \(x\) and \(y\) satisfy the equation \(y = Ae^{-kx^2}\), where \(A\) and \(k\) are constants. The graph of \(\ln y\) against \(x^2\) is a straight line passing through the points \((0.64, 0.76)\) and \((1.69, 0.32)\), as shown in the diagram. Find the values of \(A\) and \(k\) correct to 2 decimal places. [5]
Substitute values of \(\ln y\) and \(x^2\), and solve for \(k\) or \(\ln A\)
M1
Obtain \(k = 0.42\) or \(A = 2.80\)
A1
Solve for \(\ln A\) or \(k\)
M1
Obtain \(A = 2.80\) or \(k = 0.42\)
A1
OR1: State or imply \(\ln y = \ln A - kx^2\)
B1
Using values of \(\ln y\) and \(x^2\), equate gradient of line to \(-k\) and solve for \(k\)
M1
Obtain \(k = 0.42\)
A1
Solve for \(\ln A\)
M1
Obtain \(A = 2.80\)
A1
OR2: Obtain two correct equations in \(k\) and \(A\) and substituting \(y-\) and \(x^2 -\) values in \(y = Ae^{-kx^2}\)
B1
Solve for \(k\)
M1
Obtain \(k = 0.42\)
A1
Solve for \(A\)
M1
Obtain \(A = 2.80\)
A1
[5]
[SR: If unsound substitutions are made, e.g. using \(x = 0.64\) and \(y = 0.76\), give B1M0A0M1A0 in the EITHER and OR1 schemes, and B0M1A0M1A0 in the OR2 scheme.]
**EITHER:** State or imply $\ln y = \ln A - kx^2$ | B1 |
Substitute values of $\ln y$ and $x^2$, and solve for $k$ or $\ln A$ | M1 |
Obtain $k = 0.42$ or $A = 2.80$ | A1 |
Solve for $\ln A$ or $k$ | M1 |
Obtain $A = 2.80$ or $k = 0.42$ | A1 |
**OR1:** State or imply $\ln y = \ln A - kx^2$ | B1 |
Using values of $\ln y$ and $x^2$, equate gradient of line to $-k$ and solve for $k$ | M1 |
Obtain $k = 0.42$ | A1 |
Solve for $\ln A$ | M1 |
Obtain $A = 2.80$ | A1 |
**OR2:** Obtain two correct equations in $k$ and $A$ and substituting $y-$ and $x^2 -$ values in $y = Ae^{-kx^2}$ | B1 |
Solve for $k$ | M1 |
Obtain $k = 0.42$ | A1 |
Solve for $A$ | M1 |
Obtain $A = 2.80$ | A1 | [5]
[SR: If unsound substitutions are made, e.g. using $x = 0.64$ and $y = 0.76$, give B1M0A0M1A0 in the EITHER and OR1 schemes, and B0M1A0M1A0 in the OR2 scheme.]
\includegraphics{figure_3}
The variables $x$ and $y$ satisfy the equation $y = Ae^{-kx^2}$, where $A$ and $k$ are constants. The graph of $\ln y$ against $x^2$ is a straight line passing through the points $(0.64, 0.76)$ and $(1.69, 0.32)$, as shown in the diagram. Find the values of $A$ and $k$ correct to 2 decimal places. [5]
\hfill \mbox{\textit{CAIE P3 2013 Q3 [5]}}