Question

A 25 ft. ladder is leaning against a house when its base starts to slide away. By the time the base if 7 ft. from the house, the base if moving away at a rate of 72 ft/sec.(a) What is the rate of change of the height of the top of the ladder? __ ft/sec(b) At what rate is the area of the triangle formed by the ladder, wall, and ground changing then? __ ft^2/sec(c) At what rate is the angle between the ladder and the ground changing then? __ rad/sec

Answer 1

Answer:

a) -21 ft/sec

b) 790.5 $ft^2/sec$

c) -3 rad/sec

Step-by-step explanation:

The situation is depicted in the picture attached

(See picture)

 

a) What is the rate of change of the height of the top of the ladder?

We now that x'(t) = 72 when x(t) = 7

We want to compute y'(t) at that moment

By taking the implicit derivative:

$\large\bf [(x(t))^2+(y(t))^2]'=[(25)^2]\Rightarrow 2x(t)x'(t)+2y(t)y'(t)=0\Rightarrow\\\\\Rightarrow2*7*72+2y(t)y'(t)=0\Rightarrow y'(t)=-\frac{504}{y(t)}$

But,

$\large\bf y(t)=\sqrt{(25)^2-(x(t))^2}=\sqrt{(25)^2-(7)^2}=24$

and

$\large\bf y'(t)=-\frac{504}{y(t)}=-\frac{504}{24}=-21\;ft/sec$

(The minus sign means the ladder is descending so the height is decreasing.)

(b) At what rate is the area of the triangle formed by the ladder, wall, and ground changing then?

Taking the derivative on the equation of the area:

$\large\bf A'(t)=\left(\frac{x(t)y(t)}{2}\right)'=\frac{x'(t)y(t)+x(t)y'(t)}{2}=\\\\\frac{72*24+7*(-21)}{2}=790.5 \;ft^2/sec$

(The area of the triangle at that moment is increasing)

(c) At what rate is the angle between the ladder and the ground changing then?

Taking the derivative for the cosine of the angle:

$\large\bf (cos(\alpha(t)))'=\left(\frac{x(t)}{25}\right)'\Rightarrow -\alpha'(t)sin(\alpha(t))=\frac{x'(t)}{25}\Rightarrow\\\Rightarrow -\alpha'(t)\frac{y(t)}{25}=\frac{x'(t)}{25}\Rightarrow\alpha'(t)=-\frac{x'(t)}{y(t)}=-\frac{72}{24}=-3\,rad/sec$

(The angle is decreasing)