Let $g$ be a vector-valued function defined by $g(t)=(\sin (\pi t),-2 \cos (\pi t))$.

Find $g^{\prime}(t)$

Choose 1 answer:
(A) $(\pi \cos (\pi t), 2 \pi \sin (\pi t))$
(B) $\cos (\pi t)+2 \sin (\pi t))$
(C) $(-\sin (\pi t),-2 \sin (\pi t))$
(D) $(-\cos (\pi t), 2 \sin (\pi t))$

Question

Let $g$ be a vector-valued function defined by $g(t)=(\sin (\pi t),-2 \cos (\pi t))$.

Find $g^{\prime}(t)$

Choose 1 answer:
(A) $(\pi \cos (\pi t), 2 \pi \sin (\pi t))$
(B) $\cos (\pi t)+2 \sin (\pi t))$
(C) $(-\sin (\pi t),-2 \sin (\pi t))$
(D) $(-\cos (\pi t), 2 \sin (\pi t))$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find the derivative of the vector-valued function $g(t)=(\sin (\pi t), -2 \cos (\pi t))$, we differentiate each component of the vector function with respect to $t$. ‖ ‖ step 2 ⋮ The derivative of the first component $\sin (\pi t)$ with respect to $t$ is $\pi \cos (\pi t)$. This is because the derivative of $\sin$ is $\cos$, and by the chain rule, we multiply by the derivative of the inside function, which is $\pi$. ‖ ‖ step 3 ⋮ The derivative of the second component $-2 \cos (\pi t)$ with respect to $t$ is $2\pi \sin (\pi t)$. This is because the derivative of $\cos$ is $-\sin$, and by the chain rule, we multiply by the derivative of the inside function, which is $\pi$, and by the constant $-2$. ‖ ‖ step 4 ⋮ Combining the results from steps 2 and 3, we get $g'(t) = (\pi \cos (\pi t), 2\pi \sin (\pi t))$. ‖ ***A*** ∻Key Concept∻ ⚹Differentiation of vector-valued functions⚹ ∻Explanation∻ ⚹To differentiate a vector-valued function, differentiate each component function separately with respect to the variable. Apply the chain rule when necessary.⚹◊What is the formula for finding the derivative of a vector-valued function?⍭ Generate me a similar question◊

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