Find dy/dx and d2y/dx2 without eliminating the parameter. x=1-t^(2), y=1+t
The Correct Answer and Explanation is:
Given the parametric equations:x=1−t2,y=1+tx = 1 – t^2,\quad y = 1 + tx=1−t2,y=1+t
We are to find:
- dydx\frac{dy}{dx}dxdy — the first derivative of yyy with respect to xxx,
- d2ydx2\frac{d^2y}{dx^2}dx2d2y — the second derivative of yyy with respect to xxx,
without eliminating the parameter.
Step 1: First Derivative dydx\frac{dy}{dx}dxdy
Using parametric differentiation, the first derivative is given by:dydx=dydtdxdt\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}dxdy=dtdxdtdy
Differentiate xxx and yyy with respect to ttt:
- dxdt=ddt(1−t2)=−2t\frac{dx}{dt} = \frac{d}{dt}(1 – t^2) = -2tdtdx=dtd(1−t2)=−2t
- dydt=ddt(1+t)=1\frac{dy}{dt} = \frac{d}{dt}(1 + t) = 1dtdy=dtd(1+t)=1
So,dydx=1−2t=−12t\frac{dy}{dx} = \frac{1}{-2t} = -\frac{1}{2t}dxdy=−2t1=−2t1
Step 2: Second Derivative d2ydx2\frac{d^2y}{dx^2}dx2d2y
The second derivative in parametric form is given by:d2ydx2=ddt(dydx)÷dxdt\frac{d^2y}{dx^2} = \frac{d}{dt}\left( \frac{dy}{dx} \right) \div \frac{dx}{dt}dx2d2y=dtd(dxdy)÷dtdx
We already have dydx=−12t\frac{dy}{dx} = -\frac{1}{2t}dxdy=−2t1, so compute its derivative with respect to ttt:ddt(−12t)=12t2\frac{d}{dt}\left( -\frac{1}{2t} \right) = \frac{1}{2t^2}dtd(−2t1)=2t21
Now divide by dxdt=−2t\frac{dx}{dt} = -2tdtdx=−2t:d2ydx2=12t2−2t=−14t3\frac{d^2y}{dx^2} = \frac{\frac{1}{2t^2}}{-2t} = -\frac{1}{4t^3}dx2d2y=−2t2t21=−4t31
Final Answers:
dydx=−12t,d2ydx2=−14t3\frac{dy}{dx} = -\frac{1}{2t},\quad \frac{d^2y}{dx^2} = -\frac{1}{4t^3}dxdy=−2t1,dx2d2y=−4t31
Explanation
When working with parametric equations, such as x=1−t2x = 1 – t^2x=1−t2 and y=1+ty = 1 + ty=1+t, we treat both xxx and yyy as functions of a third variable ttt, known as the parameter. Instead of expressing yyy directly in terms of xxx, we differentiate each with respect to ttt, and use these derivatives to find the derivatives of yyy with respect to xxx.
To find the first derivative dydx\frac{dy}{dx}dxdy, we apply the chain rule of calculus. Since dy/dx=(dy/dt)/(dx/dt)dy/dx = (dy/dt)/(dx/dt)dy/dx=(dy/dt)/(dx/dt), we compute the derivatives:
- dy/dt=1dy/dt = 1dy/dt=1, since y=1+ty = 1 + ty=1+t,
- dx/dt=−2tdx/dt = -2tdx/dt=−2t, because x=1−t2x = 1 – t^2x=1−t2.
Substituting, we get dydx=1−2t=−12t\frac{dy}{dx} = \frac{1}{-2t} = -\frac{1}{2t}dxdy=−2t1=−2t1.
To find the second derivative d2ydx2\frac{d^2y}{dx^2}dx2d2y, we use the formula:d2ydx2=ddt(dydx)÷dxdt\frac{d^2y}{dx^2} = \frac{d}{dt}\left( \frac{dy}{dx} \right) \div \frac{dx}{dt}dx2d2y=dtd(dxdy)÷dtdx
This involves differentiating −12t-\frac{1}{2t}−2t1 with respect to ttt, which gives 12t2\frac{1}{2t^2}2t21, and then dividing this result by −2t-2t−2t, the derivative of xxx. Thus:d2ydx2=12t2−2t=−14t3\frac{d^2y}{dx^2} = \frac{\frac{1}{2t^2}}{-2t} = -\frac{1}{4t^3}dx2d2y=−2t2t21=−4t31
This process allows us to compute the derivatives directly from the parametric equations without needing to eliminate the parameter.
