Respuesta :
Answer:
The value of rate of which the base is changing [tex]\frac{dB}{dt}[/tex] = - 3 [tex]\frac{cm}{s}[/tex]
Step-by-step explanation:
Area = 20 [tex]cm^{2}[/tex]
height = 4 cm
[tex]\frac{dH}{dt} = 2 \frac{cm}{sec}[/tex]
[tex]\frac{dA}{dt} =[/tex] [tex]2 \ \frac{cm^{2} }{sec}[/tex]
we know that area of the triangle is given by
[tex]A= \frac{1}{2} B H[/tex]
[tex]B = \frac{2A}{H}[/tex]
[tex]B = \frac{2 (20)}{4}[/tex]
B = 10 cm
Rate of change of area is given by
[tex]\frac{dA}{dT} = \frac{1}{2} [B\frac{dH}{dt} + H \frac{dB}{dt} ][/tex]
4 = 0.5 [10 × 2 + 4 [tex]\frac{dB}{dt}[/tex] ]
[tex]\frac{dB}{dt}[/tex] = - 3 [tex]\frac{cm}{s}[/tex]
This is the value of rate of which the base is changing.
This question is based on the area of triangle. Therefore, the rate at which the base of the triangle is changing when the height of the triangle is 4 cm and the area is 20 [tex]cm^2[/tex] is -3 cm/sec.
Given:
A triangle has a height that is increasing at a rate of 2 cm/sec and its area is increasing at a rate of 4 [tex]cm^2/sec.[/tex]
According to the question,
Firstly, now differentiating the area and height.
[tex]\dfrac{dA}{dt} = 2cm^2/sec\\\\\dfrac{dH}{dt} = 2 cm/sec[/tex]
As we know that, area of the triangle is given by,
[tex]Area = \dfrac{1}{2} \times B \times H\\\\20 = \dfrac{1}{2} \times B \times 4\\\\B= \dfrac{20}{2} \\\\B = 10 cm[/tex]
Now, calculating the rate of change of area is given by,
[tex]\dfrac{dA}{dT} = \dfrac{1}{2} [ B \dfrac{dH}{dT} +H \dfrac{dB}{dT}]\\\\4 = 0.5 [10 \times 2 + 4 \dfrac{dB}{dT}]\\\\ \dfrac{dB}{dT} = -3 cm/sec[/tex]
Therefore, the rate at which the base of the triangle is changing when the height of the triangle is 4 cm and the area is 20 [tex]cm^2[/tex] is -3 cm/sec.
For more details, prefer this link:
https://brainly.com/question/10317896
