Q2. By the chain rule,
![\dfrac{dy}{dx} = \dfrac{dy}{dt} \cdot \dfrac{dt}{dx} = \dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}](https://tex.z-dn.net/?f=%5Cdfrac%7Bdy%7D%7Bdx%7D%20%3D%20%5Cdfrac%7Bdy%7D%7Bdt%7D%20%5Ccdot%20%5Cdfrac%7Bdt%7D%7Bdx%7D%20%3D%20%5Cdfrac%7B%5Cfrac%7Bdy%7D%7Bdt%7D%7D%7B%5Cfrac%7Bdx%7D%7Bdt%7D%7D)
We have
![x=2t \implies \dfrac{dx}{dt}=2](https://tex.z-dn.net/?f=x%3D2t%20%5Cimplies%20%5Cdfrac%7Bdx%7D%7Bdt%7D%3D2)
![y=t^4+1 \implies \dfrac{dy}{dt}=4t^3](https://tex.z-dn.net/?f=y%3Dt%5E4%2B1%20%5Cimplies%20%5Cdfrac%7Bdy%7D%7Bdt%7D%3D4t%5E3)
The slope of the tangent line to the curve at
is then
![\dfrac{dy}{dx} \bigg|_{t=1} = \dfrac{4t^3}{2} \bigg|_{t=1} = 2t^3\bigg|_{t=1} = 2](https://tex.z-dn.net/?f=%5Cdfrac%7Bdy%7D%7Bdx%7D%20%5Cbigg%7C_%7Bt%3D1%7D%20%3D%20%5Cdfrac%7B4t%5E3%7D%7B2%7D%20%5Cbigg%7C_%7Bt%3D1%7D%20%3D%202t%5E3%5Cbigg%7C_%7Bt%3D1%7D%20%3D%202)
so the slope of the normal line is
. When
, we have
![x\bigg|_{t=1} = 2t\bigg|_{t=1} = 2](https://tex.z-dn.net/?f=x%5Cbigg%7C_%7Bt%3D1%7D%20%3D%202t%5Cbigg%7C_%7Bt%3D1%7D%20%3D%202)
![y\bigg|_{t=1} = (t^4+1)\bigg|_{t=1} = 2](https://tex.z-dn.net/?f=y%5Cbigg%7C_%7Bt%3D1%7D%20%3D%20%28t%5E4%2B1%29%5Cbigg%7C_%7Bt%3D1%7D%20%3D%202)
so the curve passes through (2, 2). Using the point-slope formula for a line, the equation of the normal line is
![y - 2 = -\dfrac12 (x - 2) \implies y = -\dfrac12 x + 3](https://tex.z-dn.net/?f=y%20-%202%20%3D%20-%5Cdfrac12%20%28x%20-%202%29%20%5Cimplies%20y%20%3D%20-%5Cdfrac12%20x%20%2B%203)
Q3. Differentiating with the product, power, and chain rules, we have
![y = x(x+1)^{1/2} \implies \dfrac{dy}{dx} = \dfrac{3x+2}{2\sqrt{x+1}} \implies \dfrac{dy}{dx}\bigg|_{x=3} = \dfrac{11}4](https://tex.z-dn.net/?f=y%20%3D%20x%28x%2B1%29%5E%7B1%2F2%7D%20%5Cimplies%20%5Cdfrac%7Bdy%7D%7Bdx%7D%20%3D%20%5Cdfrac%7B3x%2B2%7D%7B2%5Csqrt%7Bx%2B1%7D%7D%20%5Cimplies%20%5Cdfrac%7Bdy%7D%7Bdx%7D%5Cbigg%7C_%7Bx%3D3%7D%20%3D%20%5Cdfrac%7B11%7D4)
The derivative vanishes when
![\dfrac{3x+2}{2\sqrt{x+1}} = 0 \implies 3x+2=0 \implies x = -\dfrac23](https://tex.z-dn.net/?f=%5Cdfrac%7B3x%2B2%7D%7B2%5Csqrt%7Bx%2B1%7D%7D%20%3D%200%20%5Cimplies%203x%2B2%3D0%20%5Cimplies%20x%20%3D%20-%5Cdfrac23)
Q4. Differentiating with the product and chain rules, we have
![y = (2x+1)e^{-2x} \implies \dfrac{dy}{dx} = -4xe^{-2x}](https://tex.z-dn.net/?f=y%20%3D%20%282x%2B1%29e%5E%7B-2x%7D%20%5Cimplies%20%5Cdfrac%7Bdy%7D%7Bdx%7D%20%3D%20-4xe%5E%7B-2x%7D)
The stationary points occur where the derivative is zero.
![-4xe^{-2x} = 0 \implies x = 0](https://tex.z-dn.net/?f=-4xe%5E%7B-2x%7D%20%3D%200%20%5Cimplies%20x%20%3D%200)
at which point we have
![y = (2x+1)e^{-2x} \bigg|_{x=0} = 1](https://tex.z-dn.net/?f=y%20%3D%20%282x%2B1%29e%5E%7B-2x%7D%20%5Cbigg%7C_%7Bx%3D0%7D%20%3D%201)
so the stationary point has coordinates (0, 1). By its "nature", I assume the question is asking what kind of local extremum this point. Compute the second derivative and evaluate it at
.
![\dfrac{d^2y}{dx^2}\bigg|_{x=0} = (8x-4)e^{-2x}\bigg|_{x=0} = -4 < 0](https://tex.z-dn.net/?f=%5Cdfrac%7Bd%5E2y%7D%7Bdx%5E2%7D%5Cbigg%7C_%7Bx%3D0%7D%20%3D%20%288x-4%29e%5E%7B-2x%7D%5Cbigg%7C_%7Bx%3D0%7D%20%3D%20-4%20%3C%200)
The negative sign tells us this stationary point is a local maximum.
Q5. Differentiating the volume equation implicitly with respect to
, we have
![V = \dfrac{4\pi}3 r^3 \implies \dfrac{dV}{dt} = 4\pi r^2 \dfrac{dr}{dt}](https://tex.z-dn.net/?f=V%20%3D%20%5Cdfrac%7B4%5Cpi%7D3%20r%5E3%20%5Cimplies%20%5Cdfrac%7BdV%7D%7Bdt%7D%20%3D%204%5Cpi%20r%5E2%20%5Cdfrac%7Bdr%7D%7Bdt%7D)
When
, and given it changes at a rate
, we have
![\dfrac{dV}{dt} = 4\pi (5\,\mathrm{cm})^2 \left(-1.5\dfrac{\rm cm}{\rm s}\right) = -150\pi \dfrac{\rm cm^3}{\rm s}](https://tex.z-dn.net/?f=%5Cdfrac%7BdV%7D%7Bdt%7D%20%3D%204%5Cpi%20%285%5C%2C%5Cmathrm%7Bcm%7D%29%5E2%20%5Cleft%28-1.5%5Cdfrac%7B%5Crm%20cm%7D%7B%5Crm%20s%7D%5Cright%29%20%3D%20-150%5Cpi%20%5Cdfrac%7B%5Crm%20cm%5E3%7D%7B%5Crm%20s%7D)
Q6. Given that
is fixed, we have
![V = \pi r^2h \implies h = \dfrac{400\pi}{\pi r^2} = \dfrac{400}{r^2}](https://tex.z-dn.net/?f=V%20%3D%20%5Cpi%20r%5E2h%20%5Cimplies%20h%20%3D%20%5Cdfrac%7B400%5Cpi%7D%7B%5Cpi%20r%5E2%7D%20%3D%20%5Cdfrac%7B400%7D%7Br%5E2%7D)
Substitute this into the area equation to make it dependent only on
.
![A = \pi r^2 + 2\pi r \left(\dfrac{400}{r^2}\right) = \pi r^2 + \dfrac{800\pi}r](https://tex.z-dn.net/?f=A%20%3D%20%5Cpi%20r%5E2%20%2B%202%5Cpi%20r%20%5Cleft%28%5Cdfrac%7B400%7D%7Br%5E2%7D%5Cright%29%20%3D%20%5Cpi%20r%5E2%20%2B%20%5Cdfrac%7B800%5Cpi%7Dr)
Find the critical points of
.
![\dfrac{dA}{dr} = 2\pi r - \dfrac{800\pi}{r^2} = 0 \implies r = \dfrac{400}{r^2} \implies r^3 = 400 \implies r = 2\sqrt[3]{50}](https://tex.z-dn.net/?f=%5Cdfrac%7BdA%7D%7Bdr%7D%20%3D%202%5Cpi%20r%20-%20%5Cdfrac%7B800%5Cpi%7D%7Br%5E2%7D%20%3D%200%20%5Cimplies%20r%20%3D%20%5Cdfrac%7B400%7D%7Br%5E2%7D%20%5Cimplies%20r%5E3%20%3D%20400%20%5Cimplies%20r%20%3D%202%5Csqrt%5B3%5D%7B50%7D)
Check the sign of the second derivative at this radius to confirm it's a local minimum (sign should be positive).
![\dfrac{d^2A}{dr^2}\bigg|_{r=2\sqrt[3]{50}} = \left(2\pi + \dfrac{1600\pi}{r^3}\right)\bigg|_{r=2\sqrt[3]{50}} = 6\pi > 0](https://tex.z-dn.net/?f=%5Cdfrac%7Bd%5E2A%7D%7Bdr%5E2%7D%5Cbigg%7C_%7Br%3D2%5Csqrt%5B3%5D%7B50%7D%7D%20%3D%20%5Cleft%282%5Cpi%20%2B%20%5Cdfrac%7B1600%5Cpi%7D%7Br%5E3%7D%5Cright%29%5Cbigg%7C_%7Br%3D2%5Csqrt%5B3%5D%7B50%7D%7D%20%3D%206%5Cpi%20%3E%200)
Hence the minimum surface area is
![A\bigg_{r=2\sqrt[3]{50}\,\rm cm} = \left(\pi r^2 + \dfrac{800\pi}r\right)\bigg|_{r=2\sqrt[3]{50}\,\rm cm} = 60\pi\sqrt[3]{20}\,\rm cm^2](https://tex.z-dn.net/?f=A%5Cbigg_%7Br%3D2%5Csqrt%5B3%5D%7B50%7D%5C%2C%5Crm%20cm%7D%20%3D%20%5Cleft%28%5Cpi%20r%5E2%20%2B%20%5Cdfrac%7B800%5Cpi%7Dr%5Cright%29%5Cbigg%7C_%7Br%3D2%5Csqrt%5B3%5D%7B50%7D%5C%2C%5Crm%20cm%7D%20%3D%2060%5Cpi%5Csqrt%5B3%5D%7B20%7D%5C%2C%5Crm%20cm%5E2)
Q7. The volume of the box is
![V = 8x^2](https://tex.z-dn.net/?f=V%20%3D%208x%5E2)
(note that the coefficient 8 is measured in cm) while its surface area is
![A = 2x^2 + 12x](https://tex.z-dn.net/?f=A%20%3D%202x%5E2%20%2B%2012x)
(there are two
-by-
faces and four 8-by-
faces; again, the coefficient 12 has units of cm).
When
, we have
![210 = 2x^2 + 12x \implies x^2 + 6x - 105 = 0 \implies x = -3 \pm\sqrt{114}](https://tex.z-dn.net/?f=210%20%3D%202x%5E2%20%2B%2012x%20%5Cimplies%20x%5E2%20%2B%206x%20-%20105%20%3D%200%20%5Cimplies%20x%20%3D%20-3%20%5Cpm%5Csqrt%7B114%7D)
This has to be a positive length, so we have
.
Given that
, differentiate the volume and surface area equations with respect to
.
![\dfrac{dV}{dt} = (16\,\mathrm{cm})x \dfrac{dx}{dt} = (16\,\mathrm{cm})(\sqrt{114}-3\,\mathrm{cm})\left(0.05\dfrac{\rm cm}{\rm s}\right) = \dfrac{4(\sqrt{114}-3)}5 \dfrac{\rm cm^3}{\rm s}](https://tex.z-dn.net/?f=%5Cdfrac%7BdV%7D%7Bdt%7D%20%3D%20%2816%5C%2C%5Cmathrm%7Bcm%7D%29x%20%5Cdfrac%7Bdx%7D%7Bdt%7D%20%3D%20%2816%5C%2C%5Cmathrm%7Bcm%7D%29%28%5Csqrt%7B114%7D-3%5C%2C%5Cmathrm%7Bcm%7D%29%5Cleft%280.05%5Cdfrac%7B%5Crm%20cm%7D%7B%5Crm%20s%7D%5Cright%29%20%3D%20%5Cdfrac%7B4%28%5Csqrt%7B114%7D-3%29%7D5%20%5Cdfrac%7B%5Crm%20cm%5E3%7D%7B%5Crm%20s%7D)
![\dfrac{dA}{dt} = 4x\dfrac{dx}{dt} + (12\,\mathrm{cm})\dfrac{dx}{dt} = \left(4(\sqrt{114}-3\,\mathrm {cm}) + 12\,\mathrm{cm}\right)\left(0.05\dfrac{\rm cm}{\rm s}\right) = \dfrac{\sqrt{114}}5 \dfrac{\rm cm^2}{\rm s}](https://tex.z-dn.net/?f=%5Cdfrac%7BdA%7D%7Bdt%7D%20%3D%204x%5Cdfrac%7Bdx%7D%7Bdt%7D%20%2B%20%2812%5C%2C%5Cmathrm%7Bcm%7D%29%5Cdfrac%7Bdx%7D%7Bdt%7D%20%3D%20%5Cleft%284%28%5Csqrt%7B114%7D-3%5C%2C%5Cmathrm%20%7Bcm%7D%29%20%2B%2012%5C%2C%5Cmathrm%7Bcm%7D%5Cright%29%5Cleft%280.05%5Cdfrac%7B%5Crm%20cm%7D%7B%5Crm%20s%7D%5Cright%29%20%3D%20%5Cdfrac%7B%5Csqrt%7B114%7D%7D5%20%5Cdfrac%7B%5Crm%20cm%5E2%7D%7B%5Crm%20s%7D)