Question

a curve is defined by parametric functions x(t) and y(t), where dxdt=3t2 and dydt=t7−2t4. for what positive value of t is d2ydx2=1 ?

Answer 1

The correct answer is d) t = 1.525

The route to this result involves a number of steps.

First dy/dx = is easily obtained from the ratio of dy/dt and dx/dt as

dy/dt = (1/3) t5 - (2/3) t2 define Q as dy/dt

What is needed is dQ/dx (the second derivative of y wrt x)

dQ/dx = (dQ/dt) ( dt/dx) but dt/dx = 1/( 3 t2) so

dQ/dx = (5/9) t2 -(4/3) /t

Setting this equal to 1 results in a cubic equation with solution t = 1.525

Answer 2

A route of a constantly moving point is described by the phrase curve, which is an abstract term, in this, the positive value is 1.525.

Positive curve:

$\frac{dx}{dt}=3t^2\\\\\frac{dy}{dt}=t^7-2t^4\\\\\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{dy}{dt} \times \frac{dt}{dx}\\\\\frac{dy}{dx}= \frac{t^7-2t^4}{3t^2}=\frac{1}{3}(\frac{t^7}{t^2}-\frac{2t^4}{t^2})\\\\\frac{dy}{dx}=\frac{1}{3}(t^5-2t^2)$

differentiate the above equation with the respect to t:

$\frac{d}{dt}(\frac{dy}{dx})=\frac{d}{dt}(\frac{1}{3}(t^5-2t^2))\\\\\frac{dx}{dt} \cdot \frac{d}{dx}(\frac{dy}{dx})=(\frac{1}{3}(5t^4-4t))\\\\\frac{d^2y}{dx^2}=\frac{(\frac{1}{3}(5t^4-4t))}{\frac{dx}{dt}}\\\\\frac{d^2y}{dx^2}=\frac{(\frac{1}{3}(5t^4-4t))}{3t^2}$

Given that

$\bold{\frac{d^2y}{dx^2}= 1}$

then

$1= \frac{(\frac{1}{3}(5t^4-4t))}{3t^2}\\\\1= \frac{1}{9} \frac{(5t^4-4t)}{t^2}\\\\9= \frac{t(5t^3-4)}{t^2}\\\\9= \frac{(5t^3-4)}{t}\\\\9t=(5t^3-4)\\\\9t=5t^3-4\\\\5t^3-9t-4=0$

calculating the factors:

$(t+1) (t+0.5246)(t-1.5246)=0\\\\t=-1, 1.525 , and -0.525$

When taking the positive value the final answer is $t=1.525$

Find out more about the curve here:

brainly.com/question/13933442