Differential and Integral Calculus (21-120) — Feedback on Homework 3

Homework 3 was due on Thursday 12th September 2013 and consisted of:

• Section 3.2 Q 50
• Section 3.4 Q 72, 84

I marked 3.2/50 (out of 3), 3.4/72 (out of 3) and 3.4/84 parts (a)(b) (out of 3). Part (c) wasn't marked because it was essentially an exercise in using graphing software. Everyone got 1 free point for submitting their homework.

Section 3.2 Q50. The most common error was not using the product and quotient rules correctly. A few people said that $P'(2)=F'(2)G'(2)$, for example; but by the product rule, it's actually equal to $F'(2)G(2)+F(2)G'(2)$. Another error made by a few people was saying that all the derivatives were zero. This may have been down to confusion with the notation. When you see something like $G'(7)$, it means first differentiate $G(t)$ and then substitute $t=7$ in the derivative. If you substitute first, you end up with a constant, so you'll always get zero... this is wrong!

Section 3.4 Q72. Common errors included:

• Not applying the product or chain rules properly. Often the computational errors in this problem would have been solved by putting in a few extra lines of working!
• Not simplifying the answer. I didn't penalise anyone for this, but seeing $2xg'(x^2)+4xg'(x^2)+4x^3g''(x^2)$ as a final answer, whilst correct, isn't really good enough: it should be $6xg'(x^2)+4x^3g''(x^2)$.

Section 3.4 Q84. Most of the errors on this problem were very small:

• Lots of people gave the wrong answer for the limit in part (a). The most common instance of this was people saying that $e^{-kt} \to \infty$ as $t \to \infty$. This is not so: if $k>0$ then $e^{-kt}$ becomes very small as $t$ becomes very large, so in fact $e^{-kt} \to 0$ as $t \to \infty$. Thus $$\lim_{t \to \infty} \frac{1}{1+ae^{-kt}} = \frac{1}{1+a \cdot 0} = 1$$
• The problems with part (b) were very similar to the problems in Q72: namely, not putting enough lines of working in. There were two paths through this problem, one using the quotient rule once and the chain rule once, and the other using the chain rule twice. The main error was common to both methods: at some point, you have to use the chain rule on $ae^{-kt}$. The best way to do this is to substitute $u=-kt$, then it's clear that $\frac{du}{dt} = -k$ and hence that $\frac{d}{dt} (ae^{-kt}) = ae^{-kt} \times (-k) = -kae^{-kt}$. Many people left out the $k$, or multiplied by $t$... going a bit slower would probably mean these silly errors are avoided. A few others decreased the exponent by $1$: this only holds for polynomials, not exponentials! Remember the difference!

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