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ehidna [41]
9 months ago
5

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

Mathematics
1 answer:
creativ13 [48]9 months ago
5 0

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).

<h3>What is the formula for differentiating an exponential function?</h3>

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

#SPJ9

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