If a = t^2 hat{i} + e^t hat{j} + hat{k} and b | vedclass

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(B) Given vectors are $a = t^2 \hat{i} + e^t \hat{j} + \hat{k}$ and $b = 2 \hat{i} + t^2 \hat{j} + \log t \hat{k}$.We know that the dot product $f(t)

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$5 + 3e$

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(B) Given vectors are $a = t^2 \hat{i} + e^t \hat{j} + \hat{k}$ and $b = 2 \hat{i} + t^2 \hat{j} + \log t \hat{k}$.We know that the dot product $f(t) = a \cdot b$ is given by the sum of the products of corresponding components:$f(t) = (t^2)(2) + (e^t)(t^2) + (1)(\log t) = 2t^2 + t^2 e^t + \log t$.Now,differentiate $f(t)$ with respect to $t$ using the product rule and the derivative of $\log t$:$f^{\prime}(t) = \frac{d}{dt}(2t^2) + \frac{d}{dt}(t^2 e^t) + \frac{d}{dt}(\log t)$$f^{\prime}(t) = 4t + (2t e^t + t^2 e^t) + \frac{1}{t}$.To find $f^{\prime}(1)$,substitute $t = 1$ into the derivative expression:$f^{\prime}(1) = 4(1) + (2(1) e^1 + (1)^2 e^1) + \frac{1}{1}$$f^{\prime}(1) = 4 + 2e + e + 1$$f^{\prime}(1) = 5 + 3e$.

If $a = t^2 \hat{i} + e^t \hat{j} + \hat{k}$ and $b = 2 \hat{i} + t^2 \hat{j} + \log t \hat{k}$,and $f(t) = a \cdot b$,then $f^{\prime}(1)$ is equal to

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