Table of contents

0. Functions7h 52m
1. Limits and Continuity2h 2m
2. Intro to Derivatives1h 33m
3. Techniques of Differentiation3h 18m
4. Applications of Derivatives2h 38m
5. Graphical Applications of Derivatives6h 2m
6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
7. Antiderivatives & Indefinite Integrals1h 26m
8. Definite Integrals4h 44m
9. Graphical Applications of Integrals2h 27m
- Area Between Curves
  1h 18m
- Introduction to Volume & Disk Method
  1h 8m
10. Physics Applications of Integrals 3h 16m
- Kinematics
  1h 13m
- Work
  2h 3m
11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 34m
12. Techniques of Integration7h 39m
13. Intro to Differential Equations2h 55m
14. Sequences & Series5h 36m
15. Power Series2h 19m
- Introduction to Power Series
  1h 9m
- Taylor Series & Taylor Polynomials
  1h 10m
16. Parametric Equations & Polar Coordinates7h 58m

2. Intro to Derivatives

Tangent Lines and Derivatives

Calculus

2. Intro to Derivatives

Tangent Lines and Derivatives

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Multiple Choice

Given the parametric equations $x = t^{2} - 4$ and $y = t^{2} - 2 t$ , find the equation of the tangent line to the curve at the point where $t = 3$ .

y = 4 x + 18

y = 4 x + 5

y = 2 x + 7

y = 6 x - 10

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Verified step by step guidance

Step 1: Recall that the slope of the tangent line to a parametric curve is given by the derivative dy/dx, which can be computed as (dy/dt) / (dx/dt). Start by finding dx/dt and dy/dt from the given parametric equations x = t^2 - 4 and y = t^2 - 2t.

Step 2: Differentiate x = t^2 - 4 with respect to t to find dx/dt. Similarly, differentiate y = t^2 - 2t with respect to t to find dy/dt. This gives dx/dt = 2t and dy/dt = 2t - 2.

Step 3: Compute dy/dx by dividing dy/dt by dx/dt. Substituting the expressions for dy/dt and dx/dt, we get dy/dx = (2t - 2) / (2t). Simplify this expression to find the slope of the tangent line in terms of t.

Step 4: Evaluate the slope of the tangent line at t = 3 by substituting t = 3 into the simplified expression for dy/dx. This gives the slope of the tangent line at the specified point.

Step 5: Find the coordinates of the point on the curve corresponding to t = 3 by substituting t = 3 into the parametric equations x = t^2 - 4 and y = t^2 - 2t. Use the point-slope form of the equation of a line, y - y1 = m(x - x1), where m is the slope and (x1, y1) is the point, to write the equation of the tangent line.