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Derivative of functions having key role in differential Calculus. Solution is an easy approach with detail of all relevant concepts, formulas and information
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A curve has parametric equation x = et^ โ 2 eโt, y = 3e^2 t^ + 1 Find equation of tangent to the curve at the point for which t= Solution. Differentiating x = et^ โ 2 eโt, y = 3e^2 t^ + 1 with respect to t. dxdt = d(etโ dt^2 eโt)= et (^) + 2eโt dxdt =^ et+2^1 eโt dydt = d(3e dt^2 t +1)= 6e 2 t dydx = dydtdxdt
dydx = dydtdxdt = (^) et (^6) +2e^2 etโt ...........eq(1) Putting t=0 in parametric equations x = et^ โ 2 eโt, y = 3e^2 t^ + 1 โ x = e^0 โ 2 e^0 = 1 โ 2(1) = โ 1 Similarly y = 3e2(0)^ + 1 = 3(1) + 1 = 4โ Putting t = 0 in eq(1) dydx = (^) e (^06) +2e2(0)eโ 0 = (^) 1+2(1)6(1) = 63 = 2 = m Where m is the slope of tangent at t=0 Or (-1,4) Equation of tangent through (โ 1 , 4) = (x 1 , y 1 )with slope m = 2 y โ y 1 = m(x โ x 1 ) Or y โ 4 = 2(x + 1) Q Find the exact coordinate of stationary points on the curve with the equation y = 5xe 12 x Differentiating with respect x dydx = d[5xe dx^12 x] dydx = 5[ d[xe dx^12 x]] dydx = 5[x de dx 12 x + e 12 x dxdx ]
โ3 sin 3x + 5 cos y dydx = 0 dydx = 3 sin 3 5 cos yx ............eq(1)
Putting x = 19 ฯ, y = 16 ฯ in equation (1) dydx = 3 sin 3(5 cos( 16 19 ฯ ฯ)) dydx = 3 sin (5 cos(^1316 ฯฯ))
Hence gradient of the curve at the point ( 19 ฯ, 16 ฯ) Gradient= 35 รร^ โ^ โ^2233 = (^35)
