Homework Soursegiven the parametric curve defined by x3t2t and yt24 find th
given the parametric curve defined by x=3t^2+t and y=t^2-4. find the point (x, y) on this curve when t=2 & the the slope of the line tangent to this curve at t=2
Solution
x=3t^2+t and y=t^2-4 When t = 2 x = 3*4+2 = 14 y = 4-4 = 0 therefore(x,y) = (14,0) differentiating x wrt to t we get dx/dt = 6t+1 ...... Equation (1) differentiating y wrt to t we get dy/dt = 2 ...... Equation (2) Dividing equation(2) by equation(1) we get dy/dx = 2/(6t+1) This is the general equation of the slope Therefore slope of the tangent at t=2 dy/dx at t = 2 = 2/(6*2+1) = 2/13 Hope this helps!!! :)
