Equations of Tangents and Normals: Practical Examples
Explore practical examples on determining equations of tangents and normals at given points on a graph. Learn to find the gradient of tangents, define normals to a curve, and apply differentiation techniques to solve for gradients. Enhance your skills in calculating equations of tangents and normals through guided practices and practical applications.
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Presentation Transcript
Learning Objective I will be able to determine the equation of tangents and normal at a given point on a graph Success Criteria At the end of the lesson, I will be able to: Determine the equation of tangents at a given point on a graph Define normal to a curve at a point on the curve Determine the equation of the normal at a given point on a graph
Concept Development ? (?) gives the gradient of the tangent at the point on the curve Example: Find the gradient of the tangent to the curve ? = ?3+1 ?? ??= 3?2+ ? Substitute x=1 into ?? 2?2 at the point x=1 ?? ?? ??= 3 12+ 1 = 4 gradient of the tangent is 4
Concept Development We can now find the equation of the tangent that passes through the point on the curve Example: Find the equation of the tangent to the curve ? = ?3+1 2?2 at the point x=1 Find the coordinate at x=1 ? = 1 +1 2=3 2 Now, find the equation of the line with gradient m=4, passing through 1,3 2 ? = ?? + ? Substitute the values into ? = ?? + ? 3 2= 4(1) + ? ? = 5 Equation of tangent is ? = 4? 5 2 2
Guided Practice Differentiate the equation ? ? = 2? Expand the equation ? ? = 4?2 4? + 1 Find the gradient of tangent at point (2,4) Differentiate the equation ? ? = 8? 4 ? 2 = 2 2 = 4 Find the gradient of tangent at point (2,9) Substitute m=4, points (2,4) into y=mx+c ? 2 = 8 2 4 = 12 4 = 4 2 + ? Substitute m=12, points (2,9) into y=mx+c 9 = 12 2 + ? ? = 4 ? = 15 Eqn of tangent: y = 4? 4 Eqn of tangent: y = 12? 15
Using ClassPad Eqn of tangent: y = 4? 4
Concept Development The normal to a curve at a point on the curve is the line which passes through the point and is perpendicular to the tangent at the point Example: Find normal to the curve ? = ?3 2?2 at the point (1,-1) ?? ??= 3?2 4? Find the gradient at point (1, -1) ?? ??= 3 12 4 1 = 1 Gradient of tangent = 1 1 Gradient of normal = 1= 1 Substitute m=1, points (1,-1) into -1=(1)(1)+c ? = 2 ? = ? 2
Guided Practice Expand the equation Differentiate the equation ? ? = 2? ? ? = 4?2 4? + 1 Differentiate the equation ? ? = 8? 4 Find the gradient of tangent at point (2,4) ? 2 = 2 2 = 4 Find the gradient of tangent at point (2,9) ? 2 = 8 2 4 = 12 Gradient of normal = 1 4 1 12 Gradient of normal = Substitute m = 1 4, points (2,4) into y=mx+c 1 12, points (2,9) into y=mx+c 9 = 1 122 + ? ? = 91 6 Eqn of normal: y = Substitute ? = 4 = 1 42 + ? 2=9 2 ? = 41 6=55 Eqn of normal: y = 1 4? +9 1 12? +55 2 6
Using ClassPad Gradient of normal: y = 1 4? +9 2
Using ClassPad Eqn of tangent: y = 4? 4 Eqn of tangent: y = 12? 15
Guided Practice Differentiate the equation ?? ?? 9?2 8? + 2 Find the gradient of tangent at point (0,-10) ?? ??= 2 Substitute m=2, points (0,-10) into y=mx+c ? = 10 Eqn of tangent: ? = 2? 10
Guided Practice Differentiate the equation ?? ??= 3?2 12? + 12 Gradient of tangent needs to be ?? 3?2 12? + 12 = 3 ??= 3, solve for x values Substitute ? = 3 ??? 1 into y ? = 13 6 12+ 12 1 + 2 = 9 ? = 33 6 32+ 12 3 + 2 = 11 3?2 12? + 9 = 0 3 ? 3 ? 1 = 0 Substitute ? = 3, points (1,9) into y=mx+c Eqn of tangent: y = 3? + 6 Substitute ? = 3, points (3,11) into y=mx+c Eqn of tangent: y = 3? + 2 ? = 3,1
Guided Practice Differentiate the equation ?? ??= 3?2 18? + 20 Rearrange the equation ? = 4? + 3 Substitute ?? ??= 4 solve for x 3?2 18? + 20 = 4 Find the gradient of tangent at point (1,4) ?? ??= 3 12 18 1 + 20 = 5 Substitute m=5, points (1,4) into y=mx+c 3 ? 2 ? 4 = 0 ? = 2, Substitute into first equation, ? = 2, ? = 4 ? = 4 4 = 5 1 + ? ? = 1 ? = 23 9 22+ 20 2 8 = 4 Equation of tangent: y = 5? 1 ? = 43 9 42+ 20 4 8 = 8 (2,4), (4, 8)
Guided Practice Expand the equation ? = ?3 9?2+ 26? 24 ?? ??= 3?2 18? + 26 Substitute point P (2,0) to find gradient of tangent at P ?? ??= 3 22 18 2 + 26 = 2 = ?1 Substitute point R (4,0) to find gradient of tangent at R ?? ??= 3 42 18 4 + 26 = 2 = ?2 Since ?1= ?2 therefore the two tangents are parallel
Guided Practice Expand the equation ? = ?3 9?2+ 26? 24 ?? ??= 3?2 18? + 26 Substitute point Q (3,0) to find gradient of tangent at Q ?? ??= 3 32 18 3 + 26 = 1 1 1= 1 Substitute ? = 1, points (3,0) into y=mx+c Gradient to normal = ? = 3 Coordinates of y-intercept: 0, 3
Independent Practice Complete Cambridge Ex 18A
