MATH6015: Week 13

December 12, 2012 in MATH6015

Test 2 Solutions

Please find the test 2 solutions and marking scheme here.

Exam Layout

Answer Q.1 [40 Marks] and two out of Q. 2, 3, 4 [30 marks each].

Q. 1 — 8 short questions each worth 5 marks each

Q. 2 — Differentiation with applications

Q. 3 — Integration with applications

Q.4 — Applications of differentiation and integration

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December 14, 2012 at 3:15 pm

James Goggin

I was wondering if you could give me a hand with a couple of questions from the 2009 exam paper Q 1 d: given that $f(x)= xe^{-2x}$ , evaluate $f'(0.2)$ and $f'(1.4)$
and Q 1 e:

$\displaystyle \int_{\pi/4}^{\pi/2}(3\sin 2x-2\cos 3x)\,dx$ .

Cheers

December 14, 2012 at 4:38 pm

J.P. McCarthy

James,

$f'(x)$ means the derivative of $f(x)=xe^{-2x}$ . Note also that

$\displaystyle \frac{dy}{dx}\equiv f'(x)\equiv f'\equiv$ ‘the derivative of $f(x)$

Here $f'(0.2)$ means find the value of the derivative when $x=0.2$ .

$f(x)=x\cdot e^{-2x}$ is the product of the functions $u(x)=x$ and $v(x)=e^{-2x}$ . The Product Rule says the derivative of a product $y=u\cdot v$ , rather than $\displaystyle \frac{dy}{dx}=\frac{du}{dx}\cdot\frac{dv}{dx}$ is

$\displaystyle \frac{dy}{dx}=u\cdot \frac{dv}{dx}+v\cdot \frac{du}{dx}$ .

In words, the first function by the derivative of the second, plus the second function times the derivative of the first. We can get the derivative $u=x$ very easily: $u'=1$ . The derivative of $v=e^{-2x}$ is actually presented in the tables as

$\displaystyle y=e^{ax}\Rightarrow \frac{dy}{dx}=ae^{ax}$ for a constant $a$ .

This comes from a Chain Rule:

$g(x)= g(h(x))=e^{ax}\Rightarrow g'(x)=g'(h(x))\cdot h'(x)$
$=e^{ax}\cdot\left(\frac{d}{dx}(ax)\right)=ae^{ax}$ .

All in all we have

$f'(x)=x(-2e^{-2x})+e^{-2x}(1)=e^{-2x}(1-2x)$ .

Now evaluate at $x=0.2$ :

$f'(0.2)=\left.e^{-2x}(1-2x)\right|_{x=0.2}=e^{-0.4}(0.6)\approx 0.402$ .

Similarly

$f'(1.4)=\left.e^{-2x}(1-2x)\right|_{x=1.4}=e^{-2.8}(-1.8)\approx -0.110$ .

Now for your second question. The first thing is to say is that integration is linear. These means that you can pull out constants and differentiate a sum of terms term-by-term. This means that what we are looking at is

$3\int_{\pi/4}^{\pi/2}\sin 2x\,dx-2\int_{\pi/4}^{\pi/2}\cos 3x\,dx$ .

Can this be done directly? I would argue no because

$\displaystyle \int \cos (ax)\,dx$ and

$\displaystyle \int \sin (ax)\,dx$ .

are not in the tables where $a$ is a constant. What about a manipulation? Well there are many but none of them actually simplify the integrand. This means that we need a substitution. Say for the first one we could pick $u=ax$ . We could do this because

$u=ax$ is a function which has derivative $a$ which is just a ‘multiple’…

LIATE — no logs, no inverse trig, algebraic; yes $u=ax$ .

$u=ax$ is ‘inside’ another function.

So we implement the substitution $u=ax$

$\displaystyle\Rightarrow\frac{du}{dx}=a\Rightarrow dx=\frac{du}{a}$ .

Now put the integral back together

$\displaystyle \int \cos (ax)\,dx=\int \cos(u)\cdot\frac{du}{a}=\frac{1}{a}\int \cos u\,du$ .

The integral of $\cos x$ with respect to $x$ is $\sin x$ — as the derivative of $\sin x$ with respect to $x$ is $\cos x$ . Hence we write

$\displaystyle\int \cos(ax)\,dx=\frac{1}{a}\sin u=\frac{1}{a}\sin(ax)+C$ .

Similarly,

$\displaystyle\int \sin(ax)\,dx=-\frac{1}{a}\cos (ax)+C$ .

Some people learn these formula by comparing with

$\displaystyle\int e^{ax}\,dx=\frac{1}{a}e^{ax}+C$ .

We will do that hear and say that these integrals can be integrated directly:

$3\int_{\pi/4}^{\pi/2}\sin 2x\,dx-2\int_{\pi/4}^{\pi/2}\cos 3x\,dx$

$=3\left[-\frac{1}{2}\cos(2x)\right]_{\pi/4}^{\pi/2}-2\left[\frac{1}{3}\sin(3x)\right]_{\pi/4}^{\pi/2}$

$=-\frac{3}{2}\left[\cos(2x)\right]_{\pi/4}^{\pi/2}-\frac{2}{3}\left[\sin(3x)\right]_{\pi/4}^{\pi/2}$

$=-\frac{3}{2}\left[\cos(\pi)-\cos(\pi/2)\right]-\frac{2}{3}\left[\sin(3\pi/2)-\sin(3\pi/4)\right]$

There are ways to evaluate the sines and cosines without a calculator but at this late date I either assume that you know how to do this or will use the calculator. I warn you that if you are using the calculator make sure that you are in radian mode:

$=-\frac{3}{2}\left[-1-0\right]-\frac{2}{3}\left[-1-\frac{1}{\sqrt{2}}\right]$

$=\frac{3}{2}+\frac{2}{3}\left[\frac{\sqrt{2}+1}{\sqrt{2}}\right]\approx 2.638$

Regards and Happy Christmas,
J.P.

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MATH6015: Week 13

Test 2 Solutions

Exam Layout

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J.P. McCarthy Maths

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MATH6015: Week 13

Test 2 Solutions

Exam Layout

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