Differentiation — practice questions

First, we need to find the gradient of the tangent, which is the value of $\frac{dy}{dx}$ at $x=0$. The function is a product of $u = e^{2x}$ and $v = \cos x$.

1. Differentiate u and v: $u = e^{2x} \implies \frac{du}{dx} = 2e^{2x}$ $v = \cos x \implies \frac{dv}{dx} = -\sin x$
2. Apply the Product Rule: $\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$ $\frac{dy}{dx} = (e^{2x})(-\sin x) + (\cos x)(2e^{2x})$ $\frac{dy}{dx} = e^{2x}(2\cos x - \sin x)$
3. Find the gradient at x=0: At $x=0$, $m = e^{2(0)}(2\cos(0) - \sin(0)) = e^0(2(1) - 0) = 1 \times 2 = 2$.
4. Find the y-coordinate at x=0: At $x=0$, $y = e^{2(0)}\cos(0) = e^0 \times 1 = 1$. The point is $(0, 1)$.
5. Find the equation of the tangent: Using $y - y_1 = m(x - x_1)$: $y - 1 = 2(x - 0)$ $y = 2x + 1$

Final Answer: The equation of the tangent is $y = 2x + 1$.

1. Identify variables and rates: Given: Rate of change of volume, $\frac{dV}{dt} = 50 \text{ cm}^3\text{s}^{-1}$. Required: Rate of change of radius, $\frac{dr}{dt}$, when $r = 10$ cm.
2. Find the connecting equation: The equation connecting volume $V$ and radius $r$ is $V = \frac{4}{3}\pi r^3$.
3. Differentiate with respect to time, t: We use the chain rule: $\frac{dV}{dt} = \frac{dV}{dr} \times \frac{dr}{dt}$. First, find $\frac{dV}{dr}$ from the volume formula: $\frac{dV}{dr} = \frac{d}{dr}(\frac{4}{3}\pi r^3) = 4\pi r^2$. Now substitute this into the chain rule expression: $\frac{dV}{dt} = (4\pi r^2) \times \frac{dr}{dt}$.
4. Substitute known values and solve: We are given $\frac{dV}{dt} = 50$ and we want to find $\frac{dr}{dt}$ when $r=10$. $50 = (4\pi (10)^2) \times \frac{dr}{dt}$ $50 = 400\pi \times \frac{dr}{dt}$ $\frac{dr}{dt} = \frac{50}{400\pi} = \frac{1}{8\pi}$
5. Final Answer with units: The rate of increase of the radius is $\frac{1}{8\pi} \text{ cm s}^{-1}$. Numerically, this is approximately $0.0398 \text{ cm s}^{-1}$ (3 s.f.).