Question

Math hw for tonight
help solve this problem! Thank you!
ap cal bc

Answer 1

`√(2)`

Explanation:

When t = 0, the velocity is:

`\begin{aligned}\vec{v}(t)&=3^t \vec{i}+e^(6t) \vec{j}\\\\t=0 \implies \vec{v}(0)&=3^0 \vec{i}+e^((6 \cdot 0)) \vec{j}\\ &=1 \vec{i}+1 \vec{j} \end{aligned}`

Speed is the magnitude of the velocity of the object.

`\boxed{\text{A vector $\binom{x}{y}$ has magnitude $√(x^2+y^2)$}}`

Therefore, the particle's speed at time t = 0 is:

`\begin{aligned}\implies |v|&=√(1^2+1^2)\\&=√(1+1)\\&=√(2)\end{aligned}`