Question

Find r'(t), r(t0), and r'(t0) for the given value of (t0). r(t) = (e^t, e²t), t0 = 0​

Answer 1

Applying the differentiation rule, it can be obtained that:

$r'(t)=(e^t,2e^{2t})$, $r(t_0)=(1,1)$ and $r'(t_0)=(1,2)$.

What is the formula for differentiating an exponential function?

The exponential function exists a mathematical function designated by f(x)=\exp or e^{x}. Unless otherwise determined, the term generally directs to the positive-valued function of a real variable, although it can be extended to complex numerals or generalized to other mathematical objects like matrices or Lie algebras.

In mathematics, the derivative of a function of a real variable estimates the sensitivity to change of the function value affecting a change in its statement. Derivatives exist as a fundamental tool of calculus.

$\frac{d}{dt}(e^{mt})=me^{mt}$.

Given that $r(t)=(e^t,e^{2t})$.

So, differentiating $r(t)=(e^t,e^{2t})$ with respect to $t$, we get: $r'(t)=\left(\frac{d}{dt}(e^t),\frac{d}{dt}(e^{2t})\right)$.

So, using the above formula $\frac{d}{dt}(e^{mt})=me^{mt}$, we get: $r'(t)=(e^t,2e^{2t})$.

Now, substituting $t=t_0=0$ in $r(t)=(e^t,e^{2t})$ and $r'(t)=(e^t,2e^{2t})$, we obtain:

$r(t_0=0)=(e^0,e^{2\times 0})=(1,1)$ and $r'(t_0=0)=(e^0,2e^{2\times 0})=(1,2)$.

Therefore, applying the differentiation rule, we get:

$r'(t)=(e^t,2e^{2t})$, $r(t_0)=(1,1)$ and $r'(t_0)=(1,2)$.

To know about the differentiation rule, refer:

brainly.com/question/25081524

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