Question 5
The table shows values of the variables $x$ and $y$, which are related by an equation of the form $y=A b^{x^2}$, where $A$ and $b$ are constants.
| $x$ | 1 | 1.5 | 2 | 2.5 |
|---|---|---|---|---|
| $y$ | 2.0 | 11.3 | 128 | 2896 |
(a) Use the data to draw a straight line graph of $\ln y$ against $x^2$. [2]
(b) Use your graph to estimate the values of $A$ and $b$. Give your answers correct to 1 significant figure. [5]
(c) Estimate the value of $y$ when $x=1.75$. [2]
(d) Estimate the positive value of $x$ when $y=20$. [2]
Solution
Step (a): Plotting the Graph of $\ln y$ Against $x^2$
We are given the equation relating $x$ and $y$:
$$ y = A b^{x^2}. $$
Taking the natural logarithm of both sides:
$$ \ln y = \ln A + x^2 \ln b. $$
This equation is in the form of a straight line $y = mx + c$, where:
- The slope $m = \ln b$
- The intercept $c = \ln A$
We are given the following data:
$$ \begin{array}{|c|c|c|c|c|} \hline x & 1 & 1.5 & 2 & 2.5 \\ \hline y & 2.0 & 11.3 & 128 & 2896 \\ \hline \end{array} $$
To create the straight-line graph:
- Calculate $x^2$ for each value of $x$.
- Calculate $\ln y$ for each corresponding value of $y$.
$$ \begin{array}{|c|c|c|c|c|} \hline x & 1 & 1.5 & 2 & 2.5 \\ \hline x^2 & 1 & 2.25 & 4 & 6.25 \\ \hline y & 2.0 & 11.3 & 128 & 2896 \\ \hline \ln y & \ln 2.0 \approx 0.693 & \ln 11.3 \approx 2.431 & \ln 128 \approx 4.855 & \ln 2896 \approx 7.975 \\ \hline \end{array} $$
\begin{tikzpicture}
\draw[step=1cm,black,very thin] (-1,-1) grid (7,8);
\draw[thick,->](-1,0) -- (7,0) node[below]{$x^2$};
\draw[thick,->](0,-1) -- (0,8) node[above]{$\ln y$};
\draw[->]
(0,-.7) -- (1,0.693) -- (2.25,2.431) -- (4,4.855) -- (6.25,7.975);
\end{tikzpicture}
Step (b): Estimating Values of $A$ and $b$
From the graph, we can obtain the slope and intercept.
- The slope of the line is $\ln b$.
- The intercept is $\ln A$.
Thus
$$ \ln b=\frac{4.855-0.693}{4-1} =\frac{4.162}{3} =1.387. $$
By inspecting the graph, the intercept is approximately $-0.7$. We can calculate the values of $A$ and $b$ as follows:
$$ b = e^{\text{slope}} \quad \text{and} \quad A = e^{\text{intercept}}. $$
$$ b \approx e^{1.387} \approx 4, \quad A \approx e^{-0.7} \approx 0.5. $$
Thus, the estimated values of $A$ and $b$ are:
$$ A \approx 0.5 \quad \text{(to 1 significant figure)} \quad \text{and} \quad b \approx 4.0 \quad \text{(to 1 significant figure)}. $$
Step (c): Estimating the Value of $y$ When $x = 1.75$
Now, using the equation $y = A b^{x^2}$, we can estimate the value of $y$ for $x = 1.75$.
First, calculate $x^2$:
$$ x^2 = (1.75)^2 = 3.0625. $$
Then, substitute the values of $A \approx 0.5$ and $b \approx 4.0$ into the equation:
$$ y = 0.5 \times 4.0^{3.0625}. $$
Estimate the value of $4.0^{3.0625}$ (using a calculator):
$$ 4.0^{3.0625} \approx 66.6. $$
Thus:
$$ y \approx 0.5 \times 66.6 \approx 33.3. $$
So, the estimated value of $y$ when $x = 1.75$ is approximately $33.3$.
Step (d): Estimating the Positive Value of $x$ When $y = 20$
To estimate the positive value of $x$ when $y = 20$, we use the equation $y = A b^{x^2}$.
First, solve for $x^2$:
$$ y = A b^{x^2} \quad \Rightarrow \quad \frac{y}{A} = b^{x^2} \quad \Rightarrow \quad x^2 = \frac{\ln \left( \frac{y}{A} \right)}{\ln b}. $$
Substitute the known values $A \approx 0.5$, $b \approx 4.0$, and $y = 20$:
$$ x^2 = \frac{\ln \left( \frac{20}{0.5} \right)}{\ln 4.0} = \frac{\ln (40)}{\ln 4.0}. $$
Using a calculator:
$$ \ln 40 \approx 3.6889, \quad \ln 4.0 \approx 1.3863. $$
Thus:
$$ x^2 = \frac{3.6889}{1.3863} \approx 2.66. $$
Finally, take the square root of both sides:
$$ x \approx \sqrt{2.66} \approx 1.63. $$
So, the positive value of $x$ when $y = 20$ is approximately $1.63$.
Question 6
Given that $ \mathrm{f}^{\prime \prime}(x)=(5 x+2)^{-\frac{2}{5}} $, $ \mathrm{f}^{\prime}(6)=\frac{17}{3} $ and $ \mathrm{f}(6)=\frac{26}{3} $, find an expression for $ \mathrm{f}(x) $. [8]
Solution
Step 1: Integrate the second derivative to find the first derivative
We are given that the second derivative of the function $f(x)$ is:
$$ f''(x) = (5x + 2)^{-\frac{2}{5}}. $$
To find $f'(x)$, we integrate $f''(x)$ with respect to $x$:
$$ f'(x) = \int (5x + 2)^{-\frac{2}{5}} \, dx. $$
Let $u = 5x + 2$, so that $du = 5 \, dx$, or $dx = \frac{du}{5}$. Now the integral becomes:
$$ f'(x) = \int u^{-\frac{2}{5}} \cdot \frac{du}{5}. $$
Simplifying:
$$ f'(x) = \frac{1}{5} \int u^{-\frac{2}{5}} \, du = \frac{1}{5} \cdot \frac{u^{\frac{3}{5}}}{\frac{3}{5}} = \frac{1}{5} \cdot \frac{5}{3} u^{\frac{3}{5}} = \frac{1}{3} (5x + 2)^{\frac{3}{5}} + C_1. $$
Thus, the first derivative is:
$$ f'(x) = \frac{1}{3} (5x + 2)^{\frac{3}{5}} + C_1. $$
Step 2: Use the initial condition $f'(6)=\frac{17}{3}$ to find $C_1$
We are given that $ f'(6)=\frac{17}{3} $. Substituting $x=6$ into the expression for $f'(x)$:
$$ f'(6) = \frac{1}{3}(5(6)+2)^{\frac{3}{5}} + C_1 = \frac{1}{3}(30+2)^{\frac{3}{5}} + C_1 = \frac{1}{3}\cdot 32^{\frac{3}{5}} + C_1. $$
We also know that:
$$ f'(6)=\frac{17}{3}. $$
So:
$$ \frac{1}{3}\cdot 32^{\frac{3}{5}} + C_1 = \frac{17}{3}. $$
Since $32 = 2^5$:
$$ 32^{\frac{3}{5}} = (2^5)^{\frac{3}{5}} = 2^3 = 8. $$
Therefore:
$$ \frac{8}{3} + C_1 = \frac{17}{3}. $$
Subtracting $\frac{8}{3}$ from both sides:
$$ C_1 = \frac{17}{3} - \frac{8}{3} = \frac{9}{3} = 3. $$
Thus, $C_1 = 3$.
Step 3: Integrate the first derivative to find the original function $f(x)$
Now we have:
$$ f'(x) = \frac{1}{3}(5x + 2)^{\frac{3}{5}} + 3. $$
To find $f(x)$, integrate $f'(x)$:
$$ f(x) = \int \left( \frac{1}{3}(5x + 2)^{\frac{3}{5}} + 3 \right) dx. $$
Split the integral into two parts:
$$ f(x) = \frac{1}{3}\int (5x + 2)^{\frac{3}{5}} \, dx + \int 3 \, dx. $$
For the first integral, let $u = 5x + 2$, so that $du = 5dx$, or $dx = \frac{du}{5}$:
$$ \frac{1}{3}\int (5x + 2)^{\frac{3}{5}} \, dx = \frac{1}{3}\cdot\frac{1}{5}\int u^{\frac{3}{5}} \, du = \frac{1}{15}\cdot\frac{u^{\frac{8}{5}}}{\frac{8}{5}} = \frac{1}{15}\cdot\frac{5}{8}u^{\frac{8}{5}} = \frac{1}{24}(5x + 2)^{\frac{8}{5}}. $$
The second integral is:
$$ \int 3 \, dx = 3x. $$
Thus, the general expression for $f(x)$ is:
$$ f(x) = \frac{1}{24}(5x + 2)^{\frac{8}{5}} + 3x + C_2. $$
Step 4: Use the initial condition $f(6)=\frac{26}{3}$ to find $C_2$
We are given that:
$$ f(6)=\frac{26}{3}. $$
Substitute $x=6$ into the expression for $f(x)$:
$$ f(6) = \frac{1}{24}(5(6)+2)^{\frac{8}{5}} + 3(6) + C_2 = \frac{1}{24}(32)^{\frac{8}{5}} + 18 + C_2. $$
Since:
$$ 32^{\frac{8}{5}} = (2^5)^{\frac{8}{5}} = 2^8 = 256, $$
we get:
$$ f(6) = \frac{1}{24}\cdot256 + 18 + C_2 = \frac{256}{24} + 18 + C_2 = \frac{64}{6} + 18 + C_2. $$
Now use $ f(6)=\frac{26}{3} $:
$$ \frac{64}{6} + 18 + C_2 = \frac{26}{3}. $$
Convert to denominator 3:
$$ \frac{64}{6} = \frac{32}{3}, \qquad 18 = \frac{54}{3}. $$
So:
$$ \frac{32}{3} + \frac{54}{3} + C_2 = \frac{26}{3}. $$
Simplifying:
$$ \frac{86}{3} + C_2 = \frac{26}{3}. $$
Subtract $\frac{86}{3}$ from both sides:
$$ C_2 = \frac{26}{3} - \frac{86}{3} = \frac{-60}{3} = -20. $$
Thus, $C_2 = -20$.
Final Answer
$$ f(x) = \frac{1}{24}(5x + 2)^{\frac{8}{5}} + 3x - 20. $$
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