{"id":1424,"date":"2026-01-28T18:53:23","date_gmt":"2026-01-28T08:53:23","guid":{"rendered":"https:\/\/archive4ones.com\/2ndstudy\/?p=1424"},"modified":"2026-01-28T18:53:23","modified_gmt":"2026-01-28T08:53:23","slug":"year11-math-3-2-3-applications-of-the-derivative-warm-up-workbook","status":"publish","type":"post","link":"https:\/\/archive4ones.com\/2ndstudy\/?p=1424","title":{"rendered":"Year11 MATH 3-2-3 Applications of the Derivative warm-up workbook"},"content":{"rendered":"\n

Unit1<\/h4>\n\n\n\n

Focus: Tangents, normals, and stationary points.<\/em><\/p>\n\n\n\n

    \n
  1. Find the gradient of the tangent to y<\/mi>=<\/mo>x<\/mi>2<\/mn><\/msup>+<\/mo>2<\/mn>x<\/mi><\/mrow>y = x^2 + 2x<\/annotation><\/semantics><\/math> at the point (<\/mo>1<\/mn>,<\/mo>3<\/mn>)<\/mo><\/mrow>(1, 3)<\/annotation><\/semantics><\/math>.<\/li>\n\n\n\n
  2. Find the equation of the tangent to y<\/mi>=<\/mo>x<\/mi>2<\/mn><\/msup><\/mrow>y = x^2<\/annotation><\/semantics><\/math> at x<\/mi>=<\/mo>2<\/mn><\/mrow>x = 2<\/annotation><\/semantics><\/math>.<\/li>\n\n\n\n
  3. Determine the coordinates of the stationary point<\/strong> for y<\/mi>=<\/mo>x<\/mi>2<\/mn><\/msup>\u2212<\/mo>4<\/mn>x<\/mi>+<\/mo>5<\/mn><\/mrow>y = x^2 – 4x + 5<\/annotation><\/semantics><\/math>.<\/li>\n\n\n\n
  4. Determine the nature of the stationary point for y<\/mi>=<\/mo>x<\/mi>2<\/mn><\/msup><\/mrow>y = x^2<\/annotation><\/semantics><\/math> (Max, Min, or Inflexion).<\/li>\n\n\n\n
  5. If the displacement of a particle is s<\/mi>(<\/mo>t<\/mi>)<\/mo>=<\/mo>t<\/mi>2<\/mn><\/msup>+<\/mo>4<\/mn>t<\/mi><\/mrow>s(t) = t^2 + 4t<\/annotation><\/semantics><\/math>, find the velocity at t<\/mi>=<\/mo>2<\/mn><\/mrow>t = 2<\/annotation><\/semantics><\/math>.<\/li>\n\n\n\n
  6. Find the acceleration if velocity is v<\/mi>(<\/mo>t<\/mi>)<\/mo>=<\/mo>3<\/mn>t<\/mi>2<\/mn><\/msup>\u2212<\/mo>5<\/mn><\/mrow>v(t) = 3t^2 – 5<\/annotation><\/semantics><\/math>.<\/li>\n\n\n\n
  7. At what value of x<\/mi>x<\/annotation><\/semantics><\/math> is the tangent to y<\/mi>=<\/mo>x<\/mi>2<\/mn><\/msup>\u2212<\/mo>8<\/mn>x<\/mi><\/mrow>y = x^2 – 8x<\/annotation><\/semantics><\/math> horizontal?<\/li>\n\n\n\n
  8. Find the equation of the normal<\/strong> to y<\/mi>=<\/mo>2<\/mn>x<\/mi>2<\/mn><\/msup><\/mrow>y = 2x^2<\/annotation><\/semantics><\/math> at x<\/mi>=<\/mo>1<\/mn><\/mrow>x = 1<\/annotation><\/semantics><\/math>.<\/li>\n\n\n\n
  9. A curve has d<\/mi>y<\/mi><\/mrow>d<\/mi>x<\/mi><\/mrow><\/mfrac>=<\/mo>2<\/mn>x<\/mi>\u2212<\/mo>4<\/mn><\/mrow>\\frac{dy}{dx} = 2x – 4<\/annotation><\/semantics><\/math>. Find the intervals where the function is increasing<\/strong>.<\/li>\n\n\n\n
  10. Find the average rate of change of f<\/mi>(<\/mo>x<\/mi>)<\/mo>=<\/mo>x<\/mi>2<\/mn><\/msup><\/mrow>f(x) = x^2<\/annotation><\/semantics><\/math> between x<\/mi>=<\/mo>1<\/mn><\/mrow>x = 1<\/annotation><\/semantics><\/math> and x<\/mi>=<\/mo>3<\/mn><\/mrow>x = 3<\/annotation><\/semantics><\/math>.<\/li>\n<\/ol>\n\n\n\n

    Answers:<\/strong><\/p>\n\n\n\n

      \n
    1. 4 | 2. y<\/mi>=<\/mo>4<\/mn>x<\/mi>\u2212<\/mo>4<\/mn><\/mrow>y = 4x – 4<\/annotation><\/semantics><\/math> | 3. (<\/mo>2<\/mn>,<\/mo>1<\/mn>)<\/mo><\/mrow>(2, 1)<\/annotation><\/semantics><\/math> | 4. Minimum | 5. 8 units\/s | 6. 6<\/mn>t<\/mi><\/mrow>6t<\/annotation><\/semantics><\/math> | 7. x<\/mi>=<\/mo>4<\/mn><\/mrow>x = 4<\/annotation><\/semantics><\/math> | 8. y<\/mi>=<\/mo>\u2212<\/mo>1<\/mn>4<\/mn><\/mfrac>x<\/mi>+<\/mo>9<\/mn>4<\/mn><\/mfrac><\/mrow>y = -\\frac{1}{4}x + \\frac{9}{4}<\/annotation><\/semantics><\/math> | 9. x<\/mi>><\/mo>2<\/mn><\/mrow>x > 2<\/annotation><\/semantics><\/math> | 10. 4.<\/li>\n<\/ol>\n\n\n\n

      <\/p>\n\n\n\n

      <\/p>\n","protected":false},"excerpt":{"rendered":"

      Unit1 Focus: Tangents, normals, and stationary points. Answers:<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[8],"tags":[],"class_list":["post-1424","post","type-post","status-publish","format-standard","hentry","category-math"],"_links":{"self":[{"href":"https:\/\/archive4ones.com\/2ndstudy\/index.php?rest_route=\/wp\/v2\/posts\/1424","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/archive4ones.com\/2ndstudy\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/archive4ones.com\/2ndstudy\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/archive4ones.com\/2ndstudy\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/archive4ones.com\/2ndstudy\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=1424"}],"version-history":[{"count":4,"href":"https:\/\/archive4ones.com\/2ndstudy\/index.php?rest_route=\/wp\/v2\/posts\/1424\/revisions"}],"predecessor-version":[{"id":1442,"href":"https:\/\/archive4ones.com\/2ndstudy\/index.php?rest_route=\/wp\/v2\/posts\/1424\/revisions\/1442"}],"wp:attachment":[{"href":"https:\/\/archive4ones.com\/2ndstudy\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=1424"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/archive4ones.com\/2ndstudy\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=1424"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/archive4ones.com\/2ndstudy\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=1424"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}