Find dydxandd2ydx2. For which values of t is the curve concave upward? x=t2+1, y=t2+t

Question

Find dydxandd2ydx2. For which values of t is the curve concave upward? x=t2+1, y=t2+t

Witheyesse47 · Accepted Answer

Using the chain rule, we can write dydx=dydt⋅dtdx=dydtdxdt→ (Equation 1) It is given that x=t2+1 We can use the sum rule followed by the power rule to differentiate this dxdt=2t2−1+0=2t It is given that y=t2+t We can use the sum rule followed by the power rule to differentiate this dydt=2t2−1+1=2t+1 Substitute the expressions for dy/dt and dx/dt in (Equation 1), to get dydx=2t+12t=1+12t To find the concavity, we need the second derivative with respectr to x, so we will now differentiate both sides with respect to x ddx[dydx]=ddx[1+12t] In the right-hand side, we can use the chain rule to write d2ydx2=ddt[1+12t]⋅dtdx d2ydx2=[0−12t62]⋅1dxdt In the red expressiob, replace dx/dt with 2t d2ydx2=−12t2⋅12t=−14t3 Remembet that the graph is concave up/down depending on whether the second derivate is positive/negative respectively. Note that: (1)the number is never 0, so there are no sign changes that occur because of that. (2) the denominator is 0 when t = 0. This makes rational function discontinuous, so we will use that to divide the number line into intervals that need to be checked. To check thi sign on (−∞,0),we will used=−1 d2ydx2∣t=−1=−

Befoodly · Accepted Answer

x=t2+1,y=t2+t  Find the derivates with respect to t  dxdt=2tdydt=2t+1  then we can find  dydx=dydtdxdt=2t+12t  For the next one we take the derivate of dydx with respect to t  ddt(dydx)=2(2t)−(2t+1)(2)4t2=4t−4t−24t2=−12t2  then divide it by dxdt  d2ydx2=ddt(dydx)dxdt=−12t22t=−14t3  d2ydx2 is positive and the curve is concave up for t &amp;lt; 0  d2ydx2 is negative and the curve is concave down for t &amp;gt; 0  Answer  2t+12t,−14t3,C.U.fort&amp;lt;0

user_27qwe · Accepted Answer

Given that: x=t2+1 Derive x with respect to t: dxdt=2t Given that: y=t2+t Derive y with respect to t:dydt=2t=1 Using chain rule to find: dydx=dydt×dtdx=dy/dtdx/dt=2t+12t=1+12t Take the derivative of dydx with respect to t: ddt(dydx)=2(2t)−(2t+1)(2)4t2=4t−4t−24t2=−12t2 and then divide it by dxdt to find d2ydx2: d2ydx2=ddt(dydx)dx/dt=−12t22t=−14t3 The curve is concave upward when d2ydx2&amp;gt;0, that is, when t&amp;lt;0 1+12t,−14t3,C.U. for t&amp;lt;0

Find \frac{dy}{dx} and \frac{d^{2}y}{dx^2}. For which values of t is the c

Answered question

Answer & Explanation

New Questions in Other