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a. For the curve given parametrically by x=3t^3-5t, y=t^4+1, compute \frac{\mathrm{d}...

Question:

a. For the curve given parametrically by {eq}x=3t^3-5t, y=t^4+1 {/eq}, compute {eq}\frac{\mathrm{d} y}{\mathrm{d} x} {/eq} and {eq}\frac{\mathrm{d^2} y}{\mathrm{d} x^2} {/eq}

b. For the graph of the function {eq}y=\ln (\cos x) {/eq}, set up an integral to compute the arc-length from {eq}x=0 {/eq} to {eq}x=\frac{\pi }{3} {/eq}.

Arc Length


The arc length of a function {eq}\displaystyle y=f(x), a\leq x\leq b {/eq}

is {eq}\displaystyle \int_a^b \sqrt{1+(f'(x))^2}dx, {/eq} where {eq}\displaystyle '=\frac{d}{dx}. {/eq}

To calculate the derivatives {eq}\displaystyle \frac{dy}{dx}, \frac{d^2y}{dx^2} {/eq} when the curve is given parametrically {eq}\displaystyle x=x(t), y=y(t) {/eq}

we will use Chain rule, like below

{eq}\displaystyle \frac{dy}{dx}(t)=\frac{dy}{dt}\cdot \frac{dt}{dx}=\frac{y'(t)}{x'(t)}, {/eq}

and {eq}\displaystyle \frac{d^2y}{dx^2}=\frac{d}{dt}\left( \frac{dy}{dx} \right)\cdot \frac{dt}{dx}=\frac{d}{dt}\left( \frac{y'(t)}{x'(t)} \right)\cdot \frac{1}{x'(t)}. {/eq}

Answer and Explanation:


a. To find the first and second derivatives, {eq}\displaystyle \frac{dy}{dx}, \frac{d^2y}{dx^2} {/eq} for the parametric curve {eq}\displaystyle x = 3t^3 -5t, y =t^4+1 {/eq}

we will need the first derivatives of the function with respect to t,

{eq}\displaystyle \begin{align} \frac{dy}{dt}&=\frac{d}{dt}(t^4+1)=4t^3\\ \frac{dx}{dt}&=\frac{d}{dt}(3t^3-5t)=9t^2-5\\ \implies \frac{dy}{dx}&=\frac{y'(t)}{x'(t)}=\boxed{\frac{4t^3}{9t^2-5}}\\ \frac{d^2y}{dx^2}&=\frac{d}{dt}\left( \frac{dy}{dx} \right)\cdot \frac{dt}{dx}= \frac{d}{dt}\left( \frac{4t^3}{9t^2-5}\right)\cdot \frac{1}{9t^2-5}\\ &=\frac{36t^4-60t^2}{(9t^2-5)^3}\\ &= \boxed{\frac{12t^2(3t^2-5)}{(9t^2-5)^3}}. \end{align} {/eq}


b. The arc length of the curve {eq}\displaystyle y=\ln(\cos x), \text{ for } 0\leq x\leq \frac{\pi}{3} {/eq}

is given by {eq}\displaystyle \int_0^{\pi/3} \sqrt{1+(y'(x))^2}\ dx= \int_0^{\pi/3} \sqrt{1+\left(-\frac{\sin x}{\cos x}\right)^2}\ dx=\boxed{ \int_0^{\pi/3} \frac{1}{\cos x}\ dx}, {/eq}

where we used {eq}\displaystyle \sin^2 x+\cos^2 x=1 \text{ and } \sqrt{\cos^2 x}=|\cos x| =\cos x, \text{ on } \left[ 0,\frac{\pi}{3}\right]. {/eq}


Learn more about this topic:

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Parametric Equations in Applied Contexts

from Precalculus: High School

Chapter 24 / Lesson 6
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