Tangent at any point of curve| differential geometry msc mathematics lecture

show that the tangent at any point of any curve whose equation referred a rectangular axis x=3t,y=3t^2,z=2t^3 make a constant angle with the line y=o, z-x=o, Theory: Tangent to a Curve Makes a Constant Angle with a Fixed Line Given Statement Show that the tangent at any point of the curve whose equations referred to rectangular axes are x=3t,y=3t2,z=2t3x=3t,\qquad y=3t^2,\qquad z=2t^3 makes a constant angle with the line y=0,z−x=0.y=0,\qquad z-x=0. Theory If a curve is given in parametric form as x=f(t),y=g(t),z=h(t),x=f(t),\qquad y=g(t),\qquad z=h(t), then the direction ratios of the tangent to the curve at any point are dxdt,dydt,dzdt.\frac{dx}{dt},\qquad \frac{dy}{dt},\qquad \frac{dz}{dt}. If a fixed line has direction ratios a,b,ca,b,c, and the tangent to the curve has direction ratios l,m,nl,m,n, then the angle θ\theta between them is given by cos⁡θ=al+bm+cna2+b2+c2l2+m2+n2\boxed{\cos\theta= \frac{al+bm+cn} {\sqrt{a^2+b^2+c^2}\sqrt{l^2+m^2+n^2}}} If the value of cos⁡θ\cos\theta is constant, then the tangent makes a constant angle with the fixed line. Proof Step 1: Find the direction ratios of the tangent. The given curve is x=3t,y=3t2,z=2t3.x=3t,\qquad y=3t^2,\qquad z=2t^3. Differentiating with respect to tt, we get dxdt=3,\frac{dx}{dt}=3, dydt=6t,\frac{dy}{dt}=6t, dzdt=6t2.\frac{dz}{dt}=6t^2. Therefore, the direction ratios of the tangent are 3, 6t, 6t2.\boxed{3,\ 6t,\ 6t^2}. Step 2: Find the direction ratios of the fixed line. The given line is y=0,z−x=0.y=0,\qquad z-x=0. Since y=0y=0 and z=xz=x, its points can be written as (x,y,z)=(λ,0,λ).(x,y,z)=(\lambda,0,\lambda). Hence, the direction ratios of the line are 1, 0, 1.\boxed{1,\ 0,\ 1}. Step 3: Find the angle between the tangent and the line. Let θ\theta be the angle between them. Using the formula for the angle between two lines, cos⁡θ=3(1)+6t(0)+6t2(1)32+(6t)2+(6t2)212+02+12.\cos\theta= \frac{3(1)+6t(0)+6t^2(1)} {\sqrt{3^2+(6t)^2+(6t^2)^2}\sqrt{1^2+0^2+1^2}}. Simplifying, cos⁡θ=3+6t29+36t2+36t42.\cos\theta= \frac{3+6t^2} {\sqrt{9+36t^2+36t^4}\sqrt2}. Since 9+36t2+36t4=9(1+2t2)2,9+36t^2+36t^4=9(1+2t^2)^2, we obtain cos⁡θ=3(1+2t2)3(1+2t2)2.\cos\theta= \frac{3(1+2t^2)} {3(1+2t^2)\sqrt2}. Therefore, cos⁡θ=12.\boxed{\cos\theta=\frac{1}{\sqrt2}}. Thus, θ=45∘.\boxed{\theta=45^\circ}. Conclusion Since the angle between the tangent and the fixed line is constant and equal to 45∘45^\circ, the tangent to the given curve makes a constant angle with the line y=0,z−x=0.\boxed{y=0,\qquad z-x=0}.