AHL 5.14—Implicit functions, related rates, optimisation

We are given $\frac{dr}{dt}=0.3\text{ cm/s}$ and $r=5\text{ cm}$ at the instant of interest. We need $\frac{dV}{dt}$.

Starting from $V=\frac{4}{3}\pi r^3$, differentiate both sides with respect to $t$ using the chain rule: $\frac{dV}{dt}=4\pi r^2\cdot \frac{dr}{dt}$

Substitute $r=5$ and $\frac{dr}{dt}=0.3$: $\frac{dV}{dt}=4\pi (5{)}^2(0.3)=4\pi \cdot 25\cdot 0.3=30\pi {\text{ cm}}^3/\text{s}$

The rate of increase of volume at that instant is $30\pi {\text{ cm}}^3/\text{s}$.

The chain rule gives $\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$. With $r=5$ and $\frac{dr}{dt}=0.3$, the calculation is $4\pi (25)(0.3)=30\pi$.

$10\pi$ would result if one incorrectly used $r=5$ without squaring, e.g. $4\pi (5)(0.3)/0.6$. This does not match the formula $4\pi r^2\frac{dr}{dt}$.

$60\pi$ would arise from forgetting to include the factor of $0.3$ and instead using $\frac{dr}{dt}=0.6$, or using $r^2=50$ instead of $25$. Neither is correct here.

$15\pi$ might come from using $\frac{dV}{dt}=4\pi r\cdot \frac{dr}{dt}$ (forgetting to square $r$), giving $4\pi (5)(0.3)\cdot \frac{1}{2}=3\pi$... none match. This distractor reflects a coefficient error.

The only option consistent with the correct application of the chain rule is $30\pi {\text{ cm}}^3/\text{s}$.

Differentiate $x^{2}y+y^3=10$ with respect to $x$. Use the product rule on $x^{2}y$: $\frac{d}{dx}(x^{2}y)=2xy+x^2\frac{dy}{dx}$

Differentiating $y^3$ with respect to $x$ gives $3y^2\frac{dy}{dx}$. The right side differentiates to $0$. So: $2xy+x^2\frac{dy}{dx}+3y^2\frac{dy}{dx}=0$

$\frac{dy}{dx}(x^2+3y^2)=-2xy⟹\frac{dy}{dx}=-\frac{2xy}{x^2+3y^2}$

The correct derivative is $\frac{dy}{dx}=-\frac{2xy}{x^2+3y^2}$.

The product rule on $x^{2}y$ yields $2xy+x^2\frac{dy}{dx}$; the chain rule on $y^3$ yields $3y^2\frac{dy}{dx}$. The numerator of $\frac{dy}{dx}$ must come from moving $2xy$ to the right, giving $-2xy$.

This option omits $y$ from the numerator, which would be the case if the product rule was not applied to $x^{2}y$ — incorrectly treating it as just $2x\cdot y$ without noting that $y$ also depends on $x$.

This has the correct magnitude but the wrong sign. When collecting $\frac{dy}{dx}(x^2+3y^2)=-2xy$, dividing both sides must produce a negative result.

This option confuses the product rule: it appears to differentiate $x^2$ as $2x$ but treats $y$ as $y$ in the numerator rather than $2xy$, and collapses the denominator incorrectly.

Only $-\frac{2xy}{x^2+3y^2}$ correctly applies both the product rule and the chain rule.