Question

Let \(\vec{r}(t) = \langle t^3 - 3, t^4 + 5t^3, -3 \ln(4t) \rangle\) Find a parametric equation of the line tangent to \(\vec{r}(t)\) at the point \((-2, 6, -4.159)\) x(t) = -2 + 3t y(t) = z(t) =

          Let \(\vec{r}(t) = \langle t^3 - 3, t^4 + 5t^3, -3 \ln(4t) \rangle\)
Find a parametric equation of the line tangent to \(\vec{r}(t)\) at the point \((-2, 6, -4.159)\)
x(t) = -2 + 3t
y(t) =
z(t) =

Let r⃗(t) = ⟨ t^3 - 3, t^4 + 5t^3, -3 ln(4t) ⟩
Find a parametric equation of the line tangent to r⃗(t) at the point (-2, 6, -4.159)
x(t) = -2 + 3t
y(t) =
z(t) =

Added by Audrey P.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Sanchit Jain

05/16/2023

Step 1

The derivative of r(t) is given by: r'(t) = <3t^2, 4t^3 + 15t^2, -124t^3> Show more…

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Let r(t)=< t3 - 3,t4+ 5t3,-31n(4t)> Find a parametric equation of the line tangent to 7(t) at the point (--2, 6, --4.159) x(t)=-2+3t y(t) 7)2 Check Answer