| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2007 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Parametric differentiation |
| Difficulty | Moderate -0.3 This is a straightforward parametric differentiation question requiring standard application of the chain rule. Part (i) involves routine differentiation of a power function and an exponential, while part (ii) applies the chain rule formula dy/dt = (dy/dx)(dx/dt). The calculations are direct with no conceptual challenges, making it slightly easier than average for C3. |
| Spec | 1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Obtain derivative of form \(k(4t+9)^{-\frac{1}{2}}\) | M1 | any constant \(k\) |
| Obtain correct \(2(4t+9)^{-\frac{1}{2}}\) | A1 | or (unsimplified) equiv |
| Obtain derivative of form \(ke^{\frac{1}{2}x+1}\) | M1 | any constant \(k\) different from 6 |
| Obtain correct \(3e^{\frac{1}{2}x+1}\) | A1 | 4 or equiv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Either: Form product of two derivatives | M1 | numerical or algebraic |
| Substitute for \(t\) and \(x\) in product | M1 | using \(t=4\) and calculated value of \(x\) |
| Obtain 39.7 | A1 | 3 allow \(\pm 0.1\); allow greater accuracy |
| Or: Obtain \(k(4t+9)^n e^{\frac{1}{2}(4t+9)^{\frac{1}{2}}+1}\) | M1 | differentiating \(y = 6e^{\frac{1}{2}(4t+9)^{\frac{1}{2}}+1}\) |
| Obtain correct \(6(4t+9)^{-\frac{1}{2}}e^{\frac{1}{2}(4t+9)^{\frac{1}{2}}+1}\) | A1 | or equiv |
| Substitute \(t=4\) to obtain 39.7 | A1 | (3) allow \(\pm 0.1\); allow greater accuracy |
# Question 4:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Obtain derivative of form $k(4t+9)^{-\frac{1}{2}}$ | M1 | any constant $k$ |
| Obtain correct $2(4t+9)^{-\frac{1}{2}}$ | A1 | or (unsimplified) equiv |
| Obtain derivative of form $ke^{\frac{1}{2}x+1}$ | M1 | any constant $k$ different from 6 |
| Obtain correct $3e^{\frac{1}{2}x+1}$ | A1 | **4** or equiv |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Either: Form product of two derivatives | M1 | numerical or algebraic |
| Substitute for $t$ and $x$ in product | M1 | using $t=4$ and calculated value of $x$ |
| Obtain 39.7 | A1 | **3** allow $\pm 0.1$; allow greater accuracy |
| Or: Obtain $k(4t+9)^n e^{\frac{1}{2}(4t+9)^{\frac{1}{2}}+1}$ | M1 | differentiating $y = 6e^{\frac{1}{2}(4t+9)^{\frac{1}{2}}+1}$ |
| Obtain correct $6(4t+9)^{-\frac{1}{2}}e^{\frac{1}{2}(4t+9)^{\frac{1}{2}}+1}$ | A1 | or equiv |
| Substitute $t=4$ to obtain 39.7 | A1 | **(3)** allow $\pm 0.1$; allow greater accuracy |
---4 (i) Given that $x = ( 4 t + 9 ) ^ { \frac { 1 } { 2 } }$ and $y = 6 \mathrm { e } ^ { \frac { 1 } { 2 } x + 1 }$, find expressions for $\frac { \mathrm { d } x } { \mathrm {~d} t }$ and $\frac { \mathrm { d } y } { \mathrm {~d} x }$.\\
(ii) Hence find the value of $\frac { \mathrm { d } y } { \mathrm {~d} t }$ when $t = 4$, giving your answer correct to 3 significant figures.
\hfill \mbox{\textit{OCR C3 2007 Q4 [7]}}