Sheridan College · MATH15382D

Sheridan MATH15382D tutoring

Help with derivatives, integrals, and parametric and polar curves in Sheridan MATH15382D.

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Course overview

MATH15382D brings differentiation and integration together in Sheridan’s engineering degree curriculum at Davis Campus in Brampton. Its single-variable scope includes limits, applied derivatives, integration techniques and parametric and polar curves. A useful habit is to identify the requested quantity first: a tangent slope, accumulated area and curve length require different setups even when the same curve appears.

MATH15382D topics to review

Limits, continuity and derivative applicationsIntegration methods, areas, volumes and workTangents and integrals for parametric and polar curves

Common difficulties

  • Using dy/dt as the slope dy/dx
  • Integrating with respect to t while leaving out dx/dt

Preparing for this course

Review function composition, trigonometric identities and algebraic simplification. For a parametrized curve, calculate both coordinate derivatives and mark how the parameter moves through the curve. Before integrating, identify the differential and translate the endpoint coordinates into parameter values.

Registration requirements

The guidance above is study preparation, not an eligibility decision. Confirm current prerequisites, restrictions and availability with Sheridan College.

Check the official course catalogue

Use one parametrization for a slope and an area

A curve is given by x=t2x=t^2, y=t3y=t^3, with 0≤t≤20\le t\le2. Find its tangent slope at t=1t=1 and the area below the curve and above the x-axis from x=0x=0 to x=4x=4.

Show the reasoning
  1. Divide coordinate rates

    For t>0t>0, dy/dx=(dy/dt)/(dx/dt)=3t2/(2t)=3t/2dy/dx=(dy/dt)/(dx/dt)=3t^2/(2t)=3t/2. At t=1t=1, the slope is 3/23/2. The ratio measures vertical change per horizontal change.

  2. Transform the area integral

    The curve is above the axis and x increases along the interval. Use A=∫y dxA=\int y\,dx, with dx=2t dtdx=2t\,dt. Hence A=∫02t3(2t) dtA=\int_0^2 t^3(2t)\,dt; integrating only t3t^3 would accumulate against the wrong variable.

  3. Evaluate and check another representation

    A=[2t5/5]02=64/5A=[2t^5/5]_0^2=64/5 square units. Eliminating t gives y=x3/2y=x^{3/2}, whose integral from 0 to 4 is also 64/564/5. This independently checks the differential and bounds.

Try a changed problem

Now use x=t2x=t^2, y=2t3y=2t^3, 0≤t≤10\le t\le1. Find the slope at t=1t=1 and the area under this curve.

Check your reasoning

The slope is 6t2/(2t)=3t6t^2/(2t)=3t, hence 3. The area is ∫01(2t3)(2t) dt=4/5\int_0^1(2t^3)(2t)\,dt=4/5 square units. Both the height and the parameter interval changed.

How tutoring can help

Share a question, whether or not you have tried it. A tutor can help you decide whether to differentiate or integrate, explain the factors in a derivative and choose integration bounds. We can work through the solution together before practising the method on another curve.

Exam preparation for MATH15382D

Mix differentiation and integration questions from the announced assessment scope. Explain each substitution before calculating, and verify an antiderivative by differentiation. With parametric or polar questions, sketch the path and check whether the interval traces the region once.

For a further integration-by-parts challenge, reduce a polynomial or solve for an integral that returns. The worked example and three optional questions build on the basic method.

Use your section’s outline to choose the chapters to review. We work through permitted practice and comparable problems, without completing graded assignments or exams on your behalf.

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MATH15382D · official course reference