The number of minutes taken for a chemical reaction if f(t,x). it depends on the temperature t degrees celcius, and the quantity, x grams, of a catalyst present. when the temperature is 30 degrees celcius and there are 5 grams of catalyst, the reaction takes 50 minutes. increasing the temperature by 3 degrees reduces the time taken by 5 minutes. increasing the amount of catalyst by 2 grams decreases the time taken by 3 minutes. use this information to find the partial derivatives fx(30,5) and ft(30,5). use the tangent plane approximation to find f(33,4).

The partial derivatives fx(30,5) and ft(30,5) can be calculated using the given information, and the tangent plane approximation can be used to estimate f(33,4).

The chemical reaction time is represented by the function f(t,x), where t is the temperature in degrees Celsius and x is the quantity of a catalyst present in grams. We are given that when the temperature is 30 degrees Celsius and there are 5 grams of catalyst, the reaction takes 50 minutes. Additionally, increasing the temperature by 3 degrees reduces the time taken by 5 minutes, and increasing the amount of catalyst by 2 grams decreases the time taken by 3 minutes.

To find the partial derivative fx(30,5), we need to find the rate of change of the reaction time with respect to the quantity of catalyst, while holding the temperature constant. Using the given information, we can calculate:

fx(30,5) = (f(30,5+2)-f(30,5))/2 = (-3)/2 = -1.5

To find the partial derivative ft(30,5), we need to find the rate of change of the reaction time with respect to the temperature, while holding the quantity of catalyst constant. Using the given information, we can calculate:

ft(30,5) = (f(30+3,5)-f(30,5))/3 = (-5)/3 = -1.67

Using the tangent plane approximation, we can estimate f(33,4) by calculating the change in reaction time by adjusting the temperature to 33 degrees Celsius and the quantity of catalyst to 4 grams.

f(33,4) = f(30,5) + ft(30,5) * (33-30) + fx(30,5) * (4-5)

f(33, 4) = 50 + (-1.67) * 3 + (-1.5) * -1 = 50 - 5.01 + 1.5 = 46.49

[tex]\( f_t(30,5) = -\frac{5}{3} \)[/tex], [tex]\( f_x(30,5) = -1.5 \)[/tex], [tex]\( f(33,4) \approx 46.5 \)[/tex] minutes.

ft (30, 5) =_[tex]\(-\frac{5}{3} \)[/tex]_ fx (30,5) =_-1.5_ f (33,4)=__46.5 minutes__

To find the partial derivatives [tex]\( f_t(30,5) \)[/tex] and [tex]\( f_x(30,5) \)[/tex], we'll use the given information about how changes in [tex]\( t \) and \( x \)[/tex] affect the time [tex]\( f(t,x) \)[/tex].

Given:

- [tex]\( f(30,5) = 50 \)[/tex] minutes

- Increasing [tex]\( t \)[/tex] by 3 degrees Celsius reduces [tex]\( f \)[/tex] by 5 minutes.

- Increasing [tex]\( x \)[/tex] by 2 grams decreases [tex]\( f \)[/tex] by 3 minutes.

1. Partial derivative with respect to [tex]\( t \) at \( (30,5) \)[/tex]:

  - Change in [tex]\( t \): \( \Delta t = 3 \)[/tex] degrees Celsius

  - Change in [tex]\( f \): \( \Delta f = -5 \)[/tex] minutes

  - Using the definition of partial derivative:

    [tex]\[ f_t(30,5) = \frac{\Delta f}{\Delta t} = \frac{-5}{3} = -\frac{5}{3} \text{ minutes/degree Celsius} \][/tex]

2. Partial derivative with respect to [tex]\( x \) at \( (30,5) \)[/tex]:

  - Change in [tex]\( x \): \( \Delta x = 2 \)[/tex] grams

  - Change in [tex]\( f \): \( \Delta f = -3 \)[/tex] minutes

  - Using the definition of partial derivative:

    [tex]\[ f_x(30,5) = \frac{\Delta f}{\Delta x} = \frac{-3}{2} = -1.5 \text{ minutes/gram} \][/tex]

Now, to find [tex]\( f(33,4) \)[/tex] using the tangent plane approximation, we'll use the partial derivatives we found and the point [tex]\( (30,5) \)[/tex] as a reference:

The tangent plane approximation formula is:

[tex]\[ f(t,x) \approx f(30,5) + f_t(30,5)(t-30) + f_x(30,5)(x-5) \][/tex]

Substituting the values:

[tex]\[ f(33,4) \approx 50 + \left(-\frac{5}{3}\right)(33-30) + (-1.5)(4-5) \][/tex]

[tex]\[ f(33,4) \approx 50 - 5 + 1.5 \][/tex]

[tex]\[ f(33,4) \approx 46.5 \text{ minutes} \][/tex]

So, [tex]\( f(33,4) \approx 46.5 \)[/tex] minutes.

Thus:

- [tex]\( f_t(30,5) = -\frac{5}{3} \)[/tex] minutes/degree Celsius

- [tex]\( f_x(30,5) = -1.5 \)[/tex] minutes/gram

- [tex]\( f(33,4) \approx 46.5 \)[/tex] minutes

The correct question is:

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